Kernel of group homomorphism

The kernel of a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \to H$ is the collection of all elements $g$ in $G$ such that $f(g) = e_H$ the identity of $H$.

It is important to note that the kernel of any group homomorphism $G \to H$ is always a subgroup of $G$.
Indeed:

- if $f(g_1) = e_H$ and $f(g_2) = e_H$ then $e_H = f(g_1) f(g_2) = f(g_1 g_2)$, so the kernel is closed under $G$'s operation;
- if $f(x) = e_H$ then $e_H = f(e_G) = f(x^{-1} x) = f(x^{-1}) f(x) = f(x^{-1})$ (where we have used that [the image of the identity is the identity](https://arbital.com/p/49z)), so inverses are contained in the putative subgroup;
- $f(e_G) = e_H$ because the image of the identity is the identity, so the identity is contained in the putative subgroup.

It turns out that the notion of "[https://arbital.com/p/-4h6](https://arbital.com/p/-4h6)" coincides exactly with the notion of "kernel of homomorphism". ([Proof.](https://arbital.com/p/4h7))
The "kernel of homomorphism" viewpoint of normal subgroups is much more strongly motivated from the point of view of [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7); Timothy Gowers [considers this to be the correct way](https://gowers.wordpress.com/2011/11/20/normal-subgroups-and-quotient-groups/) to introduce the teaching of normal subgroups in the first place.


Image of the identity under a group homomorphism is the identity

For any [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \to H$, we have $f(e_G) = e_H$ where $e_G$ is the identity of $G$ and $e_H$ the identity of $H$.

Indeed, $f(e_G) f(e_G) = f(e_G e_G) = f(e_G)$, so premultiplying by $f(e_G)^{-1}$ we obtain $f(e_G) = e_H$.


Under a group homomorphism, the image of the inverse is the inverse of the image

For any [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $f: G \to H$, we have $f(g^{-1}) = f(g)^{-1}$.

Indeed, $f(g^{-1}) f(g) = f(g^{-1} g) = f(e_G) = e_H$, and similarly for multiplication on the left.


The image of a group under a homomorphism is a subgroup of the codomain

Let $f: G \to H$ be a [https://arbital.com/p/-47t](https://arbital.com/p/-47t), and write $f(G)$ for the set $\{ f(g) : g \in G \}$.
Then $f(G)$ is a group under the operation inherited from $H$.

# Proof

To prove this, we must verify the group axioms.
Let $f: G \to H$ be a group homomorphism, and let $e_G, e_H$ be the identities of $G$ and of $H$ respectively.
Write $f(G)$ for the image of $G$.

Then $f(G)$ is closed under the operation of $H$: since $f(g) f(h) = f(gh)$, so the result of $H$-multiplying two elements of $f(G)$ is also in $f(G)$.

$e_H$ is the identity for $f(G)$: it is $f(e_G)$, so it does lie in the image, while it acts as the identity because $f(e_G) f(g) = f(e_G g) = f(g)$, and likewise for multiplication on the right.

Inverses exist, by "the inverse of the image is the image of the inverse".

The operation remains associative: this is inherited from $H$.

Therefore, $f(G)$ is a group, and indeed is a subgroup of $H$.


Expected value

The expected value of an action is the [https://arbital.com/p/-mean](https://arbital.com/p/-mean) numerical outcome of the possible results weighted by their [https://arbital.com/p/-1rf](https://arbital.com/p/-1rf). It may actually be impossible to get the expected value, for example, if a coin toss decides between you getting \$0 and \$10, then we say you get "\$5 in expectation" even though there is no way for you to get \$5.

The expectation of V (often shortened to "the expected V") is how much V you expect to get on average. For example, the expectation of a payoff, or an expected payoff, is how much money you will get on average; the expectation of the duration of a speech, or an expected duration, is how long the speech will last "on average."

Suppose V has discrete possible values, say $V = x_{1},$ or $V = x_{2}, ..., $ or $V = x_{k}$. Let $P(x_{i})$ refer to the probability that $V = x_{i}$. Then the expectation of V is given by:

$$\sum_{i=1}^{k}x_{i}P(x_{i})$$

Suppose V has continuous possible values, x. For instance, let $x \in \mathbb{R}$. Let $P(x)$ be the continuous probability distribution, or $\lim_{dx \to 0}$ of the probability that $x<V<(x+dx)$ divided by $dx$. Then the expectation of V is given by:

$$\int_{-∞}^{∞}xP(x)dx$$

## Importance ##

A common principle of reasoning under uncertainty is that if you are trying to achieve a good G, you should choose the act that maximizes the expectation of G.

Eric Bruylant
Michael Cohen
Alexei Andreev


The composition of two group homomorphisms is a homomorphism

Given two [group homomorphisms](https://arbital.com/p/47t) $f: G \to H$ and $g: H \to K$, the composition $gf: G \to K$ is also a homomorphism.

To prove this, note that $g(f(x)) g(f(y)) = g(f(x) f(y))$ since $g$ is a homomorphism; that is $g(f(xy))$ because $f$ is a homomorphism.

Eric Rogstad
Mark Chimes
Patrick Stevens


Injective function

A [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) $f: X \to Y$ is *injective* if it has the property that whenever $f(x) = f(y)$, it is the case that $x=y$. Given an element in the [image](https://arbital.com/p/3lm), it came from applying $f$ to exactly one element of the [domain](https://arbital.com/p/3js).

This concept is also commonly called being "one-to-one".
That can be a little misleading to someone who does not already know the term, however, because many people's natural interpretation of "one-to-one" (without otherwise having learnt the term) is that every element of the domain is matched up in a one-to-one way with *every* element of the domain, rather than simply with *some* element of the domain.
That is, a rather natural way of interpreting "one-to-one" is as "[bijective](https://arbital.com/p/499)" rather than "injective".

# Examples

- The function $\mathbb{N} \to \mathbb{N}$ (where $\mathbb{N}$ is the set of [natural numbers](https://arbital.com/p/-45h)) given by $n \mapsto n+5$ is injective: since $n+5 = m+5$ implies $n = m$. Note that this function is not [surjective](https://arbital.com/p/4bg): there is no natural number $k$ such that $k+5 = 2$, for instance, so $2$ is not in the range of the function.
- The function $f: \mathbb{N} \to \mathbb{N}$ given by $f(n) = 6$ for all $n$ is not injective: since $f(1) = f(2)$ but $1 \not = 2$, for instance.


Complexity theory: Complexity zoo

Welcome, visitor! Get ready to exercise your sense of wonder and expand your mind on the **zoo of complexity**.

That where are the cages of the animals, you ask? No cage could contain our specimens, I must say. But you can get a glimpse of their nature through the chalky definitions in each of those blackboards. Where shall we go first?

----

Oh yes! [$P$](https://arbital.com/p/), the class of problems solvable in polynomial time. A popular creature in our installations. It is commonly accepted that this  class encompasses the problems which can be solved efficiently. Now, bear in mind that the classes in this zoo only classify problems attending to how fast the computational power you need to solve them grows compared to the size of the problem you are solving! That means, even if all grow at the same polynomial pace, some of the problems who lurk inside this class need a huge overload of resources from the very beginning, which makes them intractable! Not to talk about the multiplicative factor, which also gets ignored in making this distinction between classes, or the degree of the polynomial which upper bounds the function! Beliefs like that $x$ grows as fast as $1000 x^{42}+10^{100}$ can be misleading if not properly understood.

However, in practice such instances are rare, and $P$ is still used thorough all computer science as a synonym of efficiency - though they pay special attention to the degree and distinguish between functions in $\mathcal{O}(n)$ and functions in $\mathcal{O}(n*log(n))$, even though all programs associated fall within $P$.

----

The next creature is without a doubt the most famous of the zoo! Meet [$NP$](https://arbital.com/p/), the class of problems whose solution are easily checkable. This little creature is the responsible of modern cryptography, which relies on [functions easy to compute but hard to invert](https://arbital.com/p/one_way_functions).

$NP$ also is the main star, together with $P$, of one great show in the zoo. Oh, look! It is about to start!...

*Announcer*: Ladies, gentlemen and children who are today with us! We present to you the main attraction of our bestiary, the infamous [$P$ vs $NP$](https://arbital.com/p/)!

*Announcer*: You may already have seen both of those creatures, but are they two different beasts? Or only two aspects of the same thing?

*Announcer*: Look again at them, and realize. If finding a solution to a problem is easy, then checking whether a given solution is true is also easy! You can always just find the solution again from scratch and check that both coincide. So we conclude that $P\subset NP$. Can the reverse inclusion be true as well?

*Announcer*: Imagine for a moment! The hard problems of creativity, solved at once through the clever application of an efficient algorithm! [Automated short proof writing](https://arbital.com/p/Entscheidungsproble)! Instant reversal of one way functions!

*Announcer*: However, we suspect that they are different, though a proof has remained elusive through the years.

(the public cries: If we have been searching for decades for such a proof and none has been found then you should revise your beliefs and update against that! Why suspect otherwise?)

*Announcer*: We seem to have an exceptionally well educated audience today! Then again, we have been also on the search for a proof of the opposite without fruition. Why give this fact more weight in the Bayesian scheme of the things? There happens to be a somewhat convoluted reason for this. For if $P=NP$, then we would expect proofs to be easier to find, while if $P!=NP$ the opposite is true! So it only makes sense that is more probable that we live in the world where $P!=NP$%%note:For a formalization of this notion check [natural proofs](https://arbital.com/p/). Also, see [Arguments against $P=NP$](https://arbital.com/p/4bf)%%!

*Announcer*: And that's all for today, folks. Keep enjoying your visit though the zoo!

---

(To be continued)

Eric Rogstad
Jaime Sevilla Molina
Duncan Wilson


Real number

A **real number** is any number that can be used to represent a physical quantity.

Intuitively, real numbers are any number that can be found between two [integers](https://arbital.com/p/48l), such as $0,$ $1,$ $-1,$ $\frac{3}{2},$ $\frac{-7}{2},$ [$\pi,$](https://arbital.com/p/49r) [$e$](https://arbital.com/p/e), $100 \cdot \sqrt{2},$ and so on. The set of real numbers is written $\mathbb R.$ You can think of $\mathbb R$ as [$\mathbb Q$](https://arbital.com/p/4zq) extended to include the [irrational numbers](https://arbital.com/p/54z) like $\pi$ and $e$ which can be found between rational numbers but which cannot be completely written out in [https://arbital.com/p/-4sl](https://arbital.com/p/-4sl).

## Definitions of the real numbers

The most commonly used definitions of the real numbers are constructions as extensions of the [rational numbers](https://arbital.com/p/4zq), which involve either [Cauchy sequences](https://arbital.com/p/53b) or [Dedekind cuts](https://arbital.com/p/dedekind_cut).

