There is only one logarithm

When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others? On this page, we'll answer that question, among others.

In [https://arbital.com/p/4bz](https://arbital.com/p/4bz), we saw that any $f$ with [domain](https://arbital.com/p/3js) [$\mathbb R^+$](https://arbital.com/p/positive_reals) that satisfies the equation $f(x \cdot y) = f(x) + f(y)$ for all $x$ and $y$ in its domain is either trivial (i.e., it sends all inputs to 0), or it is [isomorphic](https://arbital.com/p/4f4) to [$\log_b$](https://arbital.com/p/3nd) for some base $b$. Thus, if we want a function's output to change by a constant that depends on $y$ every time its input changes by a factor of $y$, we only have one meaningful degree of freedom, and that's the choice of $b \neq 1$ such that $f(b) = 1$. Once we choose which value of $b$ $f$ sends to 1, the entire behavior of the function is fully defined.

How much freedom does this choice give us? Almost none! To see this, let's consider the difference between choosing base $b$ as opposed to an alternative base $c.$ Say we have an input $x \in \mathbb R^+$ — what's the difference between $\log_b(x)$ and $\log_c(x)?$

Well, $x = c^y$ for some $y$, because $x$ is positive. Thus, $\log_c(x)$ $=$ $\log_c(c^y)$ $=$ $y.$ By contrast, $\log_b(x)$ $=$ $\log_b(c^y)$ $=$ $y \log_b(c).$ ([Refresher](https://arbital.com/p/4bz).) Thus, $\log_c$ and $\log_b$ disagree on $x$ only by a constant factor — namely, $\log_b(c).$ And this is true for _any_ $x$ — you can get $\log_c(x)$ by calculating $\log_b(x)$ and dividing by $\log_b(c):$ $$\log_c(x) = \frac{\log_b(x)}{\log_b(c)}.$$

This is a remarkable equation, in the sense that it's worth remarking on. No matter what base $b$ we choose for $f$, and no matter what input $x$ we put into $f$, if we want to figure out what we would have gotten if we chose base $c$ instead, all we need to do is calculate $f(c)$ and divide $f(x)$ by $f(c).$ In other words, the different logarithm functions actually aren't very different at all — each one has all the same information as all the others, and you can recover the behavior of $\log_c$ using $\log_b$ and a simple calculation!

(By a symmetric argument, you can show that $\log_b(x) = \frac{\log_c(x)}{\log_c(b)},$ which implies that $\log_b(c) = \frac{1}{\log_c(b)}$, a fact we already knew from from studying [https://arbital.com/p/-44l](https://arbital.com/p/-44l).)

Above, we asked:

> When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others?

From the fact that the [length of a written number grows logarithmically with the magnitude of the number](https://arbital.com/p/416) and the above equation, we can see that, no matter how large a number is, its base 10 representation differs in length from its base 12 representation only by a factor of $\log_{10}(12) \approx 1.08$. Similarly, the [binary](https://arbital.com/p/binary_notation) representation of a number is always about $\log_2(10) \approx 3.32$ times longer than its decimal representation. Because there is only one logarithm function (up to a multiplicative constant), which number base you use to represent numbers only affects the size of the representation by a multiplicative constant.

Similarly, if you ever want to convert between logarithmic measurements in different bases, you only need to perform a single multiplication (or division). For example, if someone calculated how many hours it took for the bacteria colony to triple, and you want to know how long it took to double, all you need to do is multiply by $\log_3(2) \approx 0.63.$ There is essentially only one logarithm function; the base merely defines the unit of measure. Given measurements taken in one base, you can easily convert to another.

Which base of the logarithm is best to use? Well, by the arguments above, it doesn't really matter which base you pick — it's easy enough to convert between bases if you need to. However, in computer science, [https://arbital.com/p/-3qq](https://arbital.com/p/-3qq), and many parts of physics, it's common to choose base 2. Why? Two reasons. First, because if you're designing a physical system to store data, a system that only has two stable states is the simplest available. Second, because if you need to pick an arbitrary amount that a thing has grown to call "one growth," then doubling (i.e., "it's now once again as large as it was when it started") is a pretty natural choice. (For example, many people find the 'half life' of a radioactive element to be a more intuitive notion than its 'fifth life,' even though both are equally valid measures of radioactivity.)

In other parts of physics and engineering, the log base 10 is more common, because it has a natural relationship to the way we represent numbers (using a base 10 representation). That is, $\log_{10}$ is one of the easiest logarithms to calculate in your head (because you can just [count the number of digits in a number's representation](https://arbital.com/p/416)), so it's often used by engineers.

Among mathematicians, the most common base of the logarithm is [$e$](https://arbital.com/p/e) $\approx 2.718,$ a [https://arbital.com/p/-trancendental_number](https://arbital.com/p/-trancendental_number) that cannot be fully written down. Why this strange base? That's an [advanced topic](https://arbital.com/p/base_e_is_natural) not covered in this tutorial. In brief, though, if you look at the [derivative](https://arbital.com/p/47d) of the logarithm (i.e., the way the function changes if you perturb the input), if the input was $x$ and you perturb it a tiny amount, the output changes by about a factor of $\frac{1}{x}$ times a constant $y$ that depends on the base of the logarithm, and $e$ is the base that causes that constant to be $1.$ For this reason and a handful of others, there is a compelling sense in which $\log_e$ is the "true form" of the logarithm. (So much so that it has its own symbol, $\ln,$ which is called the "natural logarithm.")

Regardless of which base you choose, converting between bases is easy — and as equation (1) shows, you only need to have one logarithm function in your toolkit in order to understand any other logarithm. Because if you have $\log_b$ and you want to know $\log_c(x)$, you can just calculate $\log_b(x)$ and divide it by $\log_b(c)$, and then you're done. In other words, there is really only one logarithm function, and it doesn't matter which one you pick.

[A visualization of the graph of the logarithm where you can change the base, but it changes only the labels on the axes while the curve itself remains completely static.](https://arbital.com/p/fixme:)

Eric Rogstad
Nate Soares
Patrick Stevens


The log lattice

[https://arbital.com/p/45q](https://arbital.com/p/45q) and other pages give physical interpretations of what logarithms are really doing. Now it's time to understand the raw numerics. Logarithm functions often output irrational, [https://arbital.com/p/-transcendental](https://arbital.com/p/-transcendental) numbers. For example, $\log_2(3)$ starts with

1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360

What do these long strings of digits mean? Why these numbers, in particular? We already have the tools to answer those questions, all that remains is to put them together into a visualization. In [https://arbital.com/p/44l](https://arbital.com/p/44l), we saw why it is that $\log_2(3)$ must be this number: The cost of a [3-digit](https://arbital.com/p/4sj) in terms of 2-digits is more than 1 and less than 2; and the cost of ten 3-digits = one $3^{10}$-digit in terms of 2-digits is more than 15 and less than 16, and the cost of a hundred 3-digits = one $3^{100}$-digit in terms of 2-digits is more than 158 and less than 159, and so on. The giant number above is telling us about the _entire sequence_ of $3^{n}$-digit costs, for every $n$. If we double it, we get the cost of a 9-digit in terms of 2-digits. If we triple it, we get the cost of a 27-digit in terms of 2-digits.

In fact, this number also interacts correctly with all the other outputs of $\log_2.$ $\log_2(2)=1,$ and if we add 1 to the number above, we get the cost of a 6-digit in terms of 2-digits.

That is, the outputs of the log base 2 form a gigantic lattice, with an often-irrational output corresponding to each input. The gigantic lattice is set up just so, such that whenever you start at $x$ on the left and go to $x \cdot y,$ the output you get on the right is the output to that corresponds to $x$ plus the output that corresponds to $y$.

![The values of log2 on integers from 1 to 10](http://i.imgur.com/FvRymOk.png)

_The values of the logarithm base 2 on integers from 1 to 10._

The log lattice is this giant lattice assigning a specific number to each number, such that multiplying on the left corresponds to adding on the right.

![A demonstration that log2(3^10) is 10 log2(3)](http://i.imgur.com/tuTYlge.png)

_An illustration of some of the connections between the logs base 2 of powers of 3._

It's an intricate lattice that preservers a huge amount of structure — all the structure of the numbers, in fact, given that $\log$ is invertible. $\log_2(3)$ is a number that is simultaneously $1$ less than $\log_2(6),$ and half of $\log_2(9)$, and a tenth of $\log_2(3^{10}),$ and $\log_2(3^9)$ less than $\log_2(3^{10}).$ The value of $\log_2(3)$ has to satisfy a _massive_ number of constraints, in order to be precisely the number such that multiplication on the left corresponds to addition on the right. It's no surprise, then, that [it has no concise decimal expansion](https://arbital.com/p/4n8).

![A demonstration of the relationships between the log2 powers of 3](http://i.imgur.com/e650J5j.png)

_Notice the connections between the logs base 2 of the powers of 3. log 2 of 3 is a tenth of log 2 of 310 is a tenth of the log 2 of 3100, which is twice the log 2 of 350 which is twice the log 2 of 325. And, of course, the log 2 of 3101 is about 1.58 units larger than the log 2 of 3100._

In fact, the constraints on this log lattice are so tight that there's only one way to do it, up to a multiplicative constant. You can multiply everything on the right side by a constant, and that's the only thing you can do without disturbing the structure. For example, here's part of the log lattice viewed in base 3:

![The base 3 log lattice](http://i.imgur.com/yhcy9BZ.png)

It's exactly the same as the log lattice viewed in base 2, except that everything on the right has been divided by about 1.58 (i.e., multiplied by about 0.63).

What are logarithms doing? Well, fundamentally, there is a way to transform numbers such that what was once multiplication is now addition. If you were going to do multiplication to some numbers, you can transform them into this lattice, and then do addition, and then transform them back, and you'll get the same result. There's only one way to preserve all that structure, although if you want to view the structure, you need to (arbitrarily) choose some number on the left to call "1" on the right. Given that choice $b$ of translation, the log base $b$ taps into that intricate lattice.

[https://arbital.com/p/visualization](https://arbital.com/p/visualization)

Eric Rogstad
Nate Soares
Malcolm McCrimmon
Alexei Andreev


Formal definition

Meta tag for pages which give formal or brief jargon-heavy technical definitions of a concept. Formal definitions should be secondary [lenses](https://arbital.com/p/17b) when explanations are available.

Eric Bruylant
Jessica Taylor
Patrick Stevens
Alexei Andreev


Life in logspace

[https://arbital.com/p/4gp](https://arbital.com/p/4gp) hints at the reason that engineers, scientists, and AI researchers find logarithms so useful.

At this point, you may have a pretty good intuition for what logarithms are doing, and you may be saying to yourself

> Ok, I understand logarithms, and I see how they could be a pretty useful concept sometimes. Especially if, like, someone happens to have an out-of-control bacteria colony lying around. I can also see how they're useful for measuring representations of things (like the length of a number when you write it down, or the cost of communicating a message). If I squint, I can _sort of_ see why it is that the brain encodes many percepts logarithmically (such that the tone that I perceive is the $\log_2$ of the frequency of vibration in the air), as it's not too surprising that evolution occasionally hit upon this natural tool for representing things. I guess logs are pretty neat.

Then you start seeing logarithms in other strange places — say, you learn that [AlphaGo](https://en.wikipedia.org/wiki/AlphaGo) primarily manipulated [logarithms of the probability](https://arbital.com/p/log_probability) that the human would take a particular action instead of manipulating the probabilities directly. Then maybe you learn that scientists, engineers, and navigators of yore used to pay large sums of money for giant, pre-computed tables of logarithm outputs.

At that point, you might say

> Hey, wait, why do these logs keep showing up all over the place? Why are they _that_ ubiquitous? What's going on here?

The reason here is that _addition is easier than multiplication,_ and this is true both for navigators of yore (who had to do lots of calculations to keep their ships on course), and for modern AI algorithms.

