nyuuzyou/wikis huggingface dataset:
Group of Order 35 is Cyclic Group/Proof 2

== Theorem ==
== Proof ==
Let $G$ be of order $35$.
We have that $35 = 5 \times 7$ where both $5$ and $7$ are prime.
Hence from the First Sylow Theorem, $G$ has:
:at least one Sylow $5$-subgroup
and:
:at least one Sylow $7$-subgroup
Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.
Let $n_7$ denote the number of Sylow $7$-subgroups of $G$.
By the Fourth Sylow Theorem:
:$n_5 \equiv 1 \pmod 5$
and from the Fifth Sylow Theorem:
:$n_5 \divides 35$
where $\divides$ denotes divisibility.
It follows that $n_5 = 1$.
Let $P$ denote this unique Sylow $5$-subgroup.
By the Fourth Sylow Theorem:
:$n_7 \equiv 1 \pmod 7$
and from the Fifth Sylow Theorem:
:$n_7 \divides 35$
where $\divides$ denotes divisibility.
It follows that $n_7 = 1$.
Let $Q$ denote this unique Sylow $7$-subgroup.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ and $Q$ are normal subgroups of $G$.
Consider $P \cap Q$.
By Intersection of Subgroups is Subgroup:
:$P \cap Q$ is a subgroup of $P$
:$P \cap Q$ is a subgroup of $Q$
and:
:$P \cap Q$ is a subgroup of $G$
By Lagrange's Theorem:
:$\order {P \cap Q} \divides \order P = 5$
:$\order {P \cap Q} \divides \order Q = 7$
where:
:$\order {P \cap Q}$ denotes the order of $P \cap Q$
:$\divides$ denotes divisibility.
Thus:
:$\order {P \cap Q} = 1$
and so:
:$P \cap Q = \set e$
where $e$ is the identity of $G$.
From Prime Group is Cyclic, both $P$ and $Q$ are cyclic.
Let:
:$P = \gen a$
:$Q = \gen b$
where $\gen a$ denotes the subgroup of $G$ generated by $a$.
As $P$ is a normal subgroup of $G$:
:$b^{-1} a^{-1} b \in P$
and so as $a \in P$, from :
:$b^{-1} a^{-1} b a \in P$
As $Q$ is a normal subgroup of $G$:
:$a^{-1} b a \in Q$
and so $b^{-1} \in Q$, from :
:$b^{-1} a^{-1} b a \in Q$
Thus by definition of intersection:
:$b^{-1} a^{-1} b a \in P \cap Q$
But $P \cap Q = \set e$.
So:
:$b^{-1} a^{-1} b a = e$
and so from Product of Commuting Elements with Inverses
:$a b = b a$
From Power of Product of Commutative Elements in Group:
:$\paren {a b}^7 = a^7 b^7 = a^7 \ne e$
and:
:$\paren {a b}^5 = a^5 b^5 = b^5 \ne e$
By Order of Element Divides Order of Finite Group:
:$\order {a b} \divides 35$
where $\order {a b}$ denotes the order of $a b$.
Thus $\order {a b} \in \set {1, 5, 7, 35}$
As it has been established that $\order {a b}$ is not $1$, $5$ or $7$, it follows that:
:$\order {a b} = 35$
Hence from Group whose Order equals Order of Element is Cyclic, $G$ is a cyclic group.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: Exercise $1$
Category:Group of Order 35 is Cyclic Group
Group of Order 36 is not Simple

== Theorem ==
Let $G$ be of order $36$.
Then $G$ is not simple.
== Proof ==
 $G$ is simple.
We have that:
:$36 = 2^2 \times 3^2$
Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_3 \equiv 1 \pmod 3$
and from the Fifth Sylow Theorem:
:$n_3 \divides 4$
where $\divides$ denotes divisibility.
Hence $n_3$ must be either $1$ or $4$.
Let $P$ be a Sylow $3$-subgroup of $G$. 
Let $n_3 = 4$.
By Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup, the normalizer of $P$ has index $4$.
By Order of Simple Group divides Factorial of Index of Subgroup:
:$\order G \divides 4!$
But $36$ does not divide $24$.
Hence it cannot be the case that $n_3 = 4$.
Hence $n_3 = 1$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
This contradicts the assumption that $G$ is simple.
Hence $G$ is not simple.
Category:Groups of Order 36
Group of Order 3 is Unique