### Cauchy sequences

Broadly speaking, a Cauchy sequence is a sequence where as the sequence goes on, all the elements past that point get closer and closer together. In the real numbers, every Cauchy sequence [converges](https://arbital.com/p/convergence_analysis) to a real number. However, in the set of rational numbers, not all Cauchy sequences converge to a rational number.
In the set of rationals, a Cauchy sequence which does not converge to a rational number cannot really be said to "converge" at all: the set of rationals is "missing some of the points" that would be required to make every Cauchy sequence converge.

For example, the sequence of fractions of consecutive Fibonacci numbers $1/1, 2/1, 3/2, 5/3, 8/5, \ldots$ gets closer and closer to $\frac{1 + \sqrt{5}}{2}$, but cannot be said to converge to that number because it is not in the set of rational numbers.

For each of these non-convergent Cauchy sequences, we define a new irrational number to "fill in the gap", and for the Cauchy sequences that do converge, we define a real number equal to that rational number.

### Dedekind cuts

A Dedekind cut of a [https://arbital.com/p/-540](https://arbital.com/p/-540) is a [https://arbital.com/p/-partition](https://arbital.com/p/-partition) of that set into two sets so that every element in the first set is [less than](https://arbital.com/p/) every element in the second set, and the second set has no smallest element. The latter restriction requires that the set also be a [perfect set](https://arbital.com/p/) (have no [isolated points](https://arbital.com/p/isolated_point)), in the sense used in topology.

In the real numbers, such a partition will always have the first set having a greatest element, which is known as the least-upper-bound property. However, in the rational numbers, we might come across a partition where the first set does not have such an element.

For example, define a Dedekind cut $(A, B)$ of the rational numbers such that $B = \{x \in \mathbb{Q} \ | \ x > 0 \wedge x^2 > 2\}$ and $A$ is the complement of $B$. In plainer language, $B$ consists of all the numbers greater than $\sqrt{2}$, but because $\sqrt{2}$ doesn't exist in the space of rational numbers, we can't use that to formulate our definition. Obviously every element of $A$ is less than every element of $B$, but $A$ has no greatest element either, because we can create a sequence of numbers in $A$ that gets bigger and bigger (as it approaches $\sqrt{2}$) but never stops at a maximum value.

For each of these "strict cuts" where neither set has a "boundary element", we define a new irrational number to "fill in the gap", just like with the Cauchy sequences. For the Dedekind cuts where one of the sets does have a least or greatest element, we define a real number equal to that rational number.

This definition has the advantage that each real number is represented by a unique Dedekind cut, unlike the Cauchy sequences where multiple sequences can converge to the same number.

Kevin Clancy
M Yass
Michael Cohen
Alexei Andreev
Eric Rogstad
Patrick Stevens
Nate Soares
Eric Bruylant
Joe Zeng


P vs NP

The greatest currently open problem in complexity theory and computer science in general, which talks about the relationship between the class of efficiently solvable problems, [$P$](https://arbital.com/p/), and the class of problems with easily checkable solutions, [$NP$](https://arbital.com/p/).

It is a basic result that [$P\subseteq NP$](https://arbital.com/p/). However, we still do not know whether the reverse implication, $P \supseteq NP$, holds.

A positive, constructive result would have profound implications through all the fields of science and math, not to mention the whole of society.

Mark Chimes
Jaime Sevilla Molina
Alexei Andreev

Are problems with solutions easy to check also easy to solve?


P vs NP: Arguments against P=NP

#Natural proofs
If $P \neq NP,$ then there are results showing that lower bounds in complexity are inherently harder to prove in a [technical yet natural sense](https://arbital.com/p/natural_proof). In other words, if $P \neq NP$ then proving $P \neq NP$ is hard.

The opposite proposition for $P=NP$ is also expected to be true. That is, it would make sense that if $P=NP,$ then it should be easier to prove, since we could build far more efficient [theorem provers](https://arbital.com/p/).

#Empirical argument
We have been trying to get a fast algorithm for [$NP$-complete](https://arbital.com/p/) problems since the 70s, without success. And take into account that "we" does not only comprise a minor group of theoretical computer scientists, but a whole industry trying to get faster algorithms for commercial purposes.

One could also argue more weakly that if $P=NP$, then evolution could have made use of this advantage to sped up its search process, or create more efficient minds to solve $NP$-complete problems at thoughtspeed.

#The arithmetical hierarchy argument
Some authors have drawn an analogy between the [polynomial hierarchy](https://arbital.com/p/) and the [arithmetical hierarchy](https://arbital.com/p/1mg).

There are results showing that the [arithmetical hierarchy is strict](https://arbital.com/p/), and if the analogy holds then we would have that the polynomial hierarchy is strict as well, [which automatically implies $P \neq NP$](https://arbital.com/p/collapse_of_the_polynomial_hierarchy).

Eric Rogstad
Eric Bruylant
Nate Soares
Jaime Sevilla Molina

- Proving $P \neq NP$ is harder if true than proving $P=NP$ if true.
 - After 50 years we haven't found any efficient algorithm for $NP$-complete problems.
 - The polynomial hierarchy is compared to the arithmetical hierarchy, and the analogy implies $P \neq NP$


Surjective function

A function $f:A \to B$ is *surjective* if every $b \in B$ has some $a \in A$ such that $f(a) = b$.
That is, its [codomain](https://arbital.com/p/3lg) is equal to its [image](https://arbital.com/p/3lh).

This concept is commonly referred to as being "onto", as in "The function $f$ is onto."

# Examples

- The function $\mathbb{N} \to \{ 6 \}$ (where $\mathbb{N}$ is the set of [natural numbers](https://arbital.com/p/45h)) given by $n \mapsto 6$ is surjective. However, the same function viewed as a function $\mathbb{N} \to \mathbb{N}$ is not surjective, because it does not hit the number $4$, for instance.
- The function $\mathbb{N} \to \mathbb{N}$ given by $n \mapsto n+5$ is *not* surjective, because it does not hit the number $2$, for instance: there is no $a \in \mathbb{N}$ such that $a+5 = 2$.


Conjugacy class is cycle type in symmetric group

In the [https://arbital.com/p/-497](https://arbital.com/p/-497) on a finite set, the [https://arbital.com/p/-4bj](https://arbital.com/p/-4bj) of an element is determined exactly by its [cycle type](https://arbital.com/p/4cg).

More precisely, two elements of $S_n$ are [conjugate](https://arbital.com/p/group_conjugate) in $S_n$ if and only if they have the same cycle type.

# Proof

## Same conjugacy class implies same cycle type
Suppose $\sigma$ has the cycle type $n_1, \dots, n_k$; write $$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$$
Let $\tau \in S_n$.

Then $\tau \sigma \tau^{-1}(\tau(a_{ij})) = \tau \sigma(a_{ij}) = a_{i (j+1)}$, where $a_{i (n_i+1)}$ is taken to be $a_{i 1}$.

Therefore $$\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))$$
which has the same cycle type as $\sigma$ did.

## Same cycle type implies same conjugacy class

Suppose $$\pi = (b_{11} b_{12} \dots b_{1 n_1})(b_{21} \dots b_{2 n_2}) \dots (b_{k 1} b_{k 2} \dots b_{k n_k})$$
so that $\pi$ has the same cycle type as the $\sigma$ from the previous direction of the proof.

Then define $\tau(a_{ij}) = b_{ij}$, and insist that $\tau$ does not move any other elements.

Now $\tau \sigma \tau^{-1} = \pi$ by the final displayed equation of the previous direction of the proof, so $\sigma$ and $\pi$ are conjugate.

# Example
This result makes it rather easy to [list the conjugacy classes](https://arbital.com/p/4bk) of $S_5$.


Conjugacy class

Given an element $g$ of a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$, the conjugacy class of $g$ in $G$ is $\{ x g x^{-1} : x \in G \}$.
It is the collection of elements to which $g$ is [conjugate](https://arbital.com/p/4gk).


Conjugacy classes of the symmetric group on five elements

The [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$ on five generators has size $5! = 120$ (where the exclamation mark denotes the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function).
By the result that [in a symmetric group, conjugacy classes coincide with cycle types](https://arbital.com/p/4bh), we can list the conjugacy classes of $S_5$ easily.

# The table

$$\begin{array}{|c|c|c|c|}
\hline
\text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline
(12345) & 24 & 5 & 5 \\ \hline
(1234) & 30 & 4,1 & 4 \\ \hline
(123) & 20 & 3,1,1 & 3 \\ \hline
(123)(45) & 20 & 3,2 & 6 \\ \hline
(12)(34) & 15 & 2,2,1 & 2 \\ \hline
(12) & 10 & 2,1,1,1 & 2 \\ \hline
e & 1 & 1,1,1,1,1 & 1 \\ \hline
\end{array}$$

# Determining the list of cycle types and sizes

There are five elements to permute; we need to find all the possible ways of grouping them up into disjoint cycles.
We will go through this systematically.
Note first that since there are only five elements to permute, there cannot be any element with a $6$-cycle or higher.

## Those with largest cycle of length $5$

If there is a $5$-cycle, then it permutes everything, so its [cycle type](https://arbital.com/p/4cg) is $5$.
That is, we can take a representative $(12345)$, and this is the only conjugacy class with a $5$-cycle.

Recall that every cycle of length $5$ may be written in five different ways: $(12345)$ or $(23451)$ or $(34512)$, and so on.
Without loss of generality, we may assume $1$ comes at the start (by cycling round the permutation if necessary).

Then there are $4!$ ways to fill in the remaining slots of the cycle (where the exclamation mark denotes the [https://arbital.com/p/-factorial](https://arbital.com/p/-factorial) function).

Hence there are $24$ elements of this conjugacy class.

## Those with largest cycle of length $4$

If there is a $4$-cycle, then it permutes everything except one element, so its cycle type must be $4,1$ (abbreviated as $4$).
That is, we can take a representative $(1234)$, and this is the only conjugacy class with a $4$-cycle.

Either the element $1$ is permuted by the $4$-cycle, or it is not.

- In the first case, we have $4$ possible ways to pick the image $a$ of $1$; then $3$ possible ways to pick the image $b$ of $a$; then $2$ possible ways to pick the image $c$ of $b$; then $c$ must be sent back to $1$.
That is, there are $4 \times 3 \times 2 = 24$ possible $4$-cycles containing $1$.
- In the second case, the cycle does not contain $1$, so there are $3$ possible ways to pick the image $a$ of $2$; then $2$ possible ways to pick the image $b$ of $a$; then $1$ possible way to pick the image $c$ of $b$; then $c$ must be sent back to $2$.
So there are $3 \times 2 \times 1 = 6$ possible $4$-cycles not containing $1$.