Let's say you have a whole bunch of numbers, and you know you're going to have to perform lots and lots of operations on them that involve taking some numbers you already have and multiplying them together. There are two ways you could do this: One is that you could bite the bullet and multiply them all, which gets pretty tiresome (if you're doing it by hand) and/or pretty expensive (if you're training a gigantic neural network on millions of games of go). Alternatively, you could put all your numbers through a log function (a one-time cost), and then perform all your operations using addition (much cheaper!), and then transform them back using [https://arbital.com/p/-exponentiation](https://arbital.com/p/-exponentiation) when you're done (another one-time cost).

Empirically, the second method tends to be faster, cheaper, and more convenient (at the cost of some precision, given that most log outputs are [transcendental](https://arbital.com/p/transcendental_number)).

This is the last insight about logarithms contained in this tutorial, and it is a piece of practical advice: If you have to perform a large chain of multiplications, you can often do it cheaper by first transforming your problem into "log space", where multiplication acts like addition (and exponentiation acts like multiplication). Then you can perform a bunch of additions instead, and transform the answer back into normal space at the end.

Malcolm McCrimmon
Nate Soares
Eric Rogstad
Alexei Andreev


The End (of the basic log tutorial)

That concludes our introductory tutorial on logarithms! You have made it to the end.

Throughout this tutorial, we saw that the logarithm base $b$ of $x$ calculates the number of $b$-factors in $x.$ Hopefully, this claim now means more to you than it once did. We've seen a number of different ways of interpreting what logarithms are doing, including:

- $\log_b(x) = y$ means ["it takes about $y$ digits to write $x$ in base $b$."](https://arbital.com/p/416)
- $\log_b(x) = y$ means ["it takes about $y$ $b$-digits to emulate an $x$-digit."](https://arbital.com/p/44l)
- $\log_b(x) = y$ means ["if the space of possible messages to send goes up by a factor of $x$, then the cost, in $b$-digits, goes up by a factor of $y$](https://arbital.com/p/45q)
- And, simply, $\log_b(x) = y$ means that if you start with 1 and grow it by factors of $b$, then after $y$ iterations of this your result will be $x.$

For example, $\log_2(100)$ counts the number of doublings that constitute a factor-of-100 increase. (The answer is more than 6 doublings, but slightly less than 7 doublings).

We've also seen that any function $f$ whose output grows by a constant (that depends on $y$) every time its input grows by a factor of $y$ is [very likely a logarithm function](https://arbital.com/p/4bz), and that, in essence, [https://arbital.com/p/-4gm](https://arbital.com/p/-4gm) function.

We've glanced at the [underlying structure](https://arbital.com/p/4gp) that all logarithm functions tap into, and we've briefly discussed [what makes working with logarithms so dang useful](https://arbital.com/p/4h0).

There are also a huge number of questions about, applications for, and extensions of the logarithm that we _didn't_ explore. Those include, but are not limited to:

- Why is $e$ the natural base of the logarithm?
- What is up with the link between logarithms, exponentials, and roots?
- What is the derivative of $\log_b(x)$ and why is it proportional to $\frac{1}{x}$?
- How can logarithms be efficiently calculated?
- What happens when we extend logarithms to complex numbers, and why is the result a [https://arbital.com/p/-multifunction](https://arbital.com/p/-multifunction)?

Answering these questions will require an advanced tutorial on logarithms. Such a thing does not exist yet, but you can help make it happen.


Normal subgroup

A *normal subgroup* $N$ of group $G$ is one which is closed under [conjugation](https://arbital.com/p/4gk): for all $h \in G$, it is the case that $\{ h n h^{-1} : n \in N \} = N$.
In shorter form, $hNh^{-1} = N$.

Since [conjugacy is equivalent to "changing the worldview"](https://arbital.com/p/4gk), a normal subgroup is one which "looks the same from the point of view of every element of $G$".

A subgroup of $G$ is normal if and only if it is the [kernel](https://arbital.com/p/49y) of some [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $G$ to some group $H$. ([Proof.](https://arbital.com/p/4h7))

%%%knows-requisite([https://arbital.com/p/category_theory_equaliser](https://arbital.com/p/category_theory_equaliser)):
From a category-theoretic point of view, the kernel of $f$ is an equaliser of an arrow $f$ with the zero arrow; it is therefore universal such that composition with $f$ yields zero.
%%%


Subgroup is normal if and only if it is the kernel of a homomorphism

Let $N$ be a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$.
Then $N$ is [normal](https://arbital.com/p/4h6) in $G$ if and only if there is a group $H$ and a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $\phi:G \to H$ such that the [kernel](https://arbital.com/p/49y) of $\phi$ is $N$.

# Proof

## "Normal" implies "is a kernel"
Let $N$ be normal, so it is closed under [conjugation](https://arbital.com/p/4gk).
Then we may define the [https://arbital.com/p/-quotient_group](https://arbital.com/p/-quotient_group) $G/N$, whose elements are the [left cosets](https://arbital.com/p/group_coset) of $N$ in $G$, and where the operation is that $gN + hN = (g+h)N$.
This group is well-defined ([proof](https://arbital.com/p/quotient_by_subgroup_is_well_defined_iff_normal)).

Now there is a homomorphism $\phi: G \to G/N$ given by $g \mapsto gN$.
This is indeed a homomorphism, by definition of the group operation $gN + hN = (g+h)N$.

The kernel of this homomorphism is precisely $\{ g : gN = N \}$; that is simply $N$:

- Certainly $N \subseteq \{ g : gN = N \}$ (because $nN = N$ for all $n$, since $N$ is closed as a subgroup of $G$);
- We have $\{ g : gN = N \} \subseteq N$ because if $gN = N$ then in particular $g e \in N$ (where $e$ is the group identity) so $g \in N$.

## "Is a kernel" implies "normal"
Let $\phi: G \to H$ have kernel $N$, so $\phi(n) = e$ if and only if $n \in N$.
We claim that $N$ is closed under conjugation by members of $G$.

Indeed, $\phi(h n h^{-1}) = \phi(h) \phi(n) \phi(h^{-1}) = \phi(h) \phi(h^{-1})$ since $\phi(n) = e$.
But that is $\phi(h h^{-1}) = \phi(e)$, so $hnh^{-1} \in N$.

That is, if $n \in N$ then $hnh^{-1} \in N$, so $N$ is normal.


Quotient by subgroup is well defined if and only if subgroup is normal

Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $N$ a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $G$.
Then we may define the *quotient group* $G/N$ to be the set of [left cosets](https://arbital.com/p/4j4) $gN$ of $N$ in $G$, with the group operation that $gN + hN = (gh)N$.
This is well-defined if and only if $N$ is normal.

# Proof

## $N$ normal implies $G/N$ well-defined

Recall that $G/N$ is well-defined if "it doesn't matter which way we represent a coset": whichever coset representatives we use, we get the same answer.

Suppose $N$ is a normal subgroup of $G$.
We need to show that given two representatives $g_1 N = g_2 N$ of a coset, and given representatives $h_1 N = h_2 N$ of another coset, that $(g_1 h_1) N = (g_2 h_2)N$.

So given an element of $g_1 h_1 N$, we need to show it is in $g_2 h_2 N$, and vice versa.

Let $g_1 h_1 n \in g_1 h_1 N$; we need to show that $h_2^{-1} g_2^{-1} g_1 h_1 n \in N$, or equivalently that $h_2^{-1} g_2^{-1} g_1 h_1 \in N$.

But $g_2^{-1} g_1 \in N$ because $g_1 N = g_2 N$; let $g_2^{-1} g_1 = m$.
Similarly $h_2^{-1} h_1 \in N$ because $h_1 N = h_2 N$; let $h_2^{-1} h_1 = p$.

Then we need to show that $h_2^{-1} m h_1 \in N$, or equivalently that $p h_1^{-1} m h_1 \in N$.

Since $N$ is closed under conjugation and $m \in N$, we must have that $h_1^{-1} m h_1 \in N$;
and since $p \in N$ and $N$ is closed under multiplication, we must have $p h_1^{-1} m h_1 \in N$ as required.

## $G/N$ well-defined implies $N$ normal

Fix $h \in G$, and consider $hnh^{-1} N + hN$.
Since the quotient is well-defined, this is $(hnh^{-1}h) N$, which is $hnN$ or $hN$ (since $nN = N$, because $N$ is a subgroup of $G$ and hence is closed under the group operation).
But that means $hnh^{-1}N$ is the identity element of the quotient group, since when we added it to $hN$ we obtained $hN$ itself.

That is, $hnh^{-1}N = N$.
Therefore $hnh^{-1} \in N$.

Since this reasoning works for any $h \in G$, it follows that $N$ is closed under conjugation by elements of $G$, and hence is normal.


Alternating group

The *alternating group* $A_n$ is defined as a certain subgroup of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$: namely, the collection of all elements which can be made by multiplying together an even number of [transpositions](https://arbital.com/p/4cn).
This is a well-defined notion ([proof](https://arbital.com/p/4hg)).

%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):
$A_n$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $S_n$; it is the [quotient](https://arbital.com/p/quotient_group) of $S_n$ by the [sign homomorphism](https://arbital.com/p/4hk).
%%%

# Examples

- A [cycle](https://arbital.com/p/49f) of even length is an odd permutation in the sense that it can only be made by multiplying an odd number of transpositions.
For example, $(132)$ is equal to $(13)(23)$.
- A cycle of odd length is an even permutation, in that it can only be made by multiplying an even number of transpositions.
For example, $(1354)$ is equal to $(54)(34)(14)$.

- The alternating group $A_4$ consists precisely of twelve elements: the identity, $(12)(34)$, $(13)(24)$, $(14)(23)$, $(123)$, $(124)$, $(134)$, $(234)$, $(132)$, $(143)$, $(142)$, $(243)$.

# Properties

%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):
The alternating group $A_n$ is of [index](https://arbital.com/p/index_of_a_subgroup) $2$ in $S_n$.
Therefore $A_n$ is [normal](https://arbital.com/p/4h6) in $S_n$ ([proof](https://arbital.com/p/4hl)).
Alternatively we may give the homomorphism explicitly of which $A_n$ is the [kernel](https://arbital.com/p/49y): it is the [sign homomorphism](https://arbital.com/p/4hk).
%%%

- $A_n$ is generated by its $3$-cycles. ([Proof.](https://arbital.com/p/4hs))
- $A_n$ is [simple](https://arbital.com/p/4jc). ([Proof.](https://arbital.com/p/4jb))
- The [conjugacy classes](https://arbital.com/p/4bj) of $A_n$ are [easily characterised](https://arbital.com/p/alternating_group_five_conjugacy_classes).


The collection of even-signed permutations is a group

The collection of elements of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ which are made by multiplying together an even number of permutations forms a subgroup of $S_n$.

This proves that the [https://arbital.com/p/-alternating_group](https://arbital.com/p/-alternating_group) $A_n$ is well-defined, if it is given as "the subgroup of $S_n$ containing precisely that which is made by multiplying together an even number of transpositions".

# Proof

Firstly we must check that "I can only be made by multiplying together an even number of transpositions" is a well-defined notion; this [is in fact true](https://arbital.com/p/4hh).

We must check the group axioms.

- Identity: the identity is simply the product of no transpositions, and $0$ is even.
- Associativity is inherited from $S_n$.
- Closure: if we multiply together an even number of transpositions, and then a further even number of transpositions, we obtain an even number of transpositions.
- Inverses: if $\sigma$ is made of an even number of transpositions, say $\tau_1 \tau_2 \dots \tau_m$, then its inverse is $\tau_m \tau_{m-1} \dots \tau_1$, since a transposition is its own inverse.


The sign of a permutation is well-defined

The [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ contains elements which are made up from [transpositions](https://arbital.com/p/4cn) ([proof](https://arbital.com/p/4cp)).
It is a fact that if $\sigma \in S_n$ can be made by multiplying together an even number of transpositions, then it cannot be made by multiplying an odd number of transpositions, and vice versa.