== Theorem ==
There exists exactly $1$ group of order $3$, up to isomorphism:
:$C_3$, the cyclic group of order $3$.
== Proof ==
From Existence of Cyclic Group of Order n we have that one such group of order $3$ is the cyclic group of order $3$.
This is exemplified by the additive group of integers modulo $3$, whose Cayley table can be presented as:
Consider an arbitrary group $\struct {G, \circ}$ whose identity element is $e$.
Let the underlying set of $G$ be:
:$G = \set {e, a, b}$
where $a, b \in G$ are arbitrary.
Since $e$ is the identity, we can start off the Cayley table for $G$ as:
:$\begin{array}{c|ccc}
\circ & e & a & b \\
\hline
e & e & a & b \\
a & a & & \\
b & b & & \\
\end{array}$
Consider the element at $a \circ a$.
We have that $a \circ a$ must be either $e$ or $b$, as from Group has Latin Square Property it cannot be $a$.
If $a \circ a = e$, then that leaves only $b$ to complete the middle row.
But $b$ already exists in the final column.
So $a \circ a$ cannot be $e$, so must be $b$.
Hence the Cayley table for $G$ so far is:
:$\begin{array}{c|ccc}
\circ & e & a & b \\
\hline
e & e & a & b \\
a & a & b & \\
b & b & & \\
\end{array}$
The rest of the table is completed by following the result that Group has Latin Square Property:
:$\begin{array}{c|ccc}
\circ & e & a & b \\
\hline
e & e & a & b \\
a & a & b & e \\
b & b & e & a \\
\end{array}$
and it is seen by inspection that $G$ is indeed the cyclic group of order $3$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.4$
* : Chapter $3$: Elementary consequences of the definitions: Groups with three elements
Category:Groups of Order 3
Group of Order 40 has Normal Subgroup of Order 5

== Theorem ==
Let $G$ be of order $40$.
Then $G$ has a normal subgroup of order $5$.
== Proof ==
We have that:
:$40 = 2^3 \times 5$
From the First Sylow Theorem, $G$ has at least one Sylow $5$-subgroup, which is of order $5$.
Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_5 \equiv 1 \pmod 5$
and from the Fifth Sylow Theorem:
:$n_5 \divides 8$
where $\divides$ denotes divisibility.
It follows that $n_5 = 1$.
From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $5$-subgroup is normal.
Category:Groups of Order 40
Group of Order 42 has Normal Subgroup of Order 7

== Theorem ==
Let $G$ be of order $42$.
Then $G$ has a normal subgroup of order $7$.
== Proof ==
We have that:
:$42 = 2 \times 3 \times 7$
From the First Sylow Theorem, $G$ has at least one Sylow $7$-subgroup, which is of order $7$.
Let $n_7$ denote the number of Sylow $7$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_7 \equiv 1 \pmod 7$
and from the Fifth Sylow Theorem:
:$n_7 \divides 6$
where $\divides$ denotes divisibility.
It follows that $n_7 = 1$.
From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $7$-subgroup is normal.
Category:Groups of Order 42
Group of Order 48 is not Simple

== Theorem ==
Let $G$ be of order $48$.
Then $G$ is not simple.
== Proof ==
 $G$ is simple.
We have that:
:$48 = 2^4 \times 3$
Let $n_2$ denote the number of Sylow $2$-subgroups of $G$.
From Sylow $2$-Subgroups in Group of Order $48$, $n_2$ is either $1$ or $3$.
Let $P$ be a Sylow $2$-subgroup of $G$.
 
Let $n_2 = 3$.
By Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup, the normalizer of $P$ has index $3$.
By Order of Simple Group divides Factorial of Index of Subgroup:
:$\order G \divides 3!$
But $48$ does not divide $6$.
Hence it cannot be the case that $n_2 = 3$.
Hence $n_2 = 1$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
This contradicts the assumption that $G$ is simple.
Hence $G$ is not simple.
Category:Groups of Order 48
Group of Order 54 has Normal Subgroup of Order 27

== Theorem ==
Let $G$ be of order $54$.
Then $G$ has a normal subgroup of order $27$.
== Proof ==
We have that:
:$54 = 2 \times 3^3$
From the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup, which is of order $3^3 = 27$.
Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_3 \equiv 1 \pmod 3$
and from the Fifth Sylow Theorem:
:$n_3 \divides 2$
where $\divides$ denotes divisibility.
It follows that $n_3 = 1$.
From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $3$-subgroup is normal.
Category:Groups of Order 54
Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup

== Theorem ==
Let $G$ be a group of order $56$.
Then $G$ has either:
:exactly one normal Sylow $2$-subgroup
or:
:exactly one normal Sylow $7$-subgroup.
== Proof ==
Let $G$ be a group of order $56$ whose identity is $e$.
We have that:
:$56 = 2^3 \times 7$
From the First Sylow Theorem, $G$ has at least one Sylow $2$-subgroup and Sylow $7$-subgroup.
Let:
:$n_2$ denote the number of Sylow $2$-subgroups of $G$
:$n_7$ denote the number of Sylow $7$-subgroups of $G$.
From Sylow p-Subgroup is Unique iff Normal:
:if $n_2 = 1$ then the unique Sylow $2$-subgroup is normal
and:
:if $n_7 = 1$ then the unique Sylow $7$-subgroup is normal.
It remains to be shown that either $n_2 = 1$ or $n_7 = 1$.
By the Fourth Sylow Theorem:
:$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd)
and from the Fifth Sylow Theorem:
:$n_2 \divides 56$
where $\divides$ denotes divisibility.
It follows that $n_2 \in \set {1, 7}$.
By the Fourth Sylow Theorem:
:$n_7 \equiv 1 \pmod 7$
and from the Fifth Sylow Theorem:
:$n_7 \divides 56$
where $\divides$ denotes divisibility.
It follows that $n_7 \in {1, 8}$.
If either $n_2 = 1$ or $n_7 = 1$ the proof is finished.
Suppose $n_7 = 8$.
As all of these Sylow $7$-subgroups are prime, the intersection of any $2$ of them is $\set e$.
Thus, these $8$ Sylow $7$-subgroups contribute $6 \times 8 = 48$ distinct elements to $G$.
This leaves $7$ elements still to be accounted for.
A Sylow $2$-subgroup is of order $2^3 = 8$.
Thus all $7$ of these remaining elements must all be in that one Sylow $2$-subgroup
So if $n_7 \ne 1$, then $n_2 = 1$.
The result follows.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: Exercise $3$
Category:Groups of Order 56
Category:Sylow p-Subgroups
Category:Examples of Normal Subgroups
Group of Order Prime Squared is Abelian

== Theorem ==
A group whose order is the square of a prime is abelian.
== Proof ==
Let $G$ be a group of order $p^2$, where $p$ is prime.
Let $\map Z G$ be the center of $G$.
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
By Lagrange's Theorem:
:$\order {\map Z G} \divides \order G$
It follows that $\order {\map Z G} = 1, p$ or $p^2$.
By Center of Group of Prime Power Order is Non-Trivial:
:$\order {\map Z G} \ne 1$
Suppose $\order {\map Z G} = p$.
Then:
So $G / \map Z G$ is non-trivial, and of prime order.
From Prime Group is Cyclic, $G / \map Z G$ is a cyclic group.
But by Quotient of Group by Center Cyclic implies Abelian, that cannot be the case.
Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$.
Therefore $G$ is abelian.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51 \beta$
* : $\S 51.2$ The quotient group $G / Z$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Corollary $10.22$
Category:Order of Groups
Category:Abelian Groups
Category:P-Groups
Category:Groups of Order p^2
Group of Order less than 6 is Abelian

== Theorem ==
All groups with less than $6$ elements are abelian.
== Proof ==
Let $G$ be a non-abelian group.
From Non-Abelian Group has Order Greater than 4, the order of $G$ must be at least $5$.
But $5$ is a prime number.
By Prime Group is Cyclic it follows that a group of order $5$ is cyclic.
By Cyclic Group is Abelian this group is abelian.
Hence the result.
== Also see ==
* Symmetric Group on 3 Letters
* Definition:Symmetry Group of Equilateral Triangle
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 29 \beta$
* : $\S 44.3$ Some consequences of Lagrange's Theorem
Category:Abelian Groups
Group of Order p^2 q has Normal Sylow p-Subgroup