This comes to a total of $30$ possible $4$-cycles in this conjugacy class.

## Those with largest cycle of length $3$

Now we have two possible conjugacy classes: $3,1,1$ and $3,2$.

### The $3,1,1$ class

An example representative for this class is $(123)$.

We proceed with a slightly different approach to the $4,1$ case.
Using the notation for the [https://arbital.com/p/-binomial_coefficient](https://arbital.com/p/-binomial_coefficient), we have $\binom{5}{3} = 10$ possible ways to select the numbers which appear in the $3$-cycle.
Each selection has two distinct ways it could appear as a $3$-cycle: the selection $\{1,2,3\}$ can appear as $(123)$ (or the duplicate cycles $(231)$ and $(312)$), or as $(132)$ (or the duplicate cycles $(321)$ or $(213)$).

That is, we have $2 \times 10 = 20$ elements of this conjugacy class.

### The $3,2$ class

An example representative for this class is $(123)(45)$.

Again, there are $\binom{5}{3} = 10$ possible ways to select the numbers which appear in the $3$-cycle; having made this selection, we have no further choice about the $2$-cycle.

Given a selection of the elements of the $3$-cycle, as before we have two possible ways to turn it into a $3$-cycle.

We are also given the selection of the elements of the $2$-cycle, but there is no choice about how to turn this into a $2$-cycle because, for instance, $(12)$ is equal to $(21)$ as cycles.
So this time the selection of the elements of the $3$-cycle has automatically determined the corresponding $2$-cycle.

Hence there are again $2 \times 10 = 20$ elements of this conjugacy class.

## Those with largest cycle of length $2$

There are two possible cycle types: $2,2,1$ and $2,1,1,1$.

### The $2,2,1$ class

An example representative for this class is $(12)(34)$.

There are $\binom{5}{2}$ ways to select the first two elements; then once we have done so, there are $\binom{3}{2}$ ways to select the second two.
Having selected the elements moved by the first $2$-cycle, there is only one distinct way to make them into a $2$-cycle, since (for example) $(12)$ is equal to $(21)$ as permutations; similarly the selection of the elements determines the second $2$-cycle.

However, this time we have double-counted each element, since (for example) the permutation $(12)(34)$ is equal to $(34)(12)$ by the result that [disjoint cycles commute](https://arbital.com/p/49g).

Therefore there are $\binom{5}{2} \times \binom{3}{2} / 2 = 15$ elements of this conjugacy class.

### The $2,1,1,1$ class

An example representative for this class is $(12)$.

There are $\binom{5}{2}$ ways to select the two elements which this cycle permutes.
Once we have selected the elements, there is only one distinct way to put them into a $2$-cycle, since (for example) $(12)$ is equal to $(21)$ as permutations.

Therefore there are $\binom{5}{2} = 10$ elements of this conjugacy class.

## Those with largest cycle of length $1$

There is only the identity in this class, so it is of size $1$.


Church-Turing thesis

The **Church-Turing thesis** (often abbreviated **CT thesis**) states:
> Every effectively computable function is Turing-computable

That is, for every function which can be computed by physical means %%note:So no [hypercomputers](https://arbital.com/p/)%% there exists a [Turing machine](https://arbital.com/p/) which computes exactly that.

The Church-Turing thesis is not a definite mathematical statement, but an inductive statement which affirms that every sensible model of computation we can come up with is equivalent or at least [reducible](https://arbital.com/p/reduction) to the model proposed by Turing. Thus, we cannot prove it in a mathematical sense, but we can [gather evidence for it](https://arbital.com/p/).

For example, this model was proven to coincide with Church's lambda calculus, another universal model of computation, and the equivalence between Church's lambda calculus and Turing's automatic machines is often taken as [evidence](https://arbital.com/p/4j7) that they correctly capture our intuition of "effectively computable".

There are many [consequences](https://arbital.com/p/) of the CT thesis for computer science in general, artificial intelligence, [epistemology](https://arbital.com/p/11w), and other fields of knowledge.

Alexei Andreev
Eric Rogstad
Patrick Stevens
Eric Bruylant
Connor Flexman
Jaime Sevilla Molina


Properties of the logarithm

With a [solid interpretation of logarithms](https://arbital.com/p/45q) under our belt, we are now in a position to look at the basic properties of the logarithm and understand what they are saying. The defining [characteristic of logarithm functions](https://arbital.com/p/47k) is that they are [https://arbital.com/p/-real_valued](https://arbital.com/p/-real_valued) functions $f$ such that

__Property 1: Multiplying inputs adds outputs.__
$$
f(x \cdot y) = f(x) + f(y) \tag{1}
$$
for all $x, y \in$ [$\mathbb R^+$](https://arbital.com/p/positive_reals). This says that whenever the input grows (or shrinks) by a factor of $y$, the output goes up (or down) by only a fixed amount, which depends on $y$. In fact, equation (1) alone tells us quite a bit about the behavior of $f,$ and from it, we can _almost_ guarantee that $f$ is a logarithm function. First, let's see how far we can get using equation (1) all by itself:

---

__Property 2: 1 is mapped to 0.__
$$
f(1) = 0. \tag{2}
$$

This says that the amount the output changes if the input grows by a factor of 1 is zero — i.e., the output does not change if the input changes by a factor of 1. This is obvious, as "the input changed by a factor of 1" means "the input did not change."

__Exercise:__ Prove (2) from (1).
%%hidden(Proof):
$$f(x) = f(x \cdot 1) = f(x) + f(1),\text{ so }f(1) = 0.$$
%%

---

__Property 3: Reciprocating the input negates the output.__
$$
f(x) = -f\left(\frac{1}{x}\right). \tag{3}
$$

This says that the way that growing the input by a factor of $x$ changes the output is exactly the opposite from the way that shrinking the input by a factor of $x$ changes the output. In terms of the "communication cost" interpretation, if doubling (or tripling, or $n$-times-ing) the possibility space increases costs by $c$, then halving (or thirding, or $n$-parts-ing) the space decreases costs by $c.$

__Exercise:__ Prove (3) from (2) and (1).
%%hidden(Proof):
$$x \cdot \frac{1}{x} = 1,\text{ so }f(1) = f\left(x \cdot \frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right).$$

$$f(1)=0,\text{ so }f(x)\text{ and }f\left(\frac{1}{x}\right)\text{ must be opposites.}$$
%%

---

__Property 4: Dividing inputs subtracts outputs.__

$$
f\left(\frac{x}{y}\right) = f(x) - f(y). \tag{4}
$$

This follows immediately from (1) and (3).

__Exercise:__ Give an interpretation of (4).
%%hidden(Interpretation):
There are at least two good interpretations:

1. $f\left(x \cdot \frac{1}{y}\right) = f(x) - f(y),$ i.e., shrinking the input by a factor of $y$ is the opposite of growing the input by a factor of $y.$
2. $f\left(z \cdot \frac{x}{y}\right) = f(z) + f(x) - f(y),$ i.e., growing the input by a factor of $\frac{x}{y}$ affects the output just like growing the input by $x$ and then shrinking it by $y.$

Try translating these into the communication cost interpretation if it is not clear why they're true.
%%

---

__Property 5: Exponentiating the input multiplies the output.__

$$
f\left(x^n\right) = n \cdot f(x). \tag{5}
$$

This says that multiplying the input by $x$, $n$ times incurs $n$ identical changes to the output. In terms of the communication cost metaphor, this is saying that you can emulate an $x^n$ digit using $n$ different $x$-digits.

__Exercise:__ Prove (5).
%%hidden(Proof):
This is easy to prove when $n \in \mathbb N:$
$$f\left(x^n\right) = f(\underbrace{x \cdot x \cdot \ldots x}_{n\text{ times}}) = \underbrace{f(x) + f(x) + \ldots f(x)}_{n\text{ times}} = n \cdot f(x).$$

For $n \in \mathbb Q,$ this is a bit more difficult; we leave it as an exercise to the reader. Hint: Use the proof of (6) below, for $n \in \mathbb N,$ to bootstrap up to the case where $n \in \mathbb Q.$

For $n \in \mathbb R,$ this is actually _not_ provable from (1) alone; we need an additional assumption (such as [https://arbital.com/p/-continuity](https://arbital.com/p/-continuity)) on $f$.
%%

Property 5 is actually _false,_ in full generality — it's possible to create a function $f$ that obeys (1), and obeys (5) for $n \in \mathbb Q,$ but which exhibits pathological behavior on irrational numbers. For more on this, see [pathological near-logarithms](https://arbital.com/p/).

This is the first place that property (1) fails us: 5 is true for $n \in \mathbb Q,$ but if we want to guarantee that it's true for $n \in \mathbb R,$ we need $f$ to be [continuous](https://arbital.com/p/continuity), i.e. we need to ensure that if $f$ follows 5 on the rationals it's not allowed to do anything insane on irrational numbers only.

---

__Property 6: Rooting the input divides the output.__

$$
f(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x}) = \frac{f(x)}{n}. \tag{6}
$$

This says that, to change the output one $n$th as much as you would if you multiplied the input by $x$, multiply the input by the $n$th root of $x$. See [https://arbital.com/p/44l](https://arbital.com/p/44l) for a physical interpretation of this fact.

__Exercise:__ Prove (6).
%%hidden(Proof):
$$(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x})^n = x,\text{ so }f\left((\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x})^n\right)\text{ has to equal }f(x).$$

$$f\left((\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x})^n\right) = n \cdot f(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x}),\text{ so }f(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x}) = \frac{f(x)}{n}.$$
%%

As with (5), (6) is always true if $n \in \mathbb Q,$ but not necessarily always true if $n \in \mathbb R.$ To prove (6) in full generality, we additionally require that $f$ be continuous.

---

__Property 7: The function is either trivial, or sends some input to 1.__

$$
\text{Either $f$ sends all inputs to $0$, or there exists a $b \neq 1$ such that $f(b)=1.$}\tag{7}
$$

This says that either $f$ is very boring (and does nothing regardless of its inputs), or there is some particular factor $b$ such that when the input changes by a factor of $b$, the output changes by exactly $1$. In the communication cost interpretation, this says that if you're measuring communication costs, you've got to pick some unit (such as $b$-digits) with which to measure.

