# Group Theory/The symmetric group

Definition (symmetric group):

Let ${\displaystyle X}$ be a set. Then the symmetric group of ${\displaystyle X}$ is defined to be

${\displaystyle \operatorname {Sym} (X):=\operatorname {Aut} _{\textbf {Set}}(X)}$;

that is, it is the set of all bijective functions from ${\displaystyle X}$ to itself with composition as operation.

Definition (permutation):

A permutation is, by definition, an element of ${\displaystyle \operatorname {Sym} (X)}$.

Proposition (symmetric group essentially depends only on the cardinality of the underlying set):

Let ${\displaystyle X,Y}$ be sets of the same cardinality. Then there exists a group isomorphism

${\displaystyle \operatorname {Sym} (X)\cong \operatorname {Sym} (Y)}$.

Proof: Suppose that ${\displaystyle \phi :X\to Y}$ is a bijective function. Then the group isomorphism is given by

${\displaystyle \operatorname {Sym} (X)\ni f\mapsto \phi \circ f\circ \phi ^{-1}}$;

indeed, an inverse is given by

${\displaystyle \operatorname {Sym} (Y)\ni g\mapsto \phi ^{-1}\circ g\circ \phi }$. ${\displaystyle \Box }$

Definition (finite symmetric group):

Let ${\displaystyle n\in \mathbb {N} }$. Then the symmetric group of order ${\displaystyle n}$, denoted ${\displaystyle S_{n}}$, is defined to be

${\displaystyle S_{n}:=\operatorname {Sym} (\{1,\ldots ,n\})}$.

Theorem (Cayley's theorem):

Let ${\displaystyle G}$ be a finite group, and set ${\displaystyle n:=|G|\in \mathbb {N} }$. Then there exists a subgroup of ${\displaystyle S_{n}}$ which is isomorphic to ${\displaystyle G}$.

Proof: ${\displaystyle G}$ acts transitively on itself by left multiplication in the category of sets. This means that we have a group homomorphism ${\displaystyle \rho :G\to \operatorname {Aut} _{\textbf {Set}}(G)\cong S_{n}}$. Moreover, this morphism is injective; indeed, only the identity element of ${\displaystyle G}$ induces the identity element in ${\displaystyle S_{n}}$. Hence, the claim follows from the first Noether isomorphism theorem. ${\displaystyle \Box }$

Definition (matrix representation of a permutation):

The representation of ${\displaystyle S_{n}}$ in the category of vector spaces over a field ${\displaystyle \mathbb {F} }$ given by

${\displaystyle \rho :S_{n}\to \operatorname {Aut} (\mathbb {F} ^{n}),\sigma \mapsto [(x_{1},\ldots ,x_{n})\mapsto (x_{\sigma (1)},\ldots ,x_{\sigma (n)})]}$

is called the matrix representation of the permutations contained within ${\displaystyle S_{n}}$.

Definition (sign):

Let ${\displaystyle \sigma \in S_{n}}$ be a permutation. Then the sign of ${\displaystyle \sigma }$, written ${\displaystyle \operatorname {sgn}(\sigma )}$, is defined to be ${\displaystyle (-1)^{k}}$, where there exist ${\displaystyle k}$ transpositions ${\displaystyle \tau _{1},\ldots ,\tau _{k}}$ such that ${\displaystyle \sigma =\tau _{1}\tau _{2}\cdots \tau _{k}}$.

The following proposition shows that this notion is well-defined:

Proposition (equivalent characterisations of the sign of a permutation):

Definition (alternating group):

Let ${\displaystyle n\in \mathbb {N} }$. Then the alternating group, a subgroup of ${\displaystyle S_{n}}$, is defined to be

${\displaystyle A_{n}:=\ker \operatorname {sgn} =\{g\in S_{n}|\operatorname {sgn}(g)=1\}}$.

Proposition (alternating group is maximal and normal in the symmetric group):

Let ${\displaystyle n\in \mathbb {N} }$. Then ${\displaystyle A_{n}\triangleleft S_{n}}$, and further ${\displaystyle A_{n}}$ is a maximal subgroup of ${\displaystyle S_{n}}$.

In particular, ${\displaystyle A_{n}}$ is a maximal normal subgroup in ${\displaystyle S_{n}}$ (ie. maximal among the normal subgroups).

Proof: Note first that ${\displaystyle A_{n}}$ is normal as the kernel of a group homomorphism. We then have that ${\displaystyle \operatorname {sgn} }$ is a group homomorphism from ${\displaystyle S_{n}}$ to ${\displaystyle \mathbb {Z} _{2}}$, and by the first Noether isomorphism theorem, ${\displaystyle Z_{2}\cong S_{n}/A_{n}}$. In particular, there are only two cosets of ${\displaystyle A_{n}}$. Suppose that there existed a subgroup ${\displaystyle A_{n}\lneq H\lneq S_{n}}$. Then by the degree formula, we would have ${\displaystyle [S_{n}:H][H:A_{n}]=2}$, so that either ${\displaystyle [S_{n}:H]=1}$ or ${\displaystyle [H:A_{n}]=1}$. In both cases, one of the inclusions is not strict, a contradiction. ${\displaystyle \Box }$

Proposition (conjugation in the symmetric group is re-labeling):

Let ${\displaystyle (a_{1}a_{2}\ldots a_{k})\in S_{n}}$ be a cycle, and let ${\displaystyle g\in S_{n}}$ be any element. Then

${\displaystyle g^{-1}(a_{1}a_{2}\cdots a_{n})g=(g(a_{1})g(a_{2})\cdots g(a_{n})}$.

Proposition (in degrees 5 or larger all three-cycles are conjugate in the alternating group):

Let ${\displaystyle n\geq 5}$, and let ${\displaystyle (abc)}$ and ${\displaystyle (def)}$ be any two three-cycles in ${\displaystyle S_{n}}$. Then there exists ${\displaystyle g\in A_{n}}$ such that ${\displaystyle g(abc)g^{-1}=(def)}$.

Proof: Since being in the same conjugacy class is an equivalence relation, assume ${\displaystyle (abc)=(123)}$. ${\displaystyle \Box }$

Theorem (in degrees 5 or larger the alternating group is simple):

Let ${\displaystyle n\geq 5}$. Then ${\displaystyle A_{n}}$ is a simple group.

Proposition (in degrees 5 or larger neither the alternating nor the symmetric group are solvable):

Let ${\displaystyle n\geq 5}$. Then ${\displaystyle A_{n}}$ and ${\displaystyle S_{n}}$ are both not solvable.

Proof: Since ${\displaystyle A_{n}}$ is a maximal normal subgroup of ${\displaystyle S_{n}}$, ${\displaystyle \Box }$