%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):
Equivalently, there is a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $S_n$ to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_2 = \{0,1\}$, taking the value $0$ on those permutations which are made from an even number of permutations and $1$ on those which are made from an odd number.
%%%

# Proof

Let $c(\sigma)$ be the number of cycles in the [disjoint cycle decomposition](https://arbital.com/p/49f) of $\sigma \in S_n$, including singletons.
For example, $c$ applied to the identity yields $n$, while $c((12)) = n-1$ because $(12)$ is shorthand for $(12)(3)(4)\dots(n-1)(n)$.
We claim that multiplying $\sigma$ by a transposition either increases $c(\sigma)$ by $1$, or decreases it by $1$.

Indeed, let $\tau = (kl)$.
Either $k, l$ lie in the same cycle in $\sigma$, or they lie in different ones.

- If they lie in the same cycle, then $$\sigma = \alpha (k a_1 a_2 \dots a_r l a_s \dots a_t) \beta$$ where $\alpha, \beta$ are disjoint from the central cycle (and [hence commute](https://arbital.com/p/49g) with $(kl)$).
Then $\sigma (kl) = \alpha (k a_s \dots a_t)(l a_1 \dots a_r) \beta$, so we have broken one cycle into two.
- If they lie in different cycles, then $$\sigma = \alpha (k a_1 a_2 \dots a_r)(l b_1 \dots b_s) \beta$$ where again $\alpha, \beta$ are disjoint from $(kl)$.
Then $\sigma (kl) = \alpha (k b_1 b_2 \dots b_s l a_1 \dots a_r) \beta$, so we have joined two cycles into one.

Therefore $c$ takes even values if there are evenly many transpositions in $\sigma$, and odd values if there are odd-many transpositions in $\sigma$.

More formally, let $\sigma = \alpha_1 \dots \alpha_a = \beta_1 \dots \beta_b$, where $\alpha_i, \beta_j$ are transpositions.
%%%knows-requisite([https://arbital.com/p/modular_arithmetic](https://arbital.com/p/modular_arithmetic)):
(The following paragraph is more succinctly expressed as: "$c(\sigma) \equiv n+a \pmod{2}$ and also $\equiv n+b \pmod{2}$, so $a \equiv b \pmod{2}$.")
%%%
Then $c(\sigma)$ flips odd-to-even or even-to-odd for each integer $1, 2, \dots, a$; it also flips odd-to-even or even-to-odd for each integer $1, 2, \dots, b$.
Therefore $a$ and $b$ must be of the same [parity](https://arbital.com/p/even_odd_parity).


Isomorphism: Intro (Math 0)

If two things are essentially the same from a certain perspective, and they only differ in unimportant details, then they are **isomorphic**.

##Comparing Amounts##
Consider the Count von Count. He cares only about counting things. He doesn't care what they are, just how many there are. He decides that he wants to collect items into plastic crates, and he considers two crates equal if both contain the same number of items.

![Equivalent Crates](http://i.imgur.com/sG7Qfyv.jpg)

Now Elmo comes to visit, and he wants to impress the Count, but Elmo is not great at counting. Without counting them explicitly, how can Elmo tell if two crates contain the same number of items?

Well, he can take one item out of each crate and put the pair to one side.

![Pairing one pair of items](https://i.imgur.com/myTqIrRl.jpg)

 He continues pairing items up in this way and when one crate runs out he checks if there are any left over in the other crate. If there aren't any left over, then he knows there were the same number of items in both crates.

![Pairing all items in one crate with items in the other](https://i.imgur.com/S87JO2nl.jpg)

Since the Count von Count only cares about counting things, the two crates are basically equivalent, and might as well be the same crate to him. Whenever two objects are the same from a certain perspective, we say that they are **isomorphic**.

In this example, the way in which the crates were the same is that each item in one crate could be paired with an item in the other.

![Pairing all items in one box with items in the other](https://i.imgur.com/S87JO2nl.jpg)

![Bijection between crates](https://i.imgur.com/53YbraFl.jpg)

This wouldn't have been possible if the crates had different numbers of items in them.

![Different numbers of items](https://i.imgur.com/F5rLwAsl.jpg)

![No way to pair items](https://i.imgur.com/KCI9UOvl.jpg)

Whenever you can match each item in one collection with exactly one item in another collection, we say that the collections are **[bijective](https://arbital.com/p/-499)** and the way you paired them is a **[bijection](https://arbital.com/p/-499)**. A bijection is a specific kind of isomorphism.

![Bijection between crates](https://i.imgur.com/53YbraFl.jpg)

Note that there might be many different bijections between two bijective things.

![Another bijection between crates](http://i.imgur.com/Q6Si1ZX.png)

In fact, all that counting involves is pairing up the things you want to count, either with your fingers or with the concepts of 'numbers' in your head. If there are as many objects in one crate as there are numbers from one to seven, and there are as many objects in another crate as numbers from one to seven, then both crates contain the same number of objects.

##Comparing Maps##
Now imagine that you have a map of the London Underground. Such a map is not to scale, nor does it even show how the  tracks bend or which station is in which direction compared to another. They only record which stations are connected.

![Map of the London Underground](http://i.imgur.com/9d8oX2K.gif)

Your Spanish friend is coming to visit and you want to get them a version of the map in Spanish. But on the Spanish maps, the shape of the tracks is different and you can't read Spanish. What's more, not all the maps are of the London Underground! What do you do? Well, given a Spanish map and your English map, you can try to match up the stations (through trial and error) and if the stations are all *connected to each other in the same ways* on both maps then you know they are both of the same train system.

More precisely, consider the following smaller (fictional) example. There are stations *Trafalgal*, *Marybone*, *Oxbridge*, *Eastweston* and *Charlesburrough* on the English map. *Marybone* is connected to all the other stations, and *Charlesburrough* is connected to *Eastweston*. (There are no other connections)

![Fictional London Underground Map](http://i.imgur.com/J3p094x.png)

%note: We don't want to worry about whether stations are connected to themselves. You can just assume no station is ever connected to itself.%

%note: If one station is connected to another, then the second station is also connected to the first. So since *Marybone* is connected to *Eastweston* then *Eastweston* is connected to *Marybone*. If it seems silly to even mention this fact, then don't worry to much. It's just that we might just have easily decided that there's a one-way train running from *Marybone* to *Eastweston* but not in the other direction.%

Assume the Spanish map has the following stations: *Patata*, *Huesto*, *Carbon*, *Esteoeste*, and *Puente de Buey*. Assume also that *Huesto* is connected to every other station, and that *Carbon* is connected to *Esteoeste*. (Again, there are no other connections).

![Fictional Spanish Map](http://i.imgur.com/nENfDqk.png)

Then the two maps are essentially the same for your purposes. They are **isomorphic** (as [graphs](https://arbital.com/p/-graph), in fact), and the way that you matched the stations on the one map with those on the other is an **isomorphism**.

## Isomorphisms##
Imagine that you have a *London Underground Official Spanish-to-English Train Station Dictionary* that tells you how to translate the names of the train stations. So, for example, you can use it to convert *Patata* to *Trafalgal*. Then this dictionary is an **isomorphism** from the Spanish map to the English one.

In particular, the dictionary translates *Huesto* to *Marybone*, *Patata* to *Tralfalgal*, *Puento de Buey* to *Oxbridge*, *Esteoeste* to *Eastweston*, and *Carbon* to *Charlesburrough*.

You could also get a *London Underground Official* **English-to-Spanish** *Train Station Dictionary*. Then, if you were to use this dictionary to translate *Tralfalgal*, you'd get back *Patata*. Hence your first original translation of that station from English to Spanish has been undone.

In particular, the dictionary translates *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon*.

![English Stations Paired Up with Spanish Stations](http://i.imgur.com/sElv72C.png)

In fact, if you take the English map and translate all of the stations into Spanish using the one dictionary and then translate back, you'd get back to where you started. Similarly, if you translated the Spanish map from Spanish into English with the one dictionary and back to Spanish with the other you'd get the Spanish map back. Hence both of these dictionaries are complete [inverses](https://arbital.com/p/-inverse) of each other.

In fact, in [category theory](https://arbital.com/p/-4c7), this is exactly the definition of an isomorphism: if you have some translation ([https://arbital.com/p/-4d8](https://arbital.com/p/-4d8)) such that you can find a backwards translation (morphism in the opposite direction), and using the one translation after the other is for all intents and purposed the same as not having translated anything at all (i.e., no important information is lost in translation), then the original translation is an **isomorphism**. (In fact, both of them are isomorphisms).

What if you had a different pair of dictionaries.

In particular, what if the Spanish-to-English dictionary translates *Huesto* to *Marybone*, *Puento de Buey* to *Tralfalgal*, *Patata* to *Oxbridge*, *Carbon* to *Eastweston*, and *Esteoeste* to *Charlesburrough*?

Then if we translate from Spanish into English, and then translate back with the original English-to-Spanish dictionary, then *Patata* is first translated to *Oxbridge*, but then it is translated back into *Puente de Buey*. So it does not reverse the translation. Hence this is not an isomorphism.

However, what if the English-to-Spanish Dictionary translates *Marybone* to *Huesto*, *Oxbridge* to *Patata*. *Eastweston* to *Puento de Buey*, *Tralfalgal* to *Esteoeste*, and *Charlesburrough* to *Carbon*? Then the translations are reverses of each other. Hence this is another isomorphism. There may be many isomorphisms between two isomorphic maps.

![Another Way of Pairing English Stations with Spanish Stations](http://i.imgur.com/X6A6oNb.png)

##Non-Isomorphic Maps##
Imagine you had the English map from above.  As a reminder the stations were:
*Marybone*, *Eastweston*, *Charlesburrough*, *Tralfalgal*, and *Oxbridge*.

If, now, you find a Spanish map with only four stations on it, it can't possibly be isomorphic to your English map; there would be some station appearing on the English map which isn't named on the Spanish one.

![Spanish Map with Only Four Stations When English Map has Five](http://i.imgur.com/BkVZBWB.png)

Similarly, if the Spanish map has six stations, then they aren't isomorphic either since there is an extra station on the Spanish map not appearing on the English one.

![Spanish Map with Six Stations When English Map has Five](http://i.imgur.com/EYYmzWi.png)

What if there are five stations on the Spanish map. Is it then definitely isomorphic to the English one?

Recall that:
*Marybone* is connected to every other station, and *Eastweston* is connected to *Charlesburrough*.

But what if now instead on the Spanish map, *Huesto* is still connected to everything, but nothing else is connected to anything else.

![Huesto is Connected to Everything With No Other Connections. English Map Still Has the Same Connections](http://i.imgur.com/4mQD0hK.png)

Then the translation taking *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon* is not an isomorphism. *Eastweston* is connected to *Charlesburrough* on the English map, but the corresponding stations on the Spanish map, *Esteoeste* and *Carbon*, are not connected to each other. Hence this translation is not an isomorphism, since under these translations, the maps represent different things.

But even though this way of pairing up the stations isn't an isomorphism, maybe there is another way of pairing them up which is? But no, even this is doomed to failure because the *number of connections* in both cases is different. For example, *Eastweston* is connected to two stations, but no station on the Spanish map is connected to two other stations. Hence there cannot be any translation that works. No isomorphism exists between the two maps.

If no isomorphism exists between two structures, then they are **non-isomorphic**.

Notice, in fact, that the two maps do not have the same total number of connections. There are five connections on the English map, but only four on the Spanish map. Hence they cannot be isomorphic.

What if the Spanish map has the following connections? *Patata* to everything except *Carbon*, and *Carbon* to *Huesto*and *Esteoeste*. Then there still cannot be an isomorphism, since, again, *Marybone* is connected to four other stations, but nothing on the Spanish map is connected to four stations. In this case, both maps have five connections. Hence even if both maps have the same total number of connections, they may still be non-isomorphic.