__Exercise:__ Prove (7).
%%%hidden(Proof):
Suppose $f$ does not send all inputs to $0$, and let $x$ be an input that $f$ sends to some $y \neq 0.$ Then $f(\sqrt[https://arbital.com/p/y](https://arbital.com/p/y){x}) = \frac{f(x)}{y} = 1.$%%note: You may be wondering, "what if $y$ is negative, or a fraction?" If so, see [Strange roots](https://arbital.com/p/). Short version: $\sqrt[https://arbital.com/p/-3/4](https://arbital.com/p/-3/4){x}$ is perfectly well-defined.%%

$b$ is $\sqrt[https://arbital.com/p/y](https://arbital.com/p/y){x}.$ We know that $b \neq 1$ because $f(b) = 1$ whereas, by (2), $f(1) = 0$.
%%%

---

__Property 8: If the function is continuous, it is either trivial or a logarithm.__

$$
\text{If $f(b)=1$ then } f(b^x) = x. \tag{8}
$$

This property follows immediately from (5). Thus, (8) is always true if $x$ is a [rational](https://arbital.com/p/4zq), and if $f$ is continuous then it's also true when $x$ is irrational.

Property (8) states that if $f$ is non-trivial, then it inverts [exponentials](https://arbital.com/p/4ts) with base $b.$ In other words, $f$ counts the number of $b$-factors in $x$. In other words, $f$ counts how many times you need to multiply $1$ by $b$ to get $x$. In other words, $f = \log_b$!

---

Many texts take (8) to be the defining characteristic of the logarithm. As we just demonstrated, one can also define logarithms by (1) as [continuous](https://arbital.com/p/continuity) [non-trivial](https://arbital.com/p/trivial_mathematics) functions whose outputs grow by a constant (that depends on $y$) whenever their inputs grow by a factor of $y$. All other properties of the logarithm follow from that.

If you want to remove the "continuous" qualifier, you're still fine as long as you stick to [rational](https://arbital.com/p/4zq) inputs. If you want to remove the "non-trivial" qualifier, you can interpret the function $z$ that sends everything to zero as [$\log_\infty$](https://arbital.com/p/4c8). Allowing $\log_\infty$ and restricting ourselves to rational inputs, _every_ function $f$ that satisfies equation (1) is [isomorphic](https://arbital.com/p/4f4) to a logarithm function.

In other words, if you find a function whose output changes by a constant (that depends on $y$)  whenever its input grows by a factor of $y$, there is basically only one way it can behave. Furthermore, that function only has one degree of freedom — the choice of $b$ such that $f(b)=1.$ As we will see next, even that degree of freedom is rather paltry: All logarithm functions behave in essentially the same way. As such, if we find any $f$ such that $f(x \cdot y) = f(x) + f(y)$ (or any physical process well-modeled by such an $f$), then we immediately know quite a bit about how $f$ behaves.

Eric Rogstad
Adom Hartell
Nate Soares
Alexei Andreev

- $\log_b(x \cdot y) = \log_b(x) + \log_b(y)$ for any $b$, this is the defining characteristic of logarithms.
- $\log_b(1) = 0,$ because $\log_b(1) = \log_b(1 \cdot 1) = \log_b(1) + \log_b(1).$
- $\log_b\left(\frac{1}{x}\right) = -\log_b(x),$ because $\log_b(1) = \log_b\left(x \cdot \frac{1}{x}\right) = \log_b(x) + \log_b\left(\frac{1}{x}\right) = 0.$
- $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y),$ which follows immediately from the above.
- $\log_b\left(x^n\right) = n \cdot \log_b(x),$ because $x^n$ = $\underbrace{x \cdot x \cdot \ldots x}_{n\text{ times}}$.
- $\log_b\left(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x}\right) = \frac{\log_b(x)}{n},$ because $\log_b(x) = \log_b\left((\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x})^n\right) = n \cdot \log_b(\sqrt[https://arbital.com/p/n](https://arbital.com/p/n){x}).$
- For every $f$ that satisfies $f(x \cdot y) = f(x) + f(y)$ for all $x, y \in \mathbb R^+,$ either $f$ sends every input to 0 or there exists some $b$ such that $f(b) = 1,$ in which case we call $f$ $\log_b.$ Thus, $\log_b(b) = 1.$
- $\log_b(b^n) = n,$ because $\log_b(x^n) = n \log_b(x)$ and $\log_b(b) = 1.$


Category theory

summary(technical):
A **[category](https://arbital.com/p/-4cx)** consists of a
collection of [objects](https://arbital.com/p/-object_category_theory) and a collection of [morphisms](https://arbital.com/p/-4d8) (also called arrows). To each morphism $f$ is associated a pair of objects: $\text{dom}(f)$ which is the **source** or [**domain**](https://arbital.com/p/domain_of_function) and $\text{cod}(f)$ which is the **target** or [**codomain**](https://arbital.com/p/codomain_of_function) of $f$. If $\text{dom}(f) = X$ and $\text{cod}(f) = Y$, this is written $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

summary(motivation):
Many mathematical constructions (such as [products](https://arbital.com/p/-product_mathematics)) appear across different fields of mathematics, consisting of different ingredients but nevertheless capturing a similar idea (and often even under the same name). Category theory allows one to precisely describe the property that these different constructions all at once. This allows one to prove [theorems](https://arbital.com/p/-theorem) about all these structures at once. Hence, once you prove that a specific mathematical structure is, say, a product, then all the category-theoretic theorems about products are true for that structure. In fact, sometimes there are structures which non-obviously satisfy a category-theoretic property. Especially when [category-theoretic duality](https://arbital.com/p/duality_category_theory) is involved.

In addition, category theory allows the simple description of [functors](https://arbital.com/p/-functor), [natural transformations](https://arbital.com/p/-natural_transformation) and [adjunctions](https://arbital.com/p/-adjunction). These are mathematically powerful concepts which are very difficult to describe without the language of category theory. In fact, one of the founders of category theory, [Saunders Mac Lane](https://en.wikipedia.org/wiki/Saunders_Mac_Lane), has remarked that category theory was initially developed in order to provide a language in which to speak about natural transformations.

Powerfully, functors and adjunctions between categories allow one to translate concepts from one mathematical theory to another. They provide a "translation" (either full or partial) that allows one type of object to be viewed as another, and theorems to be translated across domains. In fact, using [duality](https://arbital.com/p/-duality_cateory_theory), very non-obvious translations can be found because a theorem in one category can be translated to its "opposite theory" in the other category. Connections which are not obvious in the language of the mathematical theories themselves, become clear in the language of category theory.

**Category theory** studies the abstraction of mathematical objects (such as [sets](https://arbital.com/p/3jz), [groups](https://arbital.com/p/3gd), and [topological spaces](https://arbital.com/p/topology)) in terms of the [morphisms](https://arbital.com/p/4d8) between them. Such a collection of objects and morphisms is a [category](https://arbital.com/p/4cx). Morphisms often represent [functions](https://arbital.com/p/3jy). For example, in the category of sets, morphisms represent all functions, in the category of groups they represent group homomorphism and in the category of topological spaces, they represent [continuous](https://arbital.com/p/continuous) maps.

Categories are usually drawn as diagrams with the objects represented by variables or points with (labeled) arrows between them representing morphisms. For this reason, morphisms are also referred to as arrows.

Morphisms do not have to represent functions. For example, any [partially ordered set](https://arbital.com/p/3rb) $(P, \leq)$ may be seen as a category where the objects are the elements of the poset and there is a (unique) morphism $x \rightarrow y$ between two elements $x$ and $y$ if and only if $x \leq y$.

## Definition ##
A **category** consists of a
collection of [objects](https://arbital.com/p/-object_category_theory) and a collection of [morphisms](https://arbital.com/p/-4d8). A morphism $f$ goes from one object, say $X$, to another, say $Y$, and is drawn as an arrow from $X$ to $Y$. Note that $X$ may equal $Y$ (in which case $f$ is referred to as an [**endomorphism**](https://arbital.com/p/endomorphism)). The object $X$ is called the **source** or [**domain**](https://arbital.com/p/domain_of_function) of $f$ and $Y$ is called the **target** or [**codomain**](https://arbital.com/p/codomain_of_function) of $f$. This is written as $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

Note that composition is written 'backwards' since given an element $x \in X$ and two functions $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, the result of applying $f$ then $g$ is $g(f(x))$ which equals $(g \circ f)(x)$.

## Motivation ##

Many mathematical constructions (such as [products](https://arbital.com/p/-product_mathematics)) appear across different fields of mathematics, consisting of different ingredients but nevertheless capturing a similar idea (and often even under the same name). Category theory allows one to precisely describe the property that these different constructions all at once. This allows one to prove [theorems](https://arbital.com/p/-theorem) about all these structures at once. Hence, once you prove that a specific mathematical structure is, say, a product, then all the category-theoretic theorems about products are true for that structure. In fact, sometimes there are structures which non-obviously satisfy a category-theoretic property. Especially when [category-theoretic duality](https://arbital.com/p/duality_category_theory) is involved.

In addition, category theory allows the simple description of [functors](https://arbital.com/p/-functor), [natural transformations](https://arbital.com/p/-natural_transformation) and [adjunctions](https://arbital.com/p/-adjunction). These are mathematically powerful concepts which are very difficult to describe without the language of category theory. In fact, one of the founders of category theory, [Saunders Mac Lane](https://en.wikipedia.org/wiki/Saunders_Mac_Lane), has remarked that category theory was initially developed in order to provide a language in which to speak about natural transformations.

Powerfully, functors and adjunctions between categories allow one to translate concepts from one mathematical theory to another. They provide a "translation" (either full or partial) that allows one type of object to be viewed as another, and theorems to be translated across domains. In fact, using [duality](https://arbital.com/p/-duality_cateory_theory), very non-obvious translations can be found because a theorem in one category can be translated to its "opposite theory" in the other category. Connections which are not obvious in the language of the mathematical theories themselves, become clear in the language of category theory.

##Categories Give an External View##
Although the objects and morphisms of a category are intended to *represent* e.g. sets and functions, from the [point of view of the category](https://arbital.com/p/-personification_of_mathematics) the objects and morphisms have no internal structure. It is not possible to talk directly about the elements of an object or how a given morphism maps elements. Instead (from the viewpoint of the category) the information about the objects and morphisms are given completely by which objects are sources and targets for the morphisms and how the morphisms are composed.

In fact, this is the strength of category theory: abstracting away the internal details allows one to focus only on relevant information and also capture information about multiple similar types of structures that act in a certain way across different mathematical theories.

This is similar to the way that a group abstracts away what elements are whilst only capturing the information of how they are 'added' or 'multiplied'.

It is also somewhat similar to the concept of a program's API (or an interface in Java); we can't see inside the program or know how it implements something, but we know what kind of inputs and outputs programs have, and what kinds of inputs and outputs a composition of such programs have.

Note that since it abstracts something away, a category does not always capture enough information for one's purposes. For example, there is addition of group homomorphisms defined pointwise. For this purpose, other structures such as [enriched categories](https://arbital.com/p/enriched_category) and [n-categories](https://arbital.com/p/n_category) may be used. However, for many purposes, categories are at a very good between isomorphic objects. This is considered a feature of category theory.

