Exercise 12.1.19

Proof.

(a)

If $\mathit{hH}=h^{\prime }H$, then $\mathit{ghH}=g{h}^{\prime }H$. Indeed, if $u\in \mathit{ghH}$, then $u=\mathit{ghx}$, where $x\in H$. Since $\mathit{hH}=h^{\prime }H$, then $\mathit{hx}\in \mathit{hH}$ implies $\mathit{hx}\in h^{\prime }H$, so $\mathit{hx}=h^{\prime }{x}^{\prime }$ for some $x^{\prime }\in H$. So $u=\mathit{ghx}=g{h}^{\prime }{x}^{\prime },x^{\prime }\in H$, therefore $u\in g{h}^{\prime }H$, so $\mathit{ghH}\subset g{h}^{\prime }H$, and similarly $g{h}^{\prime }H\subset \mathit{ghH}$, so $\mathit{ghH}=g{h}^{\prime }H$, and $g\cdot \mathit{hH}=\mathit{ghH}$ is well defined.

Moreover $e\cdot \mathit{hH}=\mathit{ehH}=\mathit{hH}$ and $g\cdot (g^{\prime }\cdot H)=g\cdot g^{\prime }H=g{g}^{\prime }H=(g{g}^{\prime })\cdot H$, so $g\cdot \mathit{hH}=\mathit{ghH}$ defines a left action of $G$ on the set of left cosets.

(b)

Let $u$ any element of $G$. $$u\in {\mathrm{Stab}}_G(\mathit{eH})⟺u\cdot \mathit{eH}=\mathit{eH}⟺\mathit{ueH}=\mathit{eH}⟺\mathit{uH}=H⟺u\in H.$$

The last equivalence is true, because $\mathit{uH}=H$ implies $u=\mathit{ue}\in H$, and conversely, if $u\in H$, $\mathit{uH}\subset H$ and every element $x\in H$ satisfies $x=u(u^{-1}x)$, where $u^{-1}x\in H$, so $x\in \mathit{uH}$.

$${\mathrm{Stab}}_G(\mathit{eH})=H.$$

(c)

Let $$\psi \{\begin{array}{ccc}\hfill G\hfill & \hfill \to \hfill & S_m\hfill \\ \hfill g\hfill & \hfill \mapsto \hfill & \sigma :\forall <!--\; FUNCTION\; APPLICATION\; -->i\in \text{[[}1,m\text{]]},g\cdot g_{i}H=g_{\sigma (i)}H\hfill \end{array}$$

Let $g,g^{\prime }\in G$, $\sigma =\psi (g),{\sigma }^{\prime }=\psi (g^{\prime })$. For all $i,1\le i\le m$,

$$(g{g}^{\prime })\cdot g_{i}H=g\cdot (g^{\prime }\cdot g_{i}H)=g\cdot g_{{\sigma }^{\prime }(i)}H=g_{\sigma ({\sigma }^{\prime }(i))}H=g_{(\sigma \circ {\sigma }^{\prime })(i)}H.$$

Therefore $\psi (g{g}^{\prime })=\sigma \circ {\sigma }^{\prime }$, so $\psi :G\to S_m$ is a group homomorphism.

(d)

Let $N$ be the kernel of $\psi$. For every $g\in G$, $$\begin{array}{llll}\hfill g\in N& ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->i\in \text{[[}1,m\text{]]},g\cdot g_{i}H=g_{i}H& \hfill & \\ \hfill & ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->h\in G,\mathit{ghH}=\mathit{hH}& \hfill & \\ \hfill & ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->h\in G,h^{-1}\mathit{ghH}=H& \hfill & \\ \hfill & ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->h\in G,h^{-1}\mathit{gh}\in H& \hfill & \\ \hfill & ⟺\forall <!--\; FUNCTION\; APPLICATION\; -->h\in G,g\in \mathit{hH}{h}^{-1}& \hfill & \\ \hfill & ⟺g\in \underset{h\in G}{\bigcap }\mathit{hH}{h}^{-1}& \hfill & \end{array}$$

so

$$N=\underset{h\in G}{\bigcap }\mathit{hH}{h}^{-1}.$$

( $N$ is the core of $H$ in $G$. We write $N={\mathrm{Core}}_G(H)$.)

Since $H=\mathit{eH}{e}^{-1}\supset {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{h\in G}\mathit{hH}{h}^{-1}$, $H\supset N$. (e) The first isomorphism theorem for groups gives the isomorphism

so $[G:N]=|\mathrm{Im}(\psi )|$ divides $|S_m|=m!$ by Lagrange’s theorem.

(f) We can conclude that for any subgroup $H$ of a finite group $G$, then $H$ contains the core $N$ of $H$ in $G$, which is a normal subgroup of $G$ whose index divides $[G:H]!$. (g)

$\text{∙}$

Let $H\subset S_n$ be a subgroup of index $[S_n:H]>1$, where $n\ge 5$.

Let $N={\mathrm{Core}}_{S_n}(H)$. Then $N\subset H\subset S_n$, and $N$ is normal in $S_n$, and $N\ne S_n$ (since $[S_n:H]>1$). By Proposition 8.4.6, $N=A_n$ or $N=\{e\}$.

If $N=A_n$, then $N=A_n\subset H\subset S_n$, thus $1<[S_n:H]\le [S_n:A_n]=2$, therefore $[S_n:H]=2=[S_n:A_n]$, where $A_n\subset H$, so $H=A_n$.

In the other case, $N=\{e\}$. By part (e), $[S_n:N]\mid [S_n:H]!$, thus $n!\mid m!$, where $m=[S_n:H]$. Therefore $n\le m=[S_n:H]$. This proves part (a) of Theorem 12.1.10.

$\text{∙}$

Let $H\subset A_n$ be a subgroup of index $[A_n:H]>1$.

Let $N={\mathrm{Core}}_{A_n}(H)$. Then $N\subset H\subset A_n$ and $N$ is normal in $A_n$. Since $A_n$ is simple for $n\ge 5$, and $N\subset H\ne A_n$, $N=\{e\}$.

By part (e), $[A_n:N]\mid [A_n:H]!$, thus $n!∕2\mid m!$, where $m=[A_n:H]$.

If $m<n$ then $m\le n-1,m!\le (n-1)!<n!∕2$ (since $n>2$), in contradiction with $n!∕2\mid m!$