== Theorem ==
Let $p$ and $q$ be prime numbers such that $p \ne q$.
Let $G$ be a group of order $p^2 q$.
Then $G$ has a normal Sylow $p$-subgroup.
== Proof ==
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides p^2 q$
where $\divides$ denotes divisibility.
Thus $n_p \in \set {1, q}$.
Suppose $p > q$.
Then:
:$q \not \equiv 1 \pmod p$
and so $n_p \ne q$.
Hence $n_p = 1$.
From Sylow p-Subgroup is Unique iff Normal it follows that this Sylow $p$-subgroup is normal.
Suppose $q > p$.
Then $n_p$ could equal $q$ if $q \equiv 1 \pmod p$.
Let $n_q$ denote the number of Sylow $q$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_q \equiv 1 \pmod q$
and so $n_q$ is not a multiple of $q$.
From the Fifth Sylow Theorem:
:$n_q \divides p^2 q$
It follows that:
:$n_q \in \set {1, p, p^2}$
If $n_q = 1$ then it follows from Sylow p-Subgroup is Unique iff Normal that this Sylow $q$-subgroup is normal.
As $p \not \equiv 1 \pmod q$, it follows that $n_q \ne p$.
The final possibility to be explored is when $n_q = p^2$.
Then:
But then $p$ and $q$ are consecutive prime numbers, and so:
:$p = 2, q = 3$
This gives the specific case:
:$\order G = 2^2 \times 3$
and the result follows from Normal Sylow p-Subgroups in Group of Order 12.
== Sources ==
* : Chapter $12$: Applications of Sylow Theory: $(4)$ Groups of order $p^2 q$: Proposition $12.4$
Category:Sylow p-Subgroups
Category:Examples of Normal Subgroups
Category:Groups of Order p^2 q
Group of Order p^2 q is not Simple

== Theorem ==
Let $p$ and $q$ be prime numbers such that $p \ne q$.
Let $G$ be a group of order $p^2 q$.
Then $G$ is not simple.
== Proof ==
From Group of Order $p^2 q$ has Normal Sylow $p$-Subgroup, $G$ has a normal subgroup of order $p^2$.
Hence the result, by definition of simple group.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \zeta$
Category:Groups of Order p^2 q
Category:Simple Groups
Group of Order p q has Normal Sylow p-Subgroup

== Theorem ==
Let $p$ and $q$ be prime numbers such that $p > q$.
Let $G$ be a group of order $p q$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.
== Proof ==
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides p q$
where $\divides$ denotes divisibility.
The divisors of $p q$ are $1$, $p$, $q$ and $p q$.
Of these:
:$p$ and $p q$ are $\equiv 0 \pmod p$
and as $p > q$:
:$q \equiv q \pmod p$
So the only possibility for $n_p$ is $1$.
Let $P$ denote this unique Sylow $p$-subgroups.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \alpha$
* : Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$: Proposition $12.1$
Category:Sylow p-Subgroups
Category:Examples of Normal Subgroups
Category:Groups of Order p q
Group of Order p q is Cyclic

== Theorem ==
Let $p, q$ be primes such that $p < q$ and $p$ does not divide $q - 1$.
Let $G$ be a group of order $p q$.
Then $G$ is cyclic.
== Lemma ==
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Group Direct Product of Cyclic Groups: a similar result which can often be confused with this one.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \alpha$
Category:Group of Order p q is Cyclic
Category:Finite Cyclic Groups
Category:Groups of Order p q
Group of Order p q is Cyclic/Lemma