##Common Symbols: Convention##
Different texts make use of different conventions. This site makes use of the following common convention:

 - Categories are written in blackboard bold upper-case letters and are usually near the beginning of the alphabet. E.g. $\mathbb{A}, \mathbb{B}, \mathbb{C}$.
 - Objects are written as upper-case letters usually near the beginning or end of the alphabet. E.g. $A, B, C, W, X, Y, Z$.
 - Morphisms are labelled with lower-case letters, usually near f or near u. E.g. $e, f, g, h, u, v, w$.
 - Elements of an object, where necessary, are written as lower-case letters, usually near the beginning or end of the alphabet. E.g. $a, b, c, x, y, z$
 - Functors are written as upper-case letters usually near F. E.g. $E, F, G, H$.
 - Natural transformations are written as Greek letters, usually near the beginning of the alphabet. E.g. $\alpha, \beta, \gamma, \delta$.
 - The morphisms forming part of a cone or cocone for a limit or colimit are often written as Greek letters with subscripts, usually $\kappa$ or $\lambda$.

These conventions are merely guidelines and far from universally followed. Check the definition for the symbol in question to see what it represents

##[Isomorphisms in Category Theory](https://arbital.com/p/-isomorphism_category_theory)##
In category theory, [isomorphic](https://arbital.com/p/-4f4) objects are not distinguished. Many [https://arbital.com/p/-universal_constructions](https://arbital.com/p/-universal_constructions) do not pin down a [specific construction](https://arbital.com/p/-specific_construction_category_theory) but instead only specify it [up to isomorphism](https://arbital.com/p/-isomorphism_category_theory).

Doing something in category theory which relies on a specific construction, that is, requiring objects to be equal instead of merely isomorphic, is colloquially referred to as [evil](https://arbital.com/p/-evil_category_theory).

##[Universal Properties](https://arbital.com/p/-600)##
One of the most important concepts in category theory is that of a [universal property](https://arbital.com/p/600). An object in a category which satisfies a universal property is in a sense the 'best' (often meaning smallest or largest) object satisfying a certain property. This can often be used to describe in a universal way constructions like [products](https://arbital.com/p/4mj) which are defined for multiple distinct structures. In category theory, it is defined once without referring to a specific construction. This definition can then be applied to multiple categories.

The simplest [non-trivial](https://arbital.com/p/trivial) universal construction is the [terminal object](https://arbital.com/p/terminal_object). Given a category $\mathbb{C}$, an object $T$ in $\mathbb{C}$ is called a **terminal object** if, for any object $X$ in $\mathbb{C}$, there is a [unique](https://arbital.com/p/unique) morphism $f: X \rightarrow T$. In other words there is some $f: X \rightarrow T$ and if there is also $g: X \rightarrow T$ then $f=g$. In the category of sets, the terminal objects are exactly the one element sets. Given a one element set $\{a\}$, and any set $X$, there is a unique morphism $f: X \rightarrow \{a\}$, namely the function taking every $x$ in $X$ to $a$. In the category of groups, terminal objects are exactly one-element groups. Note that terminal objects need not exist. Consider a [poset seen as a category](https://arbital.com/p/poset_as_category). If it has a largest element $T$, then each object is less than or equal to $T$. So from each object there is a unique morphism to $T$ and hence it is terminal. If, however, there is no largest element then the category has no terminal object.

As another example, [products](https://arbital.com/p/4mj) can be defined by a universal property: Given a pair of objects $X$ and $Y$, an object $P$ *along with a pair of morphisms* $f: P \rightarrow X$ and $g: P \rightarrow Y$ is called the **product** of $X$ and $Y$ if, given any other object $W$ and morphisms $u: W \rightarrow X$ and $v:W \rightarrow Y$ there is a *unique* morphism $h: W \rightarrow P$ such that $fh = u$ and $gh = v$.

The above are both special cases of a very important and more general [universal construction](https://arbital.com/p/-universal_construction): the [limit](https://arbital.com/p/-limit_category_theory). This (along with the [colimit](https://arbital.com/p/-colimit_category_theory)) is described in more detail further below.

##[Duality](https://arbital.com/p/-duality_category_theory)##
For any notion in a category, its **dual** is obtained by `reversing all the arrows' and 'reversing the order of composition'. If a statement is true in any category, then its dual is true in any category. As a [corollary](https://arbital.com/p/-corollary), if a statement is true in some categories, its dual is true in the duals of those categories.

As an example, consider the definition of a [terminal object](https://arbital.com/p/-terminal_object) given above. A statement about terminal object is that any two terminal objects are isomorphic. Let's examine the exact statement. Assume $T$ is terminal. Then for any $X$ there is unique $f: X \rightarrow T$. If we reverse the arrows, we get that for every $X$ there is unique $f: X \leftarrow T$. This is the definition of an [initial object](https://arbital.com/p/-initial_object). Consider another terminal object $T'$. The statement that $T'$ is isomorphic to $T$ is means that there is some $f: T \rightarrow T'$ and $g: T' \rightarrow T$ such that $gf = 1_T$ and $fg = 1_{T'}$. The dual of this is just the statement that there is some $f: T \leftarrow T'$ and $g: T' \leftarrow T$ such that $fg = 1_T$ and $gf = 1_{T'}$, this is exactly the same property! (The morphisms $f$ and $g$ have just been renamed). Hence, the dual of the statement that a terminal object is unique up to isomorphism is the statement that every initial object is unique up to isomorphism.

Similarly, if something is true for every category with an initial object, its dual will be true for every category with a terminal object.

The concept of duality can be a powerful way of obtaining new results which come easily within category theory, but which are not obvious in the theory to which category theory is being applied. As an advanced example, the category of [Boolean Algebras](https://arbital.com/p/-boolean_algebra) is dual to the category of [Stone Spaces](https://arbital.com/p/-stone_space). See, [Stone Duality on Wikipedia](https://en.wikipedia.org/wiki/Stone's_representation_theorem_for_Boolean_algebras) for the motivation.

##[Functors](https://arbital.com/p/-Functor)##
A **functor** is a morphism between categories.

Given two categories $\mathbb{A}$ and $\mathbb{B}$, a functor $F$ from $\mathbb{A}$ to $\mathbb{B}$, written $F: \mathbb{A} \rightarrow \mathbb{B}$ is defined as a pair of functions:

 - $F_0:$ Objects($\mathbb{A}$) $\rightarrow$ Objects($\mathbb{B}$)
 - $F_1:$ Morphisms($\mathbb{A}$) $\rightarrow$ Morphisms($\mathbb{B}$)

Which satisfy:

 1.
Preservation of domain and codomain:
If $f: X \rightarrow Y$ then $F_1(f): F_0(X) \rightarrow F_1(Y)$. (
Put differently,
Dom($F_1(f)$) = $F_0$(Dom($f$)) and Cod($F_1(f)$) = $F_0$(Cod($f$)) for every morphism $f$. )

 2. Preservation of Identity:
If the morphism $1_X: X \rightarrow X$ is the identity on $X$, then the morphism $F_1(1_X): F_0(X) \rightarrow F_0(X)$ is the identity on $F_0(X)$.

 3. Preservation of composition:
Given morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, then the composition of their images $F_1(g) \circ F_1(f): F_0(X) \rightarrow F_0(Z)$ is the image of their composition $F_1(g \circ f): F_0(X) \rightarrow F_0(Z)$.

Instead of differentiating $F_0$ and $F_1$, they are usually both written simply as $F$. E.g. $F(f): F(X) \rightarrow F(Y)$.

##[Properties of Morphisms](https://arbital.com/p/-morphism_category_theory)##
Morphisms are the central objects of study in category theory. For this reason, properties of morphisms can be very important.
A morphism $f: X \rightarrow Y$ is called the following if it satisfies the given property:

 - **[Isomorphism](https://arbital.com/p/-isomorphism_category_theory):**
 ([Self-dual](https://arbital.com/p/-self_dual_category_theory))

There exists some $g: Y \rightarrow X$ such that $gf = 1_X$ and $fg = 1_Y$.

Intuitively, an isomorphism is a way of transforming from one object to another in a way that makes them indistinguishable using the information of the category.

 - **[Monomorphism](https://arbital.com/p/-monomorphism):** ([Dual](https://arbital.com/p/-duality_category_theory) to epimorphic)

For any object $W$ and morphisms $g,h: W \rightarrow X$, if $fg = fh$ then $g = h$.

Intuitively, $f$ being a monomorphism indicates that *all of the information captured by the collection of morphisms into $X$ is preserved when composing by $f$*. It generalizes the notion of an [injective function](https://arbital.com/p/4b7), since in most [concrete categories](https://arbital.com/p/-concrete_category) (like [sets](https://arbital.com/p/-3jz), [groups](https://arbital.com/p/-3gd), and [topological spaces](https://arbital.com/p/-topology)) every injective map is a monomorphism. However, even in concrete categories (and certainly more generally), monomorphisms need not be injective.

 - **[Epimorphism](https://arbital.com/p/-epimorphism):** (Dual to monomorphic)

For any object $Z$ and morphisms $g,h: X \rightarrow Z$, if $gf = hf$ then $g = h$.

Intuitively, $f$ being an epimorphism indicates that *all the information captured by the collection of morphisms out of $Y$ is preserved when composing by $f$.*. It generalizes the notion of a [surjective function](https://arbital.com/p/-4bg). However, in an even stronger sense than for monomorphisms, a function being epimorphic and a function being surjective are far from equivalent.

Properties that more closely match surjectivity include [Section / Split Epimorphism](https://arbital.com/p/-section_category_theory), and [regular epimorphism](https://arbital.com/p/-regular_epimorphism). [strict epimorphism](https://arbital.com/p/-strict_epimorphism), [strong epimorphism](https://arbital.com/p/-strong_epimorphism), and [extremal epimorphism](https://arbital.com/p/-extremal_epimorphism). Note that despite the names, not all of these are necessarily epimorphisms, but are epimorphisms in "nice" categories.

 - **[Endomorphism](https://arbital.com/p/-endomorphism):** (Self-dual)

$X = Y$, i.e., $f: X \rightarrow X$.

An endomorphism is a morphism from an object to itself.

 - **[Automorphism](https://arbital.com/p/-automorphism):** (Self-dual)

The morphism $f$ is both an endomorphism and an isomorphism.

An automorphism is a morphism from a structure to itself that preserves all the information of the structure that is distinguishable by the category. Intuitively, it gives "another view" of a structure (say, by moving around its elements) that leaves it essentially unchanged.