![Both Maps Have Five Connections but Are Non-Isomorphic](http://i.imgur.com/BKXLPqv.png)

##Comparing Weights##
Not all isomorphisms need be mappings between structures. Consider if you work at the post-office and must weigh packages. You do not care about the size and shape of the packages, only their weight. Then you consider two packages isomorphic if their weights are equal.

Imagine, then, that you have two packages, say one containing a book %note: *The Official London Underground History of Train Stations*% and the other is a plastic crate %note: "To the Count, with love"%.

You also have a half-broken pair of brass scales: they have a pair of pans on which items can be placed.

![Empty Balanced Scales](http://i.imgur.com/2wuFlTO.png)

However, they can only tip to the left or remain flat.

![Empty Scales Tilting Left](http://i.imgur.com/oz3m8vF.png)

If the item on the left is heavier than the one on the right, then the scales tilt left.

![Scales With Heavy Object on Left, Scales Tilt Left](http://i.imgur.com/iSmYESb.png)

Otherwise, if they are of equal weight or the item on the left is lighter than the one on the right, the scales remain level.

![Scales With Same Object on Both Sides, Scales Flat](http://i.imgur.com/NZe1uQC.png)

![Scales With Heavy Object on Right, Scales Flat](http://i.imgur.com/0BqznlG.png)

Place the book on the left pan of the scale, and the crate on the right. If the scales balance then either the book is lighter than the crate or it is the same weight as the crate. Now swap them. If they remain level, then either the crate is lighter than the book or it is the same weight as the book. Since the book cannot be lighter than the crate whilst the crate is simultaneously lighter than the book, they must be the same weight. Hence they are isomorphic.

This very act of balancing the scales is an isomorphism. It has an inverse: just swap the two packages around! Start with the book on the left pan and the crate on the right. Then place the crate on the left pan and the book on the right. The fact that the scales balance both times tells you (the obvious fact) that the book weighs the same as itself. Since you already know this,  doing this actually tells you as much about the book's weight compared to itself as doing nothing at all.

If this last part seems silly or confusing, don't worry too much about it. It's just to illustrate how the idea of an isomorphism is intricately tied with having an inverse.

##An Isomorphism Joke##
A man walks into a bar. He is surprised at how the patrons are acting.  One of them says a number, like “forty-two”, and the rest break into laughter. He asks the bartender what’s going on. The bartender explains that they all come here so often that they’ve memorized all of each other’s jokes, and instead of telling them explicitly, they just give each a number, say the number, and laugh appropriately. The man is intrigued, so he shouts “Two thousand!”. He is shocked to find everyone laughs uproariously, the loudest he's heard that evening. Perplexed, he turns to the bartender and says  “They laughed so much more at mine than at any of the others." "Well of course," the bartender answers matter-of-factly, "they've never heard that one before!”

Eric Rogstad
Eric Bruylant
Mark Chimes
Team Arbital


Sign homomorphism (from the symmetric group)

The sign [homomorphism](https://arbital.com/p/47t) is given by sending a permutation $\sigma$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ to $0$ if we can make $\sigma$ by multiplying together an even number of [transpositions](https://arbital.com/p/4cn), and to $1$ otherwise.

%%%knows-requisite([https://arbital.com/p/modular_arithmetic](https://arbital.com/p/modular_arithmetic)):
Equivalently, it is given by sending $\sigma$ to the number of transpositions making it up, [modulo](https://arbital.com/p/modular_arithmetic) $2$.
%%%

The sign homomorphism [is well-defined](https://arbital.com/p/4hh).

%%%knows-requisite([https://arbital.com/p/quotient_group](https://arbital.com/p/quotient_group)):
The [https://arbital.com/p/-alternating_group](https://arbital.com/p/-alternating_group) is obtained by taking the [quotient](https://arbital.com/p/quotient_group) of the symmetric group by the sign homomorphism.
%%%


Alternating group is generated by its three-cycles

The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$ is generated by its $3$-[cycles](https://arbital.com/p/49f).
That is, every element of $A_n$ can be made by multiplying together $3$-cycles only.

# Proof

The product of two [transpositions](https://arbital.com/p/4cn) is a product of $3$-cycles:

- $(ij)(kl) = (ijk)(jkl)$
- $(ij)(jk) = (ijk)$
- $(ij)(ij) = e$.

Therefore any permutation which is a product of evenly-many transpositions (that is, all of $A_n$) is a product of $3$-cycles, because we can group up successive pairs of transpositions.

Conversely, every $3$-cycle is in $A_n$ because $(ijk) = (ij)(jk)$.


Group coset

Given a subgroup $H$ of [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$, the *left cosets* of $H$ in $G$ are sets of the form $\{ gh : h \in H \}$, for some $g \in G$.
This is written $gH$ as a shorthand.

Similarly, the *right cosets* are the sets of the form $Hg = \{ hg: h \in H \}$.

# Examples
%%%knows-requisite([https://arbital.com/p/497](https://arbital.com/p/497)):
## Symmetric group

In $S_3$, the [https://arbital.com/p/-497](https://arbital.com/p/-497) on three elements, we can list the elements as $\{ e, (123), (132), (12), (13), (23) \}$, using [cycle notation](https://arbital.com/p/49f).
Define $A_3$ (which happens to have a name: the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf)) to be the subgroup with elements $\{ e, (123), (132) \}$.

Then the coset $(12) A_3$ has elements $\{ (12), (12)(123), (12)(132) \}$, which is simplified to $\{ (12), (23), (13) \}$.

The coset $(123)A_3$ is simply $A_3$, because $A_3$ is a subgroup so is closed under the group operation. $(123)$ is already in $A_3$.
%%%

# Properties

- The left cosets of $H$ in $G$ [partition](https://arbital.com/p/set_partition) $G$. ([Proof.](https://arbital.com/p/4j5))
- For any pair of left cosets of $H$, there is a [bijection](https://arbital.com/p/499) between them; that is, all the cosets are all the same size. ([Proof.](https://arbital.com/p/4j8))

# Why are we interested in cosets?

Under certain conditions (namely that the subgroup $H$ must be [normal](https://arbital.com/p/4h6)), we may define the [https://arbital.com/p/-quotient_group](https://arbital.com/p/-quotient_group), a very important concept; see the page on ["left cosets partition the parent group"](https://arbital.com/p/4j5) for a glance at why this is useful.

Additionally, there is a key theorem whose usual proof considers cosets ([Lagrange's theorem](https://arbital.com/p/lagrange_theorem_on_subgroup_size)) which strongly restricts the possible sizes of subgroups of $G$, and which itself is enough to classify all the groups of [order](https://arbital.com/p/3gg) $p$ for $p$ [prime](https://arbital.com/p/prime_number).
Lagrange's theorem also has very common applications in [https://arbital.com/p/-number_theory](https://arbital.com/p/-number_theory), in the form of the [Fermat-Euler theorem](https://arbital.com/p/fermat_euler_theorem).


Left cosets partition the parent group

Given a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ and a subgroup $H$, the [left cosets](https://arbital.com/p/4j4) of $H$ in $G$ [partition](https://arbital.com/p/set_partition) $G$, in the sense that every element of $g$ is in precisely one coset.

# Proof
Firstly, every element is in a coset: since $g \in gH$ for any $g$.
So we must show that no element is in more than one coset.

Suppose $c$ is in both $aH$ and $bH$.
Then we claim that $aH = cH = bH$, so in fact the two cosets $aH$ and $bH$ were the same.
Indeed, $c \in aH$, so there is $k \in H$ such that $c = ak$.
Therefore $cH = \{ ch : h \in H \} = \{ akh : h \in H \}$.

Exercise: $\{ akh : h \in H \} = \{ ar : r \in H \}$.
%%hidden(Show solution):
Suppose $akh$ is in the left-hand side.
Then it is in the right-hand side immediately: letting $r=kh$.

Conversely, suppose $ar$ is in the right-hand side.
Then we may write $r = k k^{-1} r$, so $a k k^{-1} r$ is in the right-hand side; but then $k^{-1} r$ is in $H$ so this is exactly an object which lies in the left-hand side.
%%

But that is just $aH$.

By repeating the reasoning with $a$ and $b$ interchanged, we have $cH = bH$; this completes the proof.

# Why is this interesting?

The fact that the left cosets partition the group means that we can, in some sense, "compress" the group $G$ with respect to $H$.
If we are only interested in $G$ "up to" $H$, we can deal with the partition rather than the individual elements, throwing away the information we're not interested in.

This concept is most importantly used in defining the [https://arbital.com/p/-4tq](https://arbital.com/p/-4tq).
To do this, the subgroup must be [normal](https://arbital.com/p/4h6) ([proof](https://arbital.com/p/4h9)).
In this case, the collection of cosets itself inherits a group structure from the parent group $G$, and the structure of the quotient group can often tell us a lot about the parent group.


Church-Turing thesis: Evidence for the Church-Turing thesis

As the Church-Turing thesis is not a [proper mathematical sentence](https://arbital.com/p/) we cannot prove it. However, we can collect [https://arbital.com/p/-3s](https://arbital.com/p/-3s) to increase our confidence in its correctness.

#The inductive argument
Every computational model we have seen so far is [reducible](https://arbital.com/p/) to Turing's model.

Indeed, the thesis was originally formulated independently by Church and Turing in reference to two different computational models ([https://arbital.com/p/Turing_machines](https://arbital.com/p/Turing_machines) and [https://arbital.com/p/Lambda_Calculus](https://arbital.com/p/Lambda_Calculus) respectively). When they were shown to be [equivalent](https://arbital.com/p/equivilance_relation) it was massive evidence in favor of both of them.

A non-exhaustive list of models which can be shown to be reducible to Turing machines are:

* [https://arbital.com/p/Lambda_calculus](https://arbital.com/p/Lambda_calculus)
* [https://arbital.com/p/Quantum_computation](https://arbital.com/p/Quantum_computation)
* [Non-deterministic_Turing_machines](https://arbital.com/p/Nondeterministic_Turing_machines)
* [https://arbital.com/p/Register_machines](https://arbital.com/p/Register_machines)
* The set of [https://arbital.com/p/-recursive_functions](https://arbital.com/p/-recursive_functions)

#Lack of counterexamples
Perhaps the strongest argument we have for the CT thesis is that there is not a widely accepted candidate to a counterexample of the thesis. This is unlike the [https://arbital.com/p/-4ct](https://arbital.com/p/-4ct), where quantum computation stands as a likely counterexample.

One may wonder whether the computational models which use a source of [randomness](https://arbital.com/p/) (such as quantum computation or [https://arbital.com/p/-probabilistic_Turing_machines](https://arbital.com/p/-probabilistic_Turing_machines)) are a proper counterexample to the thesis: after all, Turing machines are fully [https://arbital.com/p/-deterministic](https://arbital.com/p/-deterministic), so they cannot simulate randomness.

To properly explain this issue we have to recall what it means for a quantum computer or a probabilistic Turing machine to compute something: we say that such a device computes a function $f$ if for every input $x$ the probability of the device outputting $f(x)$ is greater or equal to some arbitrary constant greater than $1/2$. Thus we can compute $f$ in a classical machine, for there exists always the possibility of simulating every possible outcome of the randomness to deterministically compute the probability distribution of the output, and output as a result the possible outcome with greater probability.

Thus, randomness is reducible to Turing machines, and the CT thesis holds.


Left cosets are all in bijection

Let $H$ be a subgroup of $G$.
Then for any two [left cosets](https://arbital.com/p/4j4) of $H$ in $G$, there is a [https://arbital.com/p/-499](https://arbital.com/p/-499) between the two cosets.

# Proof

Let $aH, bH$ be two cosets.
Define the function $f: aH \to bH$ by $x \mapsto b a^{-1} x$.

This has the correct [codomain](https://arbital.com/p/3lg): if $x \in aH$ (so $x = ah$, say), then $ba^{-1} a x = bx$ so $f(x) \in bH$.