== Lemma for Group of Order $p q$ is Cyclic ==
Let $p, q$ be primes such that $p < q$ and $p$ does not divide $q - 1$.
Let $G$ be a group of order $p q$.
There is:
:exactly one Sylow $p$-subgroup of $G$.
:exactly one Sylow $q$-subgroup of $G$.
== Proof ==
Let $H$ be a Sylow $p$-subgroup of $G$.
Let $K$ be a Sylow $q$-subgroup of $G$. 
By the Fourth Sylow Theorem, the number of Sylow $p$-subgroups of $G$ is of the form $1 + k p$ and divides $p q$.
We have that $1 + k p$ cannot divide $p$.
Then $1 + k p$ must divide $q$.
But as $q$ is prime, either:
:$1 + k p = 1$
or:
:$1 + k p = q$
But:
which contradicts our condition that $p$ does not divide $q - 1$.
Hence $1 + k p = 1$.
Thus there is only one Sylow $p$-subgroup of $G$.
Similarly, there is only one Sylow $q$-subgroup of $G$.
Category:Group of Order p q is Cyclic
Group of Order p q is Cyclic/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
By Sylow $p$-Subgroup is Unique iff Normal, $H$ and $K$ are normal subgroups of $G$.
Let $H = \gen x$ and $K = \gen y$.
To show $G$ is cyclic, it is sufficient to show that $x$ and $y$ commute, because then:
:$\order {x y} = \order x \order y = p q$
where $\order x$ denotes the order of $x$ in $G$.
Since $H$ and $K$ are normal:
:$x y x^{-1} y^{-1} = \paren {x y x^{-1} } y^{-1} \in K y^{-1} = K$
and 
:$x y x^{-1} y^{-1} = x \paren {y x ^{-1} y^{-1} } \in x H = H$
Now suppose $a \in H \cap K$.
Then:
where $e$ is the identity of $G$.
Thus:
:$x y x^{-1}y^{-1} \in K \cap H = e$
Hence $x y = y x$ and the result follows.
Category:Group of Order p q is Cyclic
Group of Order p q is Cyclic/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
Let the Sylow $p$-subgroup of $G$ be denoted $P$.
Let the Sylow $q$-subgroup of $G$ be denoted $Q$.
We have that:
:$P \cap Q = \set e$
where $e$ is the identity element of $G$.
Hence in $P \cup Q$ there are $q + p - 1$ elements.
As $p q \ge 2 q > q + p - 1$, there exists a non- identity element in $G$ that is not in $H$ or $K$.
Its order must be $p q$.
Hence, by definition, $G$ is cyclic.
Category:Group of Order p q is Cyclic
Group of Permutations either All or Half Even

== Theorem ==
Let $G$ be a group of permutations.
Then either exactly half of the permutations in $G$ are even, or they are all even.
== Proof ==
From Parity Function is Homomorphism, the mapping:
:$\sgn : G \to \Z_2$
is a homomorphism.
Then:
:$G / \map \ker \sgn \cong \Img \sgn$
from the First Isomorphism Theorem.
The only possibilities for $\Img \sgn$ are $\set 0$ or $\Z_2$.
So either:
:$\order {G / \map \ker \sgn} = 1$
in which case all the permutations of $G$ are even, or:
:$\order {G / \map \ker \sgn} = 2$
in which case exactly half of them are.
Category:Symmetric Groups
Group of Prime Order p has p-1 Elements of Order p

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group with identity $e$ whose order is $p$.
Then $G$ has $p - 1$ elements of order $p$.
== Proof ==
Let $\order g$ denote the order of an element $g$ of $G$.
From Order of Element Divides Order of Finite Group:
:$\order g \divides p$
where $\divides$ denotes divisibility.
By definition of prime number, the only divisors of $p$ are $1$ and $p$.
From Identity is Only Group Element of Order 1, only $e$ has order $1$ in $G$.
That leaves $p - 1$ elements of order $p$.
== Sources ==
* : $\S 44.2$ Some consequences of Lagrange's Theorem
Category:Prime Groups
Category:Order of Group Elements
Group of Rationals Modulo One Induces Equivalence Relation

== Theorem ==
Define a relation $\sim$ on $\Q$ such that:
:$\forall p, q \in \Q: p \sim q \iff p - q \in \Z$
Then $\sim$ is an equivalence relation
== Proof ==
From Integers under Addition form Abelian Group, $\struct {\Z, +}$ forms an abelian group.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $p \in \Q$ be arbitrary.
Then:
:$p \sim p \iff 0 = p - p \in \Z$
So $\sim$ is a reflexive relation.
=== Symmetry ===
Let $p, q \in \Q$ be arbitrary.
Suppose:
:$p \sim q$
Then:
:$p - q \in \Z$
Since:
:$-1 \in \Z$
we have:
:$-\paren {p - q} = q - p \in \Z$
Hence:
:$q \sim p$
So $\sim$ is a symmetric relation.
=== Transitivity ===
Let $p, q, r \in \Q$ be arbitrary.
Suppose $p \sim q$ and $q \sim r$.
So:
:$p - q \in \Z$
And:
:$q - r \in \Z$
Then:
:$\paren {p - q} + \paren {q - r} \in \Z$
Thus:
:$p - r \in \Z$
Therefore:
:$p \sim r$
So $\sim$ is a transitive relation.
All the criteria are therefore seen to hold for $\sim$ to be an equivalence relation.
Category:Examples of Groups
Category:Rational Numbers
Group of Rationals Modulo One is Group