 - **[Retraction / Split Monomorphism](https://arbital.com/p/-retraction_category_theory):** (Dual to split epimorphic)

There exists some $g: Y \rightarrow X$ such that $gf = 1_X$

A morphism is a **retraction** if its effect can be "reversed" or inverted by another morphism applied after it. For example, every injective map is a retraction. The morphism which inverts the retraction is a **section**.

 - **[Section / Split Epimorphism](https://arbital.com/p/-section_category_theory):** (Dual to split monomorphic)

There exists some $g: Y \rightarrow X$ such that $fg = 1_Y$

A morphism is a **section** if it "reverses" the effect of some other morphism applied before it. The morphism which is inverted is a **retraction**.

##[Limits](https://arbital.com/p/-limit_category_theory) and [Colimits](https://arbital.com/p/-colimit_category_theory)##

##[Further Universal Constructions](https://arbital.com/p/-universal_construction)##

##[Natural Transformations](https://arbital.com/p/-natural_transformation)##

##[Constructions on Categories](https://arbital.com/p/-Constructions_on_categories)##

##[Adjunctions and Adjoint Functors](https://arbital.com/p/-adjunction_category_theory)##

Daniel Satanove
Tsvi BT
Patrick Stevens
Nate Soares
Eric Bruylant
Mark Chimes
Jaime Sevilla Molina

The composition of two group homomorphisms is a homomorphism


Log base infinity

There is no $\log_{\infty},$ because $\infty$ is not a [https://arbital.com/p/-4bc](https://arbital.com/p/-4bc). Nevertheless, the function $z$ defined as $z(x) = 0$ for all $x \in$ [$\mathbb R^+$](https://arbital.com/p/positive_reals) can pretty readily be interpreted as $\log_{\infty}$.

That is, $z$ satisfies all properties of the basic [properties of the logarithm](https://arbital.com/p/4bz) except for the one that says there exists a $b$ such that $\log(b) = 1.$ In the case of $\log_\infty,$ the logarithm base infinity claims "well, if you gave me a $b$ that was _large enough_ I might return 1, but for all measly finite numbers I return 0." In fact, if you're feeling ambitious, you can define $\log_\infty$ to be a [https://arbital.com/p/-multifunction](https://arbital.com/p/-multifunction) which allows infinite inputs, and define $\log_\infty(\infty)$ to return any positive real number that you'd like (1 included). This requires a few hijinks (like defining $\infty^0$ to also return any number that you'd like), but can be made to work and satisfy all the basic logarithm properties (if you strategically re-interpret some '$=$' signs as '[$\in$](https://arbital.com/p/set_contains)' signs).

The moral of the story is that functions that send everything to zero are _almost_ logarithm functions, with the minor caveat that they utterly destroy all the intricate structure that logarithm functions tap into. (That's what happens when you choose "0" as your arbitrary scaling factor when tapping into [https://arbital.com/p/-4gp](https://arbital.com/p/-4gp).)


Two independent events

$$
\newcommand{\bP}{\mathbb{P}}
$$

summary:
$$
\newcommand{\bP}{\mathbb{P}}
$$
We say that two [events](https://arbital.com/p/event_probability), $A$ and $B$, are *independent* when learning that $A$ has occurred does not change your probability that $B$ occurs. That is, $\bP(B \mid A) = \bP(B)$.
Another way to state independence is that $\bP(A,B) = \bP(A) \bP(B)$.

We say that two [events](https://arbital.com/p/event_probability), $A$ and $B$, are *independent* when learning that $A$ has occurred does not change your probability that $B$ occurs. That is, $\bP(B \mid A) = \bP(B)$.
Equivalently,  $A$ and $B$ are independent if $\bP(A)$ doesn't change if you condition on $B$: $\bP(A \mid B) = \bP(A)$.

Another way to state independence is that $\bP(A,B) = \bP(A) \bP(B)$.

All these definitions are equivalent:

$$\bP(A,B) = \bP(A)\;  \bP(B \mid A)$$

by the [chain rule](https://arbital.com/p/chain_rule_probability), so

$$\bP(A,B) = \bP(A)\;  \bP(B)\;\; \Leftrightarrow \;\; \bP(A)\;  \bP(B \mid A) = \bP(A)\;  \bP(B) \ ,$$

and similarly for $\bP(B)\;  \bP(A \mid B)$.


Cycle type of a permutation

Given an element $\sigma$ of a [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ on finitely many elements, we may express $\sigma$ in [cycle notation](https://arbital.com/p/49f).
The cycle type of $\sigma$ is then a list of the lengths of the cycles in $\sigma$, where conventionally we omit length-$1$ cycles from the cycle type.
Conventionally we list the lengths in decreasing order, and the list is presented as a comma-separated collection of values.

The concept is well-defined because [https://arbital.com/p/-49k](https://arbital.com/p/-49k) up to reordering of the cycles.

# Examples

- The cycle type of the element $(123)(45)$ in $S_7$ is $3,2$, or (without the conventional omission of the cycles $(6)$ and $(7)$) $3,2,1,1$.
- The cycle type of the identity element is the empty list.
- The cycle type of a $k$-cycle is $k$, the list containing a single element $k$.


Two independent events: Square visualization

$$
\newcommand{\true}{\text{True}}
\newcommand{\false}{\text{False}}
\newcommand{\bP}{\mathbb{P}}
$$

summary:
$$
\newcommand{\true}{\text{True}}
\newcommand{\false}{\text{False}}
\newcommand{\bP}{\mathbb{P}}
$$

Say $A$ and $B$ are independent [events](https://arbital.com/p/event_probability), so $\bP(A, B) = \bP(A)\bP(B).$ Then we can draw their joint probability distribution using the using the [square visualization](https://arbital.com/p/496) of probabilities:

This is what independence looks like, using the [square visualization](https://arbital.com/p/496) of probabilities:

We can see that the [events](https://arbital.com/p/event_probability) $A$ and $B$ don't interact; we say that $A$ and $B$ are *independent*. Whether we look at the whole square, or just the red part of
the square where $A$ is true, the probability of $B$ stays the same. In other words, $\bP(B \mid A) = \bP(B)$. That's what we mean by independence: the
probability of $B$ doesn't change if you condition on $A$.

Our square of probabilities can be generated by multiplying together the probability of $A$ and the probability of $B$:

This picture demonstrates another way to define what it means for $A$ and $B$ to be independent:

$$\bP(A, B) = \bP(A)\bP(B)\ .$$

In terms of factoring a joint distribution
--

Let's contrast independence with non-independence. Here's a picture of two ordinary, non-independent events $A$ and $B$:

(If the meaning of this picture isn't clear, take a look at [https://arbital.com/p/496](https://arbital.com/p/496).)

We have the red blocks for $\bP(A)$ and the blue blocks for $\bP(\neg A)$ lined up in columns. This means we've [factored](https://arbital.com/p/factoring_probability) our
probability distribution using $A$ as the first factor:

$$\bP(A,B) = \bP(A) \bP(B \mid A)\ .$$

We could just as well have factored by $B$ first: $\bP(A,B) = \bP(B) \bP( A \mid B)\ .$ Then we'd draw a picture like this:

Now, here again is the picture of [two independent events](https://arbital.com/p/4cf) $A$ and $B$:

In this picture, there's red and blue lined-up columns for $\bP(A)$ and $\bP(\neg A)$, and there's *also*  dark and light lined-up rows for $\bP(B)$ and
$\bP(\neg B)$. It looks like we somehow [factored](https://arbital.com/p/factoring_probability) our probability distribution $\bP$ using both $A$ and
$B$ as the first factor.

In fact, this is exactly what happened: since $A$ and $B$ are [independent](https://arbital.com/p/4cf), we have that $\bP(B \mid A) = \bP(B)$. So the diagram
above is actually factored according to $A$ first: $\bP(A,B) = \bP(A) \bP(B \mid A)$. It's just that $\bP(B \mid A)= \bP(B) = \bP(B \mid \neg A)$, since $B$
is independent from $A$. So we don't need to have different ratios of dark to light (a.k.a. conditional probabilities of $B$) in the left and right columns:

In this visualization, we can see what happens to the probability of $B$ when you condition on $A$ or on $\neg A$: it doesn't change at all. The ratio of
\[area where $B$ happens\](https://arbital.com/p/the) to \[whole area\](https://arbital.com/p/the), is the same as the ratio $\bP(B \mid A)$ where we only look at the area where $A$ happens, which is the
same as the ratio $\bP(B \mid \neg A)$ where we only look at the area where $\neg A$ happens. The fact that the probability of $B$ doesn't change when we
condition on $A$ is exactly what we mean when we say that $A$ and $B$ are independent.

The square diagram above is *also* factored according to $B$ first, using $\bP(A,B) = \bP(B) \bP(A \mid B)$. The red / blue ratios are the same in both rows
because $\bP(A \mid B) = \bP(A) = \bP(A \mid \neg B)$, since $A$ and $B$ are independent:

We couldn't do any of this stuff if the columns and rows didn't both line up. (Which is good, because then we'd have proved the false statement that any two
events are independent!)

In terms of multiplying marginal probabilities
---

Another way to say that $A$ and $B$ are independent variables %note:We're using  the [equivalence](https://arbital.com/p/event_variable_equivalence) between [https://arbital.com/p/event_probability
events](https://arbital.com/p/event_probability
events) and [binary variables](https://arbital.com/p/binary_random_variable).% is that for any truth values $t_A,t_B \in \{\true, \false\},$

$$\bP(A = t_A, B= t_B) = \bP(A = t_A)\bP(B = t_B)\ .$$

So the [joint probabilities](https://arbital.com/p/1rh) for $A$ and $B$ are computed by separately getting the probability of $A$ and the probability of $B$, and then
multiplying the two probabilities together. For example, say we want to compute the probability $\bP(A, \neg B) = \bP(A = \true, B = \false)$. We start with
the [marginal probability](https://arbital.com/p/marginal_probability) of $A$:

and the probability of $\neg B$:

and then we multiply them:

We can get all the joint probabilities this way. So we can visualize the whole joint distribution as the thing that you get when you multiply two independent
probability distributions together. We just overlay the two distributions:

To be a little more mathematically elegant, we'd use the [topological product of two spaces](https://arbital.com/p/topological_product) shown earlier to draw the joint distribution
as a product of the distributions of $A$ and $B$:

Eric Rogstad
Jaime Sevilla Molina
Tsvi BT


Transposition (as an element of a symmetric group)

In a [https://arbital.com/p/-497](https://arbital.com/p/-497), a transposition is a permutation which has the effect of swapping two elements while leaving everything else unchanged.
More formally, it is a permutation of [order](https://arbital.com/p/order_of_a_group_element) $2$ which fixes all but two elements.