The function is [injective](https://arbital.com/p/4b7): if $b a^{-1} x = b a^{-1} y$ then (pre-multiplying both sides by $a b^{-1}$) we obtain $x = y$.

The function is [surjective](https://arbital.com/p/4bg): given $b h \in b H $, we want to find $x \in aH$ such that $f(x) = bh$.
Let $x = a h$ to obtain $f(x) = b a^{-1} a h = b h$, as required.


The alternating groups on more than four letters are simple

Let $n > 4$ be a [https://arbital.com/p/-45h](https://arbital.com/p/-45h). Then the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$ on $n$ elements is [simple](https://arbital.com/p/simple_group).

# Proof

We go by [induction](https://arbital.com/p/mathematical_induction) on $n$.
The base case of $n=5$ [we treat separately](https://arbital.com/p/4jf).

For the inductive step: let $n \geq 6$, and suppose $H$ is a nontrivial [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $A_n$.
Ultimately we aim to show that $H$ contains the $3$-[cycle](https://arbital.com/p/49f) $(123)$; since $H$ [is a union of conjugacy classes](https://arbital.com/p/subgroup_normal_iff_union_of_conjugacy_classes), and since the $3$-cycles [form a conjugacy class](https://arbital.com/p/splitting_conjugacy_classes_in_alternating_group) in $A_n$, this means every $3$-cycle is in $H$ and [hence](https://arbital.com/p/4hs) $H = A_n$.

## Lemma: $H$ contains a member of $A_{n-1}$
To start, we will show that at least $H$ contains *something* from $A_{n-1}$ (which we view as all the elements of $A_n$ which don't change the letter $n$). %%note:Recall that $A_n$ [acts](https://arbital.com/p/3t9) naturally on the set of "letters" $\{1,2,\dots, n \}$ by permutation.%%
(This approach has to be useful for an induction proof, because we need some way of introducing the simplicity of $A_{n-1}$.)
That is, we will show that there is some $\sigma \in H$ with $\sigma \not = e$, such that $\sigma(n) = n$.

Let $\sigma \in H$, where $\sigma$ is not the identity.
$\sigma$ certainly sends $n$ somewhere; say $\sigma(n) = i$, where $i \not = n$ (since if it is, we're done immediately).

Then if we can find some $\sigma' \in H$, not equal to $\sigma$, such that $\sigma'(n) = i$, we are done: $\sigma^{-1} \sigma'(n) = n$.

$\sigma$ must move something other than $i$ (that is, there must be $j \not = i$ such that $\sigma(j) \not = j$), because if it did not, it would be the transposition $(n i)$, which is not in $A_n$ because it is an odd number of transpositions.
Hence $\sigma(j) \not = j$ and $j \not = i$; also $j \not = n$ because if it were,

Now, since $n \geq 6$, we can pick $x, y$ distinct from $n, i, j, \sigma(j)$.
Then set $\sigma' = (jxy) \sigma (jxy)^{-1}$, which must lie in $H$ because $H$ is closed under [conjugation](https://arbital.com/p/4gk).

Then $\sigma'(n) = i$; and $\sigma' \not = \sigma$, because $\sigma'(j) = \sigma(y)$ which is not equal to $\sigma(j)$ (since $y \not = j$).
Hence $\sigma'$ and $\sigma$ have different effects on $j$ so they are not equal.

## Lemma: $H$ contains all of $A_{n-1}$
Now that we have shown $H$ contains some member of $A_{n-1}$.
But $H \cap A_{n-1}$ is normal in $A_n$, because $H$ is normal in $A_n$ and it is [_intersect _subgroup _is _normal the intersection of a subgroup with a normal subgroup](https://arbital.com/p/4h6).
Therefore by the inductive hypothesis, $H \cap A_{n-1}$ is either the trivial group or is $A_{n-1}$ itself.

But $H \cap A_{n-1}$ is certainly not trivial, because our previous lemma gave us a non-identity element in it; so $H$ must actually contain $A_{n-1}$.

## Conclusion

Finally, $H$ contains $A_{n-1}$ so it contains $(123)$ in particular; so we are done by the initial discussion.

# Behaviour for $n \leq 4$
- $A_1$ is the trivial group so is vacuously not simple.
- $A_2$ is also the trivial group.
- $A_3$ is isomorphic to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_3$ on three generators, so it is simple: it has no nontrivial proper subgroups, let alone normal ones.
- $A_4$ has the following normal subgroup (the [Klein four-group](https://arbital.com/p/klein_four_group)): $\{ e, (12)(34), (13)(24), (14)(23) \}$. Therefore $A_4$ is not simple.


Simple group

A simple group is a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) with no [normal subgroups](https://arbital.com/p/4h6).


The alternating group on five elements is simple

The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements is [simple](https://arbital.com/p/4jc).

# Proof

Recall that $A_5$ has [order](https://arbital.com/p/3gg) $60$, so [Lagrange's theorem](https://arbital.com/p/4jn) states that any subgroup of $A_5$ has order dividing $60$.

Suppose $H$ is a normal subgroup of $A_5$, which is not the trivial subgroup $\{ e \}$.
If $H$ has order divisible by $3$, then by [Cauchy's theorem](https://arbital.com/p/4l6) there is a $3$-[cycle](https://arbital.com/p/49f) in $H$ (because the $3$-cycles are the only elements with order $3$ in $A_5$).
Because $H$ [is a union of conjugacy classes](https://arbital.com/p/4jw), and because the $3$-cycles [form a conjugacy class in $A_n$ for $n > 4$](https://arbital.com/p/4kv), $H$ would therefore contain *every* $3$-cycle; but then [it would be the entire alternating group](https://arbital.com/p/4hs).

If instead $H$ has order divisible by $2$, then there is a double transposition such as $(12)(34)$ in $H$, since these are the only elements of order $2$ in $A_5$.
But then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12)(34)$ and $(15)(34)$, so it contains $(15)(34)(12)(34) = (125)$.
Hence as before $H$ contains every $3$-cycle so is the entire alternating group.

So $H$ must have order exactly $5$, by [Lagrange's theorem](https://arbital.com/p/4jn); so it contains an element of order $5$ since [prime order groups are cyclic](https://arbital.com/p/4jh).

The only such elements of $A_n$ are $5$-cycles; but the conjugacy class of a $5$-cycle is of size $12$, which is too big to fit in $H$ which has size $5$.


Prime order groups are cyclic

Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) whose [order](https://arbital.com/p/3gg) is equal to $p$, a [https://arbital.com/p/-4mf](https://arbital.com/p/-4mf).
Then $G$ is [isomorphic](https://arbital.com/p/49x) to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_p$ of order $p$.

# Proof
Pick any non-identity element $g$ of the group.

By [Lagrange's theorem](https://arbital.com/p/4jn), the subgroup generated by $g$ has size $1$ or $p$ (since $p$ was prime).
But it can't be $1$ because the only subgroup of size $1$ is the trivial subgroup.

Hence the subgroup must be the entire group.


Lagrange theorem on subgroup size

Lagrange's Theorem states that if $G$ is a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $H$ a subgroup, then the [order](https://arbital.com/p/3gg) $|H|$ of $H$ divides the order $|G|$ of $G$.
It generalises to infinite groups: the statement then becomes that the [left cosets](https://arbital.com/p/4j4) form a [partition](https://arbital.com/p/set_partition), and for any pair of cosets, there is a [bijection](https://arbital.com/p/499) between them.

# Proof

In full generality, the cosets [form a partition](https://arbital.com/p/4j5) and [are all in bijection](https://arbital.com/p/4j8).

To specialise this to the finite case, we have divided the $|G|$ elements of $G$ into buckets of size $|H|$ (namely, the cosets), so $|G|/|H|$ must in particular be an integer.


The alternating group on five elements is simple: Simpler proof

If there is a non-trivial [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) $H$ of the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements, then [it is a union of conjugacy classes](https://arbital.com/p/4jw).
Additionally, by [Lagrange's theorem](https://arbital.com/p/4jn), the [order](https://arbital.com/p/3gg) of a subgroup must divide the order of the group, so the total size of $H$ must divide $60$.

We can list the [conjugacy classes of $A_5$](https://arbital.com/p/alternating_group_five_conjugacy_classes); they are of size $1, 20, 15, 12, 12$ respectively.

By a brute-force check, no sum of these containing $1$ can possibly divide $60$ (which is the size of $A_5$) unless it is $1$ or $60$.

# The brute-force check

We first list the [divisors](https://arbital.com/p/number_theory_divisor) of $60$: they are $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$$

Since the subgroup $H$ must contain the identity, it must contain the conjugacy class of size $1$.
If it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $13$ (since the smallest other class is of size $12$); so it is allowed to be one of $15$, $20$, $30$, or $60$.
Since $H$ is also assumed to be a *proper* subgroup, it cannot be $A_5$ itself, so in fact $60$ is also banned.

## The class of size $20$
If $H$ then contains the conjugacy class of size $20$, then $H$ can only be of size $30$ because we have already included $21$ elements.
But there is no way to add just $9$ elements using conjugacy classes of size bigger than or equal to $12$.

So $H$ cannot contain the class of size $20$.

## The class of size $15$

In this case, $H$ is allowed to be of size $20$ or $30$, and we have already found $16$ elements of it. So there are either $4$ or $14$ elements left to find; but we are only allowed to add classes of size exactly $12$, so this can't be done either.

## The classes of size $12$

What remains is two classes of size $12$, from which we can make $1+12 = 13$ or $1+12+12 = 25$.
Neither of these divides $60$, so these are not legal options either.

This exhausts the search, and completes the proof.


Subgroup is normal if and only if it is a union of conjugacy classes

Let $H$ be a subgroup of the [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$.
Then $H$ is [normal](https://arbital.com/p/4h6) in $G$ if and only if it can be expressed as a [union](https://arbital.com/p/set_union) of [conjugacy classes](https://arbital.com/p/4bj).

# Proof

$H$ is normal in $G$ if and only if $gHg^{-1} = H$ for all $g \in G$; equivalently, if and only if $ghg^{-1} \in H$ for all $h \in H$ and $g \in G$.

But if we fix $h \in H$, then the statement that $ghg^{-1} \in H$ for all $g \in G$ is equivalent to insisting that the conjugacy class of $h$ is contained in $H$.
Therefore $H$ is normal in $G$ if and only if, for all $h \in H$, the conjugacy class of $h$ lies in $H$.

If $H$ is normal, then it is clearly a union of conjugacy classes (namely $\cup_{h \in H} C_h$, where $C_h$ is the conjugacy class of $h$).

Conversely, if $H$ is not normal, then there is some $h \in H$ such that the conjugacy class of $h$ is not wholly in $H$; so $H$ is not a union of conjugacy classes because it contains $h$ but not the entire conjugacy class of $h$.
(Here we have used that the [https://arbital.com/p/-conjugacy_classes_partition_the_group](https://arbital.com/p/-conjugacy_classes_partition_the_group).)

# Interpretation

A normal subgroup is one which is fixed under conjugation; the most natural (and, indeed, the smallest) objects which are fixed under conjugation are conjugacy classes; so this criterion tells us that to obtain a *subgroup* which is fixed under conjugation, it is necessary and sufficient to assemble these objects (the conjugacy classes), which are themselves the smallest objects which are fixed under conjugation, into a group.


Conjugacy classes of the alternating group on five elements

This page lists the [conjugacy classes](https://arbital.com/p/4bj) of the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements.
See a [different lens](https://arbital.com/p/4l0) for a derivation of this result using less theory.

$A_5$ has size $5!/2 = 60$, where the exclamation mark denotes the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function.
We will assume access to [the conjugacy class table of $S_5$](https://arbital.com/p/4bk) the [https://arbital.com/p/-497](https://arbital.com/p/-497) on five elements; $A_5$ is a [quotient](https://arbital.com/p/quotient_group) of $S_5$ by the [sign homomorphism](https://arbital.com/p/4hk).