== Theorem ==
The set of equivalence classes $\Q/\Z$ with respect to the relation
:$\forall a, b \in \Q: a \sim b :\Longleftrightarrow a-b \mathop\in\Z$
with the binary operation
: $\Q/\Z \times \Q/\Z \to \Q/\Z, \quad \struct{[a],[b]} \mapsto [a+b]$
is an infinite abelian group.
== Proof ==
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $8 \text{(b)}$
Category:Examples of Groups
Category:Rational Numbers
Group of Reflection Matrices Order 4

== Definition ==
Consider the algebraic structure $S$ of reflection matrices:
:$R_4 = \set {\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix}, \begin {bmatrix} 1 & 0 \\ 0 & -1 \end {bmatrix}, \begin {bmatrix} -1 & 0 \\ 0 & 1 \end {bmatrix}, \begin {bmatrix} -1 & 0 \\ 0 & -1 \end {bmatrix} }$
under the operation of (conventional) matrix multiplication.
$R_4$ is the group of reflection matrices of order $4$.
=== Cayley Table ===
== Also see ==
* Group of Reflection Matrices Order 4 is Klein Four-Group
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $1 \ \text{(b)}$
Category:Klein Four-Group
Group of Reflection Matrices Order 4/Cayley Table

== Cayley Table for Group of Reflection Matrices Order $4$ ==
Consider the group of reflection matrices order $4$
:$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} }$
$R_4$ can be described completely by showing its Cayley table.
Let:
Then we have:
$\quad \begin {array} {r|rrrr}
\times & r_0 & r_1 & r_2 & r_3 \\
\hline
r_0 & r_0 & r_1 & r_2 & r_3 \\
r_1 & r_1 & r_0 & r_3 & r_2 \\
r_2 & r_2 & r_3 & r_0 & r_1 \\
r_3 & r_3 & r_2 & r_1 & r_0 \\
\end {array}$
Category:Examples of Cayley Tables
Group of Reflection Matrices Order 4 is Klein Four-Group

== Theorem ==
Let $K_4$ denote the Klein $4$-group.
Let $R_4$ be the Group of Reflection Matrices Order $4$.
Then $K_4$ and $R_4$ are isomorphic algebraic structures.
== Proof ==
Establish the mapping $\phi: K_4 \to R_4$ as follows:
From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism:
=== Cayley Table of Klein $4$-Group ===
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=== Group of Reflection Matrices Order $4$ ===
The Cayley table for $S$ is as follows:
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups: Example $2$
Category:Examples of Group Isomorphisms/Order 4
Category:Klein Four-Group
Category:Group of Reflection Matrices Order 4
Group of Rotation Matrices Order 4/Cayley Table

== Cayley Table for Group of Rotation Matrices Order $4$ ==
Consider the group of rotation matrices order $4$
:$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} }$
$R_4$ can be described completely by showing its Cayley table.
Let:
:$r_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
:$r_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
:$r_2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
:$r_3 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
Then we have:
:$\begin{array}{r|rrrr}
\times & r_0 & r_1 & r_2 & r_3 \\
\hline
r_0 & r_0 & r_1 & r_2 & r_3 \\
r_1 & r_1 & r_2 & r_3 & r_0 \\
r_2 & r_2 & r_3 & r_0 & r_1 \\
r_3 & r_3 & r_0 & r_1 & r_2 \\
\end{array}$
Category:Examples of Cayley Tables
Group of Rotations about Fixed Point is not Abelian

== Theorem ==
Let $\SS$ be a rigid body in space.
Let $O$ be a fixed point in space.
Let $\GG$ be the group of all rotations of $\SS$ around $O$.
Then $\GG$ is not an abelian group.
== Proof ==
Let $\SS$ be a square lamina.
Let $O$ be the center of $\SS$.
Recall the definition of the symmetry group of the square $D_4$:
We have that:
:Reflection $t_x$ can be achieved by a rotation of $\SS$ of $\pi$ radians about $x$.
:Reflection $t_y$ can be achieved by a rotation of $\SS$ of $\pi$ radians about $y$.
Thus $D_4$ forms a subgroup of $\GG$.
From Symmetry Group of Square is Group we have that $D_4$ is not abelian.
From Subgroup of Abelian Group is Abelian it follows that $\GG$ is also not abelian.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iii)}$
Category:Examples of Groups