%%%knows-requisite([https://arbital.com/p/4cg](https://arbital.com/p/4cg)):
A transposition is precisely an element with cycle type $2$.
%%%

# Example

In $S_5$, the permutation $(12)$ is a transposition: it swaps $1$ and $2$ while leaving all three of the elements $3,4,5$ unchanged.
However, the permutation $(124)$ is not a transposition, because it has order $3$, not order $2$.


Every member of a symmetric group on finitely many elements is a product of transpositions

Given a permutation $\sigma$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, there is a finite sequence $\tau_1, \dots, \tau_k$ of [transpositions](https://arbital.com/p/4cn) such that $\sigma = \tau_k \tau_{k-1} \dots \tau_1$.
Equivalently, symmetric groups are generated by their transpositions.

Note that the transpositions might "overlap".
For example, $(123)$ is equal to $(23)(13)$, where the element $3$ appears in two of the transpositions.

Note also that the sequence of transpositions is by no means uniquely determined by $\sigma$.

# Proof

It is enough to show that a [cycle](https://arbital.com/p/49f) is expressible as a sequence of transpositions.
Once we have this result, we may simply replace the successive cycles in $\sigma$'s disjoint cycle notation by the corresponding sequences of transpositions, to obtain a longer sequence of transpositions which multiplies out to give $\sigma$.

It is easy to verify that the cycle $(a_1 a_2 \dots a_r)$ is equal to $(a_{r-1} a_r) (a_{r-2} a_r) \dots (a_2 a_r) (a_1 a_r)$.
Indeed, that product of transpositions certainly does not move anything that isn't some $a_i$; while if we ask it to evaluate $a_i$, then the $(a_1 a_r)$ does nothing to it, $(a_2 a_r)$ does nothing to it, and so on up to $(a_{i-1} a_r)$.
Then $(a_i a_r)$ sends it to $a_r$; then $(a_{i+1} a_r)$ sends the resulting $a_r$ to $a_{i+1}$; then all subsequent transpositions $(a_{i+2} a_r), \dots, (a_{r-1} a_r)$ do nothing to the resulting $a_{i+1}$.
So the output when given $a_i$ is $a_{i+1}$.

# Why is this useful?

It can make arguments simpler: if we can show that some property holds for transpositions and that it is closed under products, then it must hold for the entire symmetric group.


Order of a group element

Given an element $g$ of group $(G, +)$ (which henceforth we abbreviate simply as $G$), the order of $g$ is the number of times we must add $g$ to itself to obtain the identity element $e$.

%%%knows-requisite([https://arbital.com/p/3gg](https://arbital.com/p/3gg)):
Equivalently, it is the order of the group $\langle g \rangle$ generated by $g$: that is, the order of $\{ e, g, g^2, \dots, g^{-1}, g^{-2}, \dots \}$ under the inherited group operation $+$.
%%%

Conventionally, the identity element itself has order $1$.

# Examples

%%%knows-requisite([https://arbital.com/p/497](https://arbital.com/p/497)):
In the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$, the order of an element is the [https://arbital.com/p/-least_common_multiple](https://arbital.com/p/-least_common_multiple) of its [cycle type](https://arbital.com/p/4cg).
%%%
%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):
In the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_6$, the order of the generator is $6$.
If we view $C_6$ as being the integers [modulo](https://arbital.com/p/modular_arithmetic) $6$ under addition, then the element $0$ has order $1$; the elements $1$ and $5$ have order $6$; the elements $2$ and $4$ have order $3$; and the element $3$ has order $2$.
%%%

In the group $\mathbb{Z}$ of [integers](https://arbital.com/p/48l) under addition, every element except $0$ has infinite order. $0$ itself has order $1$, being the identity.


Strong Church Turing thesis

> Every realistic model of computation is [polynomial time reducible](https://arbital.com/p/) to [probabilistic Turing machines](https://arbital.com/p/)

Which amounts to saying that every computable process in the universe can be efficiently simulated by a probabilistic Turing machine.

The definition of *realistic model* appeals to intuition rather than a precise definition of realistic. A rule of thumb is that a computation model is realistic if it could be used to accurately model some physical process. For example, there is a clear relation between the model of [register machines](https://arbital.com/p/) and the inner workings of personal computers. On the other hand, a computational model which can access an [NP oracle](https://arbital.com/p/) does not have a physical counterpart.

As it happened with the standard [https://arbital.com/p/4bw](https://arbital.com/p/4bw), this is an inductive rather than a mathematically defined statement. However, unlike the standard CT thesis, it is [widely believed to be wrong](https://arbital.com/p/), primarily because [quantum computation](https://arbital.com/p/) stands as a highly likely counterexample to the thesis.

We remark that non-deterministic computation is not a candidate counterexample to the strong CT thesis, since it is not realistic. That said, we can find a relation between the two concepts: if [$P=NP$](https://arbital.com/p/4bd) and [$BQP\subset NP$](https://arbital.com/p/) then $BQP = P$, so quantum computation can be efficiently simulated, and we lose the strongest contestant for a counterexample to Strong CT.

#Consequences of falsehood

The falsehood of the Strong CT thesis opens up the possibility of the existence of physical processes that, while computable, cannot be modeled in a reasonable time with a classical computer. One consequence of this, that would also mean that if quantum computing is possible the speed .up it provides is non-trivial.

Epistemic processes which assign priors using a penalty for computational time complexity are misguided if the Strong CT thesis is false. See for example [Levin search](https://arbital.com/p/).

 [probably not worth raising? human brains have crazy-low decoherence times.](https://arbital.com/p/comment:)

Eric Rogstad
Eric Bruylant
Jaime Sevilla Molina
Patrick Stevens


Category (mathematics)

A **category** consists of a collection of objects with morphisms between them. A morphism $f$ goes from one object, say $X$, to another, say $Y$, and is drawn as an arrow from $X$ to $Y$. Note that $X$ may equal $Y$ (in which case $f$ is referred to as an [https://arbital.com/p/-endomorphism](https://arbital.com/p/-endomorphism)). The object $X$ is called the source or [domain](https://arbital.com/p/3js) of $f$ and $Y$ is called the target or [codomain](https://arbital.com/p/3lg) of $f$, though note that $f$ itself need not be a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) and $X$ and $Y$ need not be sets. This is written as $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

Eric Bruylant
Mark Chimes
Patrick Stevens


Dihedral group

The dihedral group $D_{2n}$ is the group of symmetries of the $n$-vertex [https://arbital.com/p/-regular_polygon](https://arbital.com/p/-regular_polygon).

# Presentation
The dihedral groups have very simple [presentations](https://arbital.com/p/group_presentation): $$D_{2n} \cong \langle a, b \mid a^n, b^2, b a b^{-1} = a^{-1} \rangle$$
The element $a$ represents a rotation, and the element $b$ represents a reflection in any fixed axis.

# Properties

- The dihedral groups $D_{2n}$ are all non-abelian for $n > 2$. ([Proof.](https://arbital.com/p/4d0))
- The dihedral group $D_{2n}$ is a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, generated by the elements $a = (123 \dots n)$ and $b = (2, n)(3, n-1) \dots (\frac{n}{2}+1, \frac{n}{2}+3)$ if $n$ is even, $b = (2, n)(3, n-1)\dots(\frac{n-1}{2}, \frac{n+1}{2})$ if $n$ is odd.

# Examples

## $D_6$, the group of symmetries of the triangle

# Infinite dihedral group

The infinite dihedral group has presentation $\langle a, b \mid b^2, b a b^{-1} = a^{-1} \rangle$.
It is the "infinite-sided" version of the finite $D_{2n}$.

We may view the infinite dihedral group as being the subgroup of the group of [homeomorphisms](https://arbital.com/p/homeomorphism) of $\mathbb{R}^2$ generated by a reflection in the line $x=0$ and a translation to the right by one unit.
The translation is playing the role of a rotation in the finite $D_{2n}$.


Dihedral groups are non-abelian

Let $n \geq 3$. Then the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) on $n$ vertices, $D_{2n}$, is not [abelian](https://arbital.com/p/3h2).

# Proof

The most natural dihedral group [presentation](https://arbital.com/p/group_presentation) is $\langle a, b \mid a^n, b^2, bab^{-1} = a^{-1} \rangle$.
In particular, $ba = a^{-1} b = a^{-2} a b$, so $ab = ba$ if and only if $a^2$ is the identity.
But $a$ is the rotation which has order $n > 2$, so $ab$ cannot be equal to $ba$.


Needs image

Meta tag for pages which would benefit from having images.

If a specific image has been described, use the [https://arbital.com/p/5v6](https://arbital.com/p/5v6) tag.

M Yass
Alexei Andreev
Stephanie Zolayvar
Patrick Stevens
Eric Bruylant
Mark Chimes


Morphism

A morphism is the abstract representation of a [relation](https://arbital.com/p/3nt) between [mathematical objects](https://arbital.com/p/-mathematical_object).

Usually, it is used to refer to [functions](https://arbital.com/p/3jy) mapping element of one set to another, but it may represent a more general notion of a relation in [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7).

##Isomorphisms##

To understand a morphism, it is easier to first understand the concept of an [https://arbital.com/p/-4f4](https://arbital.com/p/-4f4). Two mathematical structures (say two [groups](https://arbital.com/p/-3gd)) are called **isomorphic** if they are indistinguishable using the information of the language and theory under consideration.

Imagine you are the Count von Count. You care only about counting things. You don't care what it is you count, you just care how many there are. You decide that you want to collect objects you count into boxes, and you consider two boxes equal if there are the same number of elements in both boxes. How do you know if two boxes have the same number of elements? You pair them up and see if there are any left over in either box. If there aren't any left over, then the boxes are "bijective" and the way that you paired them up is a [bijection](https://arbital.com/p/499). A bijection is a simple form of an isomorphism and the boxes are said to be isomorphic.

For example, the theory of groups only talks about the way that elements are combined via group operation (and whether they are the [https://arbital.com/p/-identity](https://arbital.com/p/-identity) or [inverses](https://arbital.com/p/-4sn), but that information is already given by the information of how elements are combined under the group operation (hereafter called multiplication). The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if their multiplication acts the same way.

Patrick Stevens
Eric Bruylant
Mark Chimes
Jaime Sevilla Molina
Team Arbital


Isomorphism

A pair of mathematical structures are **isomorphic** to each other if they are "essentially the same", even if they aren't necessarily equal.

An **isomorphism** is a [https://arbital.com/p/-4d8](https://arbital.com/p/-4d8) between isomorphic structures which translates one to the other in a way that preserves all the relevant structure. An important property of an isomorphism is that it can be 'undone' by its [inverse](https://arbital.com/p/-4sn) isomorphism.

An isomorphism from an object to itself is called an **[automorphism](https://arbital.com/p/automorphism)**. They can be thought of as symmetries: different ways in which an object can be mapped onto itself without changing it.