We have that a conjugacy class splits if and only if its [cycle type](https://arbital.com/p/49f) is all odd, all distinct. ([Proof.](https://arbital.com/p/4kv))
This makes the classification of conjugacy classes very easy.

# The table

We must remove all the lines of [$S_5$'s table](https://arbital.com/p/4bk) which correspond to odd permutations (that is, those which are the product of odd-many [transpositions](https://arbital.com/p/4cn)). Indeed, those lines are classes which are not even in $A_5$.

We are left with cycle types $(5)$, $(3, 1, 1)$, $(2, 2, 1)$, $(1,1,1,1,1)$.
Only the $(5)$ cycle type can split into two, by the splitting condition.
It splits into the class containing $(12345)$ and the class which is $(12345)$ conjugated by odd permutations in $S_5$.
A representative for that latter class is $(12)(12345)(12)^{-1} = (21345)$.

$$\begin{array}{|c|c|c|c|}
\hline
\text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline
(12345) & 12 & 5 & 5 \\ \hline
(21345) & 12 & 5 & 5 \\ \hline
(123) & 20 & 3,1,1 & 3 \\ \hline
(12)(34) & 15 & 2,2,1 & 2 \\ \hline
e & 1 & 1,1,1,1,1 & 1 \\ \hline
\end{array}$$


Splitting conjugacy classes in alternating group

Recall that in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, the notion of "[https://arbital.com/p/-4bj](https://arbital.com/p/-4bj)" coincides with that of "has the same [cycle type](https://arbital.com/p/4cg)" ([proof](https://arbital.com/p/4bh)).
It turns out to be the case that when we descend to the [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_n$, the conjugacy classes nearly all remain the same; the exception is those classes in $S_n$ which have cycle type all odd and all distinct.
Those classes split into exactly two classes in $A_n$.

# Proof

## Conjugate implies same cycle type

This direction of the proof is identical to [the same direction](https://arbital.com/p/4bh) in the proof of the corresponding result on symmetric groups.

## Converse: splitting condition

Recall before we start that an even-length cycle can only be written as the product of an *odd* number of transpositions, and vice versa.

The question is: is every $\tau \in A_n$ with the same cycle type as $\sigma \in A_n$ conjugate (in $A_n)$ to $\sigma$?
If so, then the conjugacy class in $A_n$ of $\sigma$ is just the same as that of $\sigma$ in $S_n$; if not, then the conjugacy class in $S_n$ must break into two pieces in $A_n$, namely $\{ \rho \sigma \rho^{-1} : \rho \ \text{even} \}$ and $\{ \rho \sigma \rho^{-1} : \rho \ \text{odd} \}$.
(Certainly these are conjugacy classes: they only contain even permutations so they are subsets of $A_n$, while everything in either class is conjugate in $A_n$ because the definition only depends on the [parity](https://arbital.com/p/even_odd_parity) of $\rho$.)

Let $\sigma = c_1 \dots c_k$, and $\tau = c_1' \dots c_k'$ of the same cycle type: $c_i = (a_{i1} \dots a_{i r_i})$, $c_i' = (b_{i1} \dots b_{i r_i})$.

Define $\rho$ to be the permutation which takes $a_{ij}$ to $b_{ij}$, but otherwise does not move any other letter.
Then $\rho \sigma \rho^{-1} = \tau$, so if $\rho$ lies in $A_n$ then we are done: $\sigma$ and $\tau$ are conjugate in $A_n$.

That is, we may assume without loss of generality that $\rho$ is odd.

### If any of the cycles is even in length

Suppose without loss of generality that $r_1$, the length of the first cycle, is even.
Then we can rewrite $c_1' = (b_{11} \dots b_{1r_1})$ as $(b_{12} b_{13} \dots b_{1 r_1} b_{11})$, which is the same cycle expressed slightly differently.

Now $c_1'$ is even in length, so it is an odd permutation (being a product of an odd number of transpositions, $(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \dots (b_{13} b_{11})(b_{12} b_{11})$).
Hence $\rho c_1'$ is an even permutation.

But conjugating $\tau$ by $\rho c_1'$ yields $\sigma$:
$$\sigma = \rho \tau \rho^{-1} = \rho c_1' (c_1'^{-1} \tau c_1') c_1'^{-1} \rho^{-1}$$
which is the result of conjugating $c_1'^{-1} \tau c_1' = \tau$ by $\rho c_1'$.

(It is the case that $c_1'^{-1} \tau c_1' = \tau$, because $c_1'$ commutes with all of $\tau$ except with the first cycle $c_1'$, so the expression is $c_1'^{-1} c_1' c_1' c_2' \dots c_k'$, where we have pulled the final $c_1'$ through to the beginning of the expression $\tau = c_1' c_2' \dots c_k'$.)

%%hidden(Example):
Suppose $\sigma = (12)(3456), \tau = (23)(1467)$.

We have that $\tau$ is taken to $\sigma$ by conjugating with $\rho = (67)(56)(31)(23)(12)$, which is an odd permutation so isn't in $A_n$.
But we can rewrite $\tau = (32)(1467)$, and the new permutation we'll conjugate with is $\rho c_1' = (67)(56)(31)(23)(12)(32)$, where we have appended $c_1' = (23) = (32)$.
It is the case that $\rho$ is an even permutation and hence is in $A_n$, because it is the result of multiplying the odd permutation $\rho$ by the odd permutation $c_1'$.

Now the conjugation is the composition of two conjugations: first by $(32)$, to yield $(32)\tau(32)^{-1} = (23)(1467)$ (which is $\tau$ still!), and then by $\rho$.
But we constructed $\rho$ so as to take $\tau$ to $\sigma$ on conjugation, so this works just as we needed.
%%

## If all the cycles are of odd length, but some length is repeated
Without loss of generality, suppose $r_1 = r_2$ (and label them both $r$), so the first two cycles are of the same length: say $\sigma$'s version is $(a_1 a_2 \dots a_r)(c_1 c_2 \dots c_r)$, and $\tau$'s version is $(b_1 b_2 \dots b_r)(d_1 d_2 \dots d_r)$.

Then define $\rho' = \rho (b_1 d_1)(b_2 d_2) \dots (b_r d_r)$.
Since $r$ is odd and $\rho$ is an odd permutation, $\rho'$ is an even permutation.

Now conjugating by $\rho'$ is the same as first conjugating by $(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$ and then by $\rho$.

But conjugating by $(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$ takes $\tau$ to $\tau$, because it has the effect of replacing all the $b_i$ by $d_i$ and all the $d_i$ by $b_i$, so it simply swaps the two cycles round.

Hence the conjugation of $\tau$ by $\rho'$ yields $\sigma$, and $\rho'$ is in $A_n$.

%%hidden(Example):
Suppose $\sigma = (123)(456), \tau = (154)(237)$.

Then conjugation of $\tau$ by $\rho = (67)(35)(42)(34)(25)$, an odd permutation, yields $\sigma$.

Now, if we first perform the conjugation $(12)(53)(47)$, we take $\tau$ to itself, and then performing $\rho$ yields $\sigma$.
The combination of $\rho$ and $(12)(53)(47)$ is an even permutation, so it does line in $A_n$.
%%

## If all the cycles are of odd length, and they are all distinct

In this case, we are required to check that the conjugacy class *does* split.
Remember, we started out by supposing $\sigma$ and $\tau$ have the same cycle type, and they are conjugate in $S_n$ by an odd permutation (so they are not obviously conjugate in $A_n$); we need to show that indeed they are *not* conjugate in $A_n$.

Indeed, the only ways to rewrite $\tau$ into $\sigma$ (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in $\sigma$. There is no choice about which $\tau$-cycle we take to which $\sigma$-cycle, because all the cycle lengths are distinct.
But the permutation $\rho$ which takes the $\tau$-cycles to the $\sigma$-cycles is odd, so is not in $A_n$.

Moreover, since each cycle is odd, we can't get past the problem by just cycling round the cycle (for instance, by taking the cycle $(123)$ to the cycle $(231)$), because that involves conjugating by the cycle itself: an even permutation, since the cycle length is odd.
Composing with the even permutation can't take $\rho$ from being odd to being even.

Therefore $\tau$ and $\sigma$ genuinely are not conjugate in $A_n$, so the conjugacy class splits.

%%hidden(Example):
Suppose $\sigma = (12345)(678), \tau = (12345)(687)$ in $A_8$.

Then conjugation of $\tau$ by $\rho = (87)$, an odd permutation, yields $\sigma$.
Can we do this with an *even* permutation instead?

Conjugating $\tau$ by anything at all must keep the cycle type the same, so the thing we conjugate by must take $(12345)$ to $(12345)$ and $(687)$ to $(678)$.

The only ways of $(12345)$ to $(12345)$ are by conjugating by some power of $(12345)$ itself; that is even.
The only ways of taking $(687)$ to $(678)$ are by conjugating by $(87)$, or by $(87)$ and then some power of $(678)$; all of these are odd.

Therefore the only possible ways of taking $\tau$ to $\sigma$ involve conjugating by an odd permutation $(678)^m(87)$, possibly alongside some powers of an even permutation $(12345)$; therefore to get from $\tau$ to $\sigma$ requires an odd permutation, so they are in fact not conjugate in $A_8$.
%%

# Example

In $A_7$, the cycle types are $(7)$, $(5, 1, 1)$, $(4,2,1)$, $(3,2,2)$, $(3,3,1)$, $(3,1,1,1,1)$, $(1,1,1,1,1,1,1)$, and $(2,2,1,1,1)$.
The only class which splits is the $7$-cycles, of cycle type $(7)$; it splits into a pair of half-sized classes with representatives $(1234567)$ and $(12)(1234567)(12)^{-1} = (2134567)$.


Decision problem

A **decision problem** is a subset $D$ of a set of instances $A$, where generally $A$ is the set of finite [https://arbital.com/p/-bitstrings](https://arbital.com/p/-bitstrings) $\{0,1\}^*$.

In plain English, a decision problem is composed by a number of members of a collection, which generally share a common property.

In plainer English, a decision problem is a question of the form "does bitstring $w$ have property $p$, yes or no?"

Every question that fits the set form (is this string in the set?) can be re-expressed as "does this bitstring have the property of being in the set?" Everything that fits the property form can be re-expressed in set form as well, because a property implicitly specifies a set of things that have the property. Thus, the two forms are equivalent.

It is called a decision problem because we are trying to *decide* whether a given bitstring belongs to the set or not.

A **solution** of the problem would consist in a procedure which given an instance $w$ of $A$ decides correctly in finite time whether $w$ belongs to $D$ or not. That is, a solution consists in a way of identifying the common trait shared by the members of $D$ which distinguish them from the rest of instances in $A$.

We say that a problem is [https://arbital.com/p/-decidable](https://arbital.com/p/-decidable) if there exists a solution which works for every instance. Trivially, finite decision problems are decidable, since we can have a look-up list with every element which we would consult every time we are given an instance to classify. If the instance corresponds to an element in the list, we accept it, and otherwise reject it.

We say that a decision problem is [https://arbital.com/p/-semidecidable](https://arbital.com/p/-semidecidable) if there is a procedure which, in finite time, identifies members of $D$, but fails to [https://arbital.com/p/-halt](https://arbital.com/p/-halt) and reject instances which do not belong to $D$.

Decision problems can be classified according to their difficulty of solving in [complexity classes](https://arbital.com/p/), and are a central object of study in [https://arbital.com/p/-49w](https://arbital.com/p/-49w).

#Examples
##Graph connectedness
Suppose we encode every possible finite graph, up to isomorphism, in the collection of finite bitstrings. Then we could define:
$$
CONNECTED = \{s\in\{0,1\}^*:\text{$s$ represents a connected graph}\}
$$
Which would be a [decidable decision problem](https://arbital.com/p/graph_connectedness).

