Group Theory/The symmetric group

Definition (symmetric group):

Let $X$ be a set. Then the symmetric group of $X$ is defined to be

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

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

Definition (permutation):

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

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

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

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

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

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

indeed, an inverse is given by

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

Definition (finite symmetric group):

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

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

Theorem (Cayley's theorem):

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

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

Definition (matrix representation of a permutation):

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

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

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

Definition (sign):

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

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

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

Definition (alternating group):

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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