##Equality and Identity##
The simplest isomorphism is equality: if two things are equal then they are actually the same thing (and so not actually *two* things at all). Anything is obviously indistinguishable from itself under whatever measure you might use (it has any property in common with itself) and so regardless of the theory or language, anything is isomorphic to itself. This is represented by the [identity](https://arbital.com/p/-identity_function) (iso)morphism.

%%%knows-requisite([https://arbital.com/p/3gd](https://arbital.com/p/3gd)):
##[Group Isomorphisms](https://arbital.com/p/49x)##
For a more technical example, the theory of groups only talks about the way that elements are combined via group operation. The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if you can pair up the elements such that their group operations act the same way.
%%%

##Isomorphisms in Category Theory##
In [category theory](https://arbital.com/p/-4c7), an isomorphism is a morphism which has a two-sided [https://arbital.com/p/-4sn](https://arbital.com/p/-4sn). That is to say, $f:A \to B$ is an isomorphism if there is a morphism $g: B \to A$ where $f$ and $g$ cancel each other out.

Formally, this means that both composites $fg$ and $gf$ are equal to identity morphisms (morphisms which 'do nothing' or declare an object equal to itself). That is, $gf = \mathrm {id}_A$ and $fg = \mathrm {id}_B$.

Daniel Satanove
Eric Rogstad
Dylan Hendrickson
Patrick Stevens
Eric Bruylant
Mark Chimes


Square visualization of probabilities on two events: (example) Diseasitis

$$
\newcommand{\bP}{\mathbb{P}}
$$

summary: $$ \newcommand{\bP}{\mathbb{P}} $$

We can visualize the [https://arbital.com/p/22s](https://arbital.com/p/22s):

Then we can see at a glance that the probability $\bP(D \mid B)$ of the patient having Diseasitis given that the tongue depressor is black, despite some intuitions, isn't all that big:

$$\bP(D \mid B) = \frac{\bP(B, D)}{ \bP(B, D) + \bP(B, \neg D)} = \frac{3}{7} $$

From the [https://arbital.com/p/22s](https://arbital.com/p/22s):

> You are screening a set of patients for a disease, which we'll call Diseasitis. Based on prior epidemiology, you expect that around 20% of the patients in the screening population will in fact have Diseasitis. You are testing for the presence of the disease using a tongue depressor containing a chemical strip. Among patients with Diseasitis, 90% turn the tongue depressor black. However, 30% of the patients without Diseasitis will also turn the tongue depressor black. Among all the patients with black tongue depressors, how many have Diseasitis?

It seems like, since Diseasitis very strongly predicts that the patient has a black tongue depressor, it should be the case that the [https://arbital.com/p/-1rj](https://arbital.com/p/-1rj) $\bP( \text{Diseasitis} \mid \text{black tongue depressor})$ is big. But actually, it turns out that a patient with a black tongue depressor is more likely than not to be completely Diseasitis-free.

Can we see this fact at a glance? Below, we'll use the [https://arbital.com/p/-496](https://arbital.com/p/-496) to draw pictures and use our visual intuition.

To introduce some notation: our [prior probability](https://arbital.com/p/1rm) $\bP(D)$ that the patient has Diseasitis is $0.2$. We think that if the patient is sick $(D)$, then it's 90% likely that the tongue depressor will turn black $(B)$: we assign conditional probability $\bP(B \mid D) = 0.9$. We assign conditional probability $\bP(B \mid \neg D) = 0.3$ that the tongue depressor will be black even if the patient isn't sick. We want to know $\bP(D \mid B)$, the [posterior probability](https://arbital.com/p/1rp) that the patient has Diseasitis given that we've seen a black tongue depressor.

If we wanted to, we could solve this problem precisely using [https://arbital.com/p/1lz](https://arbital.com/p/1lz):

$$
\begin{align}
\bP(D \mid B) &= \frac{\bP(B \mid D) \bP(D)}{\bP(B)}\\
&= \frac{0.9 \times 0.2}{ \bP(B, D) + \bP(B, \neg D)}\\
&= \frac{0.18}{ \bP(D)\bP(B \mid D) + \bP(\neg D)\bP(B \mid \neg D)}\\
&= \frac{0.18}{ 0.18 + 0.24}\\
&= \frac{0.18}{ 0.42} = \frac{3}{7}  \approx 0.43\ .
\end{align}
$$

So even if we've seen a black tongue depressor, the patient is more likely to be healthy than not: $\bP(D \mid B) < \bP(\neg D \mid B) \approx 0.57$.

Now, this calculation might be enlightening if you are a real expert at [https://arbital.com/p/1lz](https://arbital.com/p/1lz). A better calculation would probably be the [odds ratio form of Bayes's rule](https://arbital.com/p/1x9).

But either way, maybe there's still an intuition saying that, come on, if the tongue depressor is such a strong indicator of Diseasitis that $\bP(B \mid D) = 0.9$, it must be that $\bP(D \mid B) =big$.

Let's use the [square visualization of probabilities](https://arbital.com/p/496) to make it really visibly obvious that $\bP(D \mid B) < \bP(\neg D \mid B)$, and to figure out why $\bP(B \mid D) = big$ doesn't imply $\bP(D \mid B) =big$.

We start with the probability of $\bP(D)$ (so we're  [factoring](https://arbital.com/p/factoring_probability) our probabilities by $D$ first):

Now let's break up the red column, where $D$ is true and the patient has Diseasitis, into a block for the probability $\bP(B \mid D)$ that $B$ is also true, and a block for the probability $\bP(\neg B \mid D)$ that $B$ is false.

>Among patients with Diseasitis, 90% turn the tongue depressor black.

That is, in 90% of the  outcomes where $D$ happens, $B$ also happens. So $0.9$ of the red column will be dark ($B$), and $0.1$ will be light:

>However, 30% of the patients without Diseasitis will also turn the tongue depressor black.

So we break up the blue $\neg D$ column by $\bP(B \mid \neg D) = 0.3$ and $\bP(\neg B \mid \neg D) = 0.7$:

Now we would like to know the probability $\bP(D \mid B)$ of Diseasitis once we've observed that the tongue depressor is black. Let's break up our diagram by whether or not $B$ happens:

Conditioning on $B$ is like only looking at the part of our distribution where $B$ happens. So the probability $\bP(D \mid B)$ of $D$  conditioned on $B$ is the proportion of that area where $D$ also happens:

Here we can see why $\bP(D \mid B)$ isn't all that big. It's true that $\bP(B,D)$ is big relative to $\bP(\neg B,D)$, since we know that $\bP(B \mid D)$ is big (patients with Diseasitis almost always have black tongue depressors):

But this ratio doesn't really matter if we want to know $\bP(D \mid B)$, the probability that a patient with a black tongue depressor has Diseasitis. What matters is that we also assign a reasonably high probability $\bP(B, \neg D)$ to the patient having a black tongue depressor but nevertheless *not* suffering from Diseasitis:

So even when we see a black tongue depressor, there's still a pretty high chance the patient is healthy anyway, and our [https://arbital.com/p/-1rp](https://arbital.com/p/-1rp) $\bP(D\mid B)$ is not that high. Recall our square of probabilities:

When asked about $\bP(D\mid B)$, we think of the really high probability $\bP(B\mid D) = 0.9$:

Really, we should look at the part of our probability mass where $B$ happens, and see that a sizeable portion goes to places where $\neg D$ happens, and the patient is healthy:

Side note
---

The square visualization is very similar to [frequency diagrams](https://arbital.com/p/1x1), except we can just think in terms of probability mass rather than specifically frequency. Also, see that page for [waterfall diagrams](https://arbital.com/p/1x1), another way to visualize updating probabilities.


Computer Programming Familiarity

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Group conjugate

Two elements $x, y$ of a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ are *conjugate* if there is some $h \in G$ such that $hxh^{-1} = y$.

# Conjugacy as "changing the worldview"

Conjugating by $h$ is equivalent to "viewing the world through $h$'s eyes".
This is most easily demonstrated in the [https://arbital.com/p/-497](https://arbital.com/p/-497), where it [is a fact](https://arbital.com/p/4bh) that if $$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$$
and $\tau \in S_n$, then $$\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))$$

That is, conjugating by $\tau$ has "caused us to view $\sigma$ from the point of view of $\tau$".

Similarly, in the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) $D_{2n}$ on $n$ vertices, conjugation of the rotation by a reflection yields the inverse of the rotation: it is "the rotation, but viewed as acting on the reflected polygon".
Equivalently, if the polygon is sitting on a glass table, conjugating the rotation by a reflection makes the rotation act "as if we had moved our head under the table to look upwards first".

In general, if $G$ is a group which [acts](https://arbital.com/p/3t9) as (some of) the symmetries of a certain object $X$ %%note:Which [we can always view as being the case](https://arbital.com/p/49b).%% then conjugation of $g \in G$ by $h \in G$ produces a symmetry $hgh^{-1}$ which acts in the same way as $g$ does, but on a copy of $X$ which has already been permuted by $h$.

# Closure under conjugation

If a subgroup $H$ of $G$ is closed under conjugation by elements of $G$, then $H$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6).
The concept of a normal subgroup is extremely important in group theory.

%%%knows-requisite([https://arbital.com/p/3t9](https://arbital.com/p/3t9)):
# Conjugation action

Conjugation forms a [action](https://arbital.com/p/3t9).
Formally, let $G$ act on itself: $\rho: G \times G \to G$, with $\rho(g, k) = g k g^{-1}$.
It is an exercise to show that this is indeed an action.
%%hidden(Show solution):
We need to show that the identity acts trivially, and that products may be broken up to act individually.

- $\rho(gh, k) = (gh)k(gh)^{-1} = ghkh^{-1}g^{-1} = g \rho(h, k) g^{-1} = \rho(g, \rho(h, k))$;
- $\rho(e, k) = eke^{-1} = k$.
%%

The [stabiliser](https://arbital.com/p/group_stabiliser) of this action, $\mathrm{Stab}_G(g)$ for some fixed $g \in G$, is the set of all elements such that $kgk^{-1} = g$: that is, such that $kg = gk$.
Equivalently, it is the [centraliser](https://arbital.com/p/group_centraliser) of $g$ in $G$: it is the subgroup of all elements which commute with $G$.

The [orbit](https://arbital.com/p/group_orbit) of the action, $\mathrm{Orb}_G(g)$ for some fixed $g \in G$, is the [https://arbital.com/p/-4bj](https://arbital.com/p/-4bj) of $g$ in $G$.
By the [https://arbital.com/p/-4l8](https://arbital.com/p/-4l8), this immediately gives that the size of a conjugacy class divides the [order](https://arbital.com/p/3gg) of the parent group.
%%%