##Tautology identification
Let $TAUTOLOGY$ be the collection of formulas of [https://arbital.com/p/-first_order_logic](https://arbital.com/p/-first_order_logic) which are true for every [interpretation](https://arbital.com/p/mathematical_interpretation). Then [$TAUTOLOGY$ is a semidecidable decision problem](https://arbital.com/p/), as if a formula is [https://arbital.com/p/-valid](https://arbital.com/p/-valid) we can search for a proof, but otherwise, we do not have an effective procedure to decide that a formula is not valid.

##Primality
Let:
$$
PRIMES = \{ x\in \mathbb{N}:\text{$x$ is prime}\}
$$
Then $PRIMES$ is a decidable decision problem.

We can alternately define
$$
PRIMES = \{s\in\{0,1\}^*:\text{$s$ represent a prime number in base $2$}\}
$$
This illustrates that we can specify a particular decision problem in different ways.

Eric Rogstad
Eric Bruylant
Jaime Sevilla Molina
Alex Appel

A decision problem is a question of the form "does [https://arbital.com/p/-bitstring](https://arbital.com/p/-bitstring) $w$ have [https://arbital.com/p/-property](https://arbital.com/p/-property) $p$, yes or no?"


Conjugacy classes of the alternating group on five elements: Simpler proof

The [https://arbital.com/p/-4hf](https://arbital.com/p/-4hf) $A_5$ on five elements has [conjugacy classes](https://arbital.com/p/4bj) very similar to those of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$, but where one of the classes has split in two.

Note that $A_5$ has $60$ elements, since it is precisely half of the elements of $S_5$ which has $5! = 120$ elements (where the exclamation mark is the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function).

# Table

$$\begin{array}{|c|c|c|c|}
\hline
\text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline
(12345) & 12 & 5 & 5 \\ \hline
(21345) & 12 & 5 & 5 \\ \hline
(123) & 20 & 3,1,1 & 3 \\ \hline
(12)(34) & 15 & 2,2,1 & 2 \\ \hline
e & 1 & 1,1,1,1,1 & 1 \\ \hline
\end{array}$$

# Working

Firstly, the identity is in a class of its own, because $\tau e \tau^{-1} = \tau \tau^{-1} = e$ for every $\tau$.

Now, by the same reasoning as in [the proof](https://arbital.com/p/4bh) that conjugate elements must have the same cycle type in $S_n$, that result also holds in $A_n$.

Hence we just need to see whether any of the cycle types comprise more than one conjugacy class.

Recall that the available cycle types are $(5)$, $(3,1,1)$, $(2,2,1)$, $(1,1,1,1,1)$ (the last of which is the identity and we have already considered it).

## Double-transpositions

All the double-transpositions are conjugate (so the $(2,2,1)$ cycle type does not split):

- $(ab)(cd)$ is conjugate to $(ab)(ce)$ if we conjugate by $(ab)(de)$; symmetrically this covers all the cases where one of the two transpositions remains the same.
- $(ab)(cd)$ is conjugate to $(ac)(bd)$ by $(cba)$; this covers the case that $e$ is not introduced.
- $(ab)(cd)$ is conjugate to $(ac)(be)$ by $(bc)(de)$; this covers the remaining cases that $e$ is introduced and neither of the two transpositions remains the same.

## Three-cycles

All the three-cycles are conjugate (so the $(3,1,1)$ cycle type does not split):

- $(abc)$ is conjugate to $(acb)$ by $(bc)(de)$, so three-cycles are conjugate to their permutations.
- $(abc)$ is conjugate to $(abd)$ by $(cde)$; this covers the case of introducing a single new element to the cycle.
- $(abc)$ is conjugate to $(ade)$ by $(bd)(ce)$; this covers the case of introducing two new elements to the cycle.

## Five-cycles

This class does split: I claim that $(12345)$ and $(21345)$ are not conjugate.
(Once we have this, then the class must split into two chunks, since $\{ \rho (12345) \rho^{-1}: \rho \ \text{even} \}$ is closed under conjugation in $A_5$, and $\{ \rho (12345) \rho^{-1}: \rho \ \text{odd} \}$ is closed under conjugation in $A_5$.
The first is the conjugacy class of $(12345)$ in $A_5$; the second is the conjugacy class of $(21345) = (12)(12345)(12)^{-1}$.
The only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)

Recall that $\tau (12345) \tau^{-1} = (\tau(1), \tau(2), \tau(3), \tau(4), \tau(5))$, so we would need a permutation $\tau$ such that $\tau$ sends $1$ to $2$, $2$ to $1$, $3$ to $3$, $4$ to $4$, and $5$ to $5$.
The only such permutation is $(12)$, the transposition, but that is not actually a member of $A_5$.

Hence in fact $(12345)$ and $(21345)$ are not conjugate.


Poset: Exercises

Try these exercises to test your poset knowledge.

# Corporate Ladder

Imagine a company with five distinct roles: CEO, marketing manager, marketer, IT manager, and IT worker. At this company, if the CEO gives an order, everyone else must follow it. If an order comes from an IT manager, only IT workers are obligated to follow it. Similarly, if an order comes from a marketing manager, only marketers are obligated to follow it. Nobody is obligated to follow orders from marketers or IT workers.

Do the workers at this company form a poset under the "obligated to follow orders from" relation?

%%%hidden(Show solution):
While not technically a poset due to its lack of reflexivity, it is pretty close. It is actually a strict ordering, whose underlying partial order could be obtained by making the reasonable assumption that each worker will follow her own orders.
%%%

# Bag Inclusion

We can define a notion of [power sets](https://arbital.com/p/power_set) for [bags](https://arbital.com/p/3jk) as follows. Let $X$ be a set, then we use $\mathcal{M}(X)$ to denote the set of all bags containing elements of $X$. Let $A \in \mathcal{M}(X)$. The multiplicity function of $A$, $1_A : X \rightarrow \mathbb N$ maps each element of $X$ to the number of times that element occurs in $A$. We can use multiplicity functions to define a inclusion relation $\subseteq$ for bags. For $A, B \in \mathcal M(X)$, we write $A \subseteq B$ whenever for all $x \in X$, $1_A(x) \leq 1_B(x)$.

Does $\mathcal{M}(X)$ form a poset under the bag inclusion relation $\subseteq$? If so, prove it. Otherwise, show that it does not satisfy one of the three poset properties.

# Duality

Give the dual of the following proposition.

For all posets $P$ and all $p, q \in P$, $q \prec p$ implies that $\{ r \in P~|~r \leq p \}$ is a superset of $\{ r \in P~|~r \leq q\}$.

%%hidden(Show solution):
For all posets $P$ and all $p, q \in P$, $q \succ p$ implies that $\{ r \in P~|~r \geq p \}$ is a superset of $\{ r \in P~|~r \geq q\}$ (where $q \succ p$ means $p \prec q$).
%%

# Hasse diagrams

Let $X = \{ x, y, z \}$. Draw a Hasse diagram for the poset $\langle \mathcal P(X), \subseteq \rangle$ of the power set of $X$ ordered by inclusion.

%%hidden(Show solution):
![A Hasse diagram of the power set of X, ordered by inclusion](http://i.imgur.com/WG3OLFc.png)

%%%comment:
dot source:

digraph G {
  node [= 0.1, height = 0.1](https://arbital.com/p/width)
  edge [= "none"](https://arbital.com/p/arrowhead)
  e [= "{}"](https://arbital.com/p/label)
  x [= "{x}"](https://arbital.com/p/label)
  y [= "{y}"](https://arbital.com/p/label)
  z [= "{z}"](https://arbital.com/p/label)
  xy [= "{x,y}"](https://arbital.com/p/label)
  xz [= "{x,z}"](https://arbital.com/p/label)
  yz [= "{y,z}"](https://arbital.com/p/label)
  xyz [= "{x,y,z}"](https://arbital.com/p/label)

  rankdir = BT;
  e -> x
  e -> y
  e -> z
  x -> xy
  x -> xz
  y -> xy
  y -> yz
  z -> xz
  z -> yz
  xy -> xyz
  xz -> xyz
  yz -> xyz
}
%%%

%%

#Hasse diagrams (encore)

Is it possible to draw a Hasse diagram for any poset?

%%hidden(Show solution):
Note that our description of Hasse diagrams made use of the covers relation $\prec$. The covers relation, however, is not adequate to describe the structure of many posets. Consider the poset $\langle \mathbb R, \leq \rangle$ of the real numbers ordered by the standard comparison $\leq$. We have $0 < 1$, but how would we convey that with a Hasse diagram? The problem is that $0$ has no covers, even though it is not a maximal element in $\mathbb R$. In fact, for any $x \in \mathbb R$ such that $x > 0$, we can find a $y \in \mathbb R$ such that $0 < y < x$. This "infinite density" of $\mathbb R$ makes it impossible to depict using a Hasse diagram.
%%

Eric Bruylant
Kevin Clancy
Mark Chimes
Chris Pasek


Cauchy's theorem on subgroup existence

Cauchy's theorem states that if $G$ is a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $p$ is a [prime](https://arbital.com/p/4mf) dividing $|G|$ the [order](https://arbital.com/p/3gg) of $G$, then $G$ has a subgroup of order $p$. Such a subgroup is necessarily [cyclic](https://arbital.com/p/47y) ([proof](https://arbital.com/p/4jh)).

# Proof

The proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.

Let $$X = \{ (x_1, x_2, \dots, x_p) : x_1 x_2 \dots x_p = e \}$$ the collection of $p$-[https://arbital.com/p/-tuple](https://arbital.com/p/-tuple)s of elements of the group such that the group operation applied to the tuple yields the identity.
Observe that $X$ is not empty, because it contains the tuple $(e, e, \dots, e)$.

Now, the cyclic group $C_p$ of order $p$ [acts](https://arbital.com/p/3t9) on $X$ as follows: $$(h, (x_1, \dots, x_p)) \mapsto (x_2, x_3, \dots, x_p, x_1)$$ where $h$ is the generator of $C_p$.
So a general element $h^i$ acts on $X$ by sending $(x_1, \dots, x_p)$ to $(x_{i+1}, x_{i+2} , \dots, x_p, x_1, \dots, x_i)$.

This is indeed a group action (exercise).

%%hidden(Show solution):

- It certainly outputs elements of $X$, because if $x_1 x_2 \dots x_p = e$, then $$x_{i+1} x_{i+2} \dots x_p x_1 \dots x_i = (x_1 \dots x_i)^{-1} (x_1 \dots x_p) (x_1 \dots x_i) = (x_1 \dots x_i)^{-1} e (x_1 \dots x_i) = e$$
- The identity acts trivially on the set: since rotating a tuple round by $0$ places is the same as not permuting it at all.
- $(h^i h^j)(x_1, x_2, \dots, x_p) = h^i(h^j(x_1, x_2, \dots, x_p))$ because the left-hand side has performed $h^{i+j}$ which rotates by $i+j$ places, while the right-hand side has rotated by first $j$ and then $i$ places and hence $i+j$ in total.
%%

Now, fix $\bar{x} = (x_1, \dots, x_p) \in X$.

By the [Orbit-Stabiliser theorem](https://arbital.com/p/4l8), the [orbit](https://arbital.com/p/4v8) $\mathrm{Orb}_{C_p}(\bar{x})$ of $\bar{x}$ divides $|C_p| = p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\bar{x} \in X$.

Now, what is the size of the set $X$?
%%hidden(Show solution):
It is $|G|^{p-1}$.

Indeed, a single $p$-tuple in $X$ is specified precisely by its first $p$ elements; then the final element is constrained to be $x_p = (x_1 \dots x_{p-1})^{-1}$.
%%

Also, the orbits of $C_p$ acting on $X$ partition $X$ ([proof](https://arbital.com/p/4mg)).
Since $p$ divides $|G|$, we must have $p$ dividing $|G|^{p-1} = |X|$.
Therefore since $|\mathrm{Orb}_{C_p}((e, e, \dots, e))| = 1$, there must be at least $p-1$ other orbits of size $1$, because each orbit has size $p$ or $1$: if we had fewer than $p-1$ other orbits of size $1$, then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$, and all the remaining orbits would have to be of size $p$, contradicting that $p \mid |X|$.

Hence there is indeed another orbit of size $1$; say it is the singleton $\{ \bar{x} \}$ where $\bar{x} = (x_1, \dots, x_p)$.

Now $C_p$ acts by cycling $\bar{x}$ round, and we know that doing so does not change $\bar{x}$, so it must be the case that all the $x_i$ are equal; hence $(x, x, \dots, x) \in X$ and so $x^p = e$ by definition of $X$.


Equaliser (category theory)

In [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7), an *equaliser* of a pair of arrows $f, g: A \to B$ is an object $E$ and a universal arrow $e: E \to A$ such that $ge = fe$.
Explicitly, $ge = fe$, and for any object $X$ and arrow $x: X \to A$ such that $fx = gx$, there is a unique factorisation $\bar{x} : X \to A$ such that $e \bar{x} = x$.


Orbit-stabiliser theorem

summary(Technical): Let $G$ be a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd), [acting](https://arbital.com/p/3t9) on a set $X$. Let $x \in X$.
Writing $\mathrm{Stab}_G(x)$ for the [stabiliser](https://arbital.com/p/4mz) of $x$, and $\mathrm{Orb}_G(x)$ for the [orbit](https://arbital.com/p/4v8) of $x$, we have $$|G| = |\mathrm{Stab}_G(x)| \times |\mathrm{Orb}_G(x)|$$ where $| \cdot |$ refers to the size of a set.

Let $G$ be a finite [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd), [acting](https://arbital.com/p/3t9) on a set $X$. Let $x \in X$.
Writing $\mathrm{Stab}_G(x)$ for the [stabiliser](https://arbital.com/p/4mz) of $x$, and $\mathrm{Orb}_G(x)$ for the [orbit](https://arbital.com/p/4v8) of $x$, we have $$|G| = |\mathrm{Stab}_G(x)| \times |\mathrm{Orb}_G(x)|$$ where $| \cdot |$ refers to the size of a set.

This statement generalises to infinite groups, where the same proof goes through to show that there is a [bijection](https://arbital.com/p/499) between the [left cosets](https://arbital.com/p/4j4) of the group $\mathrm{Stab}_G(x)$ and the orbit $\mathrm{Orb}_G(x)$.

# Proof

Recall that the [https://arbital.com/p/-4lt](https://arbital.com/p/-4lt) of the parent group.

Firstly, it is enough to show that there is a bijection between the left cosets of the stabiliser, and the orbit.
Indeed, then $$|\mathrm{Orb}_G(x)| |\mathrm{Stab}_G(x)| = |\{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}| |\mathrm{Stab}_G(x)|$$
but the right-hand side is simply $|G|$ because an element of $G$ is specified exactly by specifying an element of the stabiliser and a coset.
(This follows because the [cosets partition the group](https://arbital.com/p/4j5).)

## Finding the bijection

Define $\theta: \mathrm{Orb}_G(x) \to \{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}$, by $$g(x) \mapsto g \mathrm{Stab}_G(x)$$

This map is well-defined: note that any element of $\mathrm{Orb}_G(x)$ is given by $g(x)$ for some $g \in G$, so we need to show that if $g(x) = h(x)$, then $g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)$.
This follows: $h^{-1}g(x) = x$ so $h^{-1}g \in \mathrm{Stab}_G(x)$.

The map is [injective](https://arbital.com/p/4b7): if $g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)$ then we need $g(x)=h(x)$.
But this is true: $h^{-1} g \in \mathrm{Stab}_G(x)$ and so $h^{-1}g(x) = x$, from which $g(x) = h(x)$.

The map is [surjective](https://arbital.com/p/4bg): let $g \mathrm{Stab}_G(x)$ be a left coset.
Then $g(x) \in \mathrm{Orb}_G(x)$ by definition of the orbit, so $g(x)$ gets taken to $g \mathrm{Stab}_G(x)$ as required.

Hence $\theta$ is a well-defined bijection.

Mark Chimes
Patrick Stevens
Alexei Andreev

A [group action](https://arbital.com/p/-3t9) of a group $G$ acting on a set $X$ describes how $G$ sends elements of $X$ to other elements of $X$.  Given a specific element $x \in X$, the [stabiliser](https://arbital.com/p/-4mz) is all those elements of the group which send $x$ back to itself, and the [orbit](https://arbital.com/p/-4v8) of $x$ is all the elements to which $x$ can get sent.

This theorem tells you that $G$ is divided into equal-sized pieces using $x$. Each piece "looks like" the stabilizer of $x$ (and is the same size), and the orbit of $x$ tells you how to "move the piece around" over $G$ in order to cover it.

Put another way, each element $y$ in the orbit of $x$ is transformed "in the same way" by $G$ relative to $y$.

This theorem is closely related to [Lagrange's Theorem](https://arbital.com/p/-4jn).


Needs exercises

One way the user can make sure they've understood the material is to do [exercises](https://arbital.com/p/4n2). They should usually be on their own [lens](https://arbital.com/p/17b), as additional "homework problems" that users can solve to get more practice. They can also be embedded in the explanation pages, so the user can solve them as they are learning, especially on [path](https://arbital.com/p/1rt) pages where the reader has asked for exercises.

Many pages could be improved by adding exercises, but priority should be given to:

* Pages which don't currently have exercises.
* Pages which have a full explanation of a concept (rather than just [defienitions](https://arbital.com/p/4gs)).
* Pages which are a requisite that many other concepts rely on.
* Pages which are mostly self-contained, rather than overviews of a large and multifaceted concept (exercises should go with child pages explaining the subtopics).

Eric Bruylant
Mark Chimes
Alexei Andreev


Join and meet: Exercises

Try these exercises to test your knowledge of joins and meets.

Tangled up
--------------------

![a big crazy poset](http://i.imgur.com/cYqkKm7.png)

Determine whether or not the following joins and meets exist in the poset depicted by the above Hasse diagram.

$c \vee b$

%%hidden(Show solution):
This join does *not* exist, because $i$ and $j$ are incomparable upper bounds of $\{ c, b \}$, and no smaller upper bounds of $\{ c, b \}$ exist.
%%

$g \vee e$

%%hidden(Show solution):
This join exists. $g \vee e = j$.
%%

$j \wedge f$

%%hidden(Show solution):
This meet does *not* exist, because $d$ and $b$ are incomparable lower bounds of $\{ j, f \}$, and no larger lower bounds of $\{ j, f \}$ exist.
%%

$\bigvee \{a,b,c,d,e,f,g,h,i,j,k,l\}$

%%hidden(Show solution):
This join exists. It is $l$.
%%

Join fu
------------

Let $P$ be a poset, $S \subseteq P$, and $p \in P$. Prove that if both $\bigvee S$ and $(\bigvee S) \vee p$ exist then $\bigvee (S \cup \{p\})$ exists as well, and $(\bigvee S) \vee p = \bigvee (S \cup \{p\})$.

%%hidden(Show solution):
For any $X \subset P$, let $X^U$ denote the set of upper bounds of $X$. The above proposition follows from the fact that $\{\bigvee S, p\}^U = (S \cup p)^U$, which is apparent from the following chain of bi-implications:

$q \in \{\bigvee S, p\}^U \iff$

for all $s \in S, q \geq \bigvee S \geq s$, and $q \geq p \iff$

$q \in (S \cup \{p\})^U$.

If two subsets of a poset have the same set of upper bounds, then either they both lack a least upper bound, or both have the same least upper bound.
%%

Meet fu
------------

Let $P$ be a poset, $S \subseteq P$, and $p \in P$. Prove that if both $\bigwedge S$ and $(\bigwedge S) \wedge p$ exist then $\bigwedge (S \cup \{p\})$ exists as well, and $(\bigwedge S) \wedge p = \bigwedge(S \cup \{p\})$.

%%hidden(Show solution):
Note that the proposition we are trying to prove here is the dual of the one stated in join fu. Thanks to the duality principle, this theorem therefore comes for free with our solution to join fu.
%%

Quite a big join
--------------------------

In the poset $\langle \mathbb N, | \rangle$ discussed in [https://arbital.com/p/43s](https://arbital.com/p/43s), does $\bigvee \mathbb N$ exist? If so, what is it?


Stabiliser is a subgroup

Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) which [acts](https://arbital.com/p/3t9) on the [set](https://arbital.com/p/-3jz) $X$.
Then for every $x \in X$, the [stabiliser](https://arbital.com/p/4mz) $\mathrm{Stab}_G(x)$ is a [subgroup](https://arbital.com/p/-subgroup) of $G$.

# Proof

We must check the group axioms.

- The [identity](https://arbital.com/p/-54p), $e$, is in the stabiliser because $e(x) = x$; this is part of the definition of a group action.
- [Closure](https://arbital.com/p/-3gy) is satisfied: if $g(x) = x$ and $h(x) = x$, then $(gh)(x) = g(h(x))$ by definition of a group action, but that is $g(x) = x$.
- [Associativity](https://arbital.com/p/-3h4) is inherited from the parent group.
- [Inverses](https://arbital.com/p/-inverse_mathematics): if $g(x) = x$ then $g^{-1}(x) = g^{-1} g(x) = e(x) = x$.

Dylan Hendrickson
Mark Chimes
Patrick Stevens

Given a [group](https://arbital.com/p/-3gd) $G$ [acting](https://arbital.com/p/-3t9) on a [set](https://arbital.com/p/-3jz) $X$, the [stabiliser](https://arbital.com/p/-4mz) of some [element](https://arbital.com/p/-element_mathematics) $x \in X$ is a [subgroup](https://arbital.com/p/-subgroup) of $G$.


Prime number

A [https://arbital.com/p/-45h](https://arbital.com/p/-45h) $n > 1$ is *prime* if it has no [divisors](https://arbital.com/p/divisor_number_theory) other than itself and $1$.
Equivalently, it has the property that if $n \mid ab$ %%note:That is, $n$ divides the product $ab$%% then $n \mid a$ or $n \mid b$.
Conventionally, $1$ is considered to be neither prime nor [composite](https://arbital.com/p/composite_number) (i.e. non-prime).

# Examples

- The number $2$ is prime, because its divisors are $1$ and $2$; therefore it has no divisors other than itself and $1$.
- The number $3$ is also prime, as are $5, 7, 11, 13, \dots$.
- The number $4$ is not prime; neither are $6, 8, 9, 10, 12, \dots$.

# Properties

- There are infinitely many primes. ([Proof.](https://arbital.com/p/54r))
- Every natural number may be written as a product of primes; moreover, this can only be done in one way (if we count "the same product but with the order swapped" as being the same: for example, $2 \times 3 = 3 \times 2$ is just one way of writing $6$). ([Proof.](https://arbital.com/p/fundamental_theorem_of_arithmetic))

# How to find primes

If we want to create a list of all the primes below a given number, or the first $n$ primes for some fixed $n$, then an efficient way to do it is the [Sieve of Eratosthenes](https://arbital.com/p/sieve_of_eratosthenes).
(There are other sieves available, but Eratosthenes is the simplest.)

There are many [tests](https://arbital.com/p/primality_testing) for primality and for compositeness.

# More general concept

This definition of "prime" is, in a more general [ring-theoretic](https://arbital.com/p/3gq) setting, known instead as the property of [irreducibility](https://arbital.com/p/5m1).
Confusingly, there is a slightly different notion in this ring-theoretic setting, which goes by the name of "prime"; this notion has [a separate page on Arbital](https://arbital.com/p/prime_element_ring_theory).
In the ring of integers, the two ideas of "prime" and "irreducible" actually coincide, but that is because the integers form a ring with several very convenient properties: in particular, being a [Euclidean domain](https://arbital.com/p/euclidean_domain), they are a [https://arbital.com/p/-principal_ideal_domain](https://arbital.com/p/-principal_ideal_domain) (PID), and [PIDs have unique factorisation](https://arbital.com/p/pid_implies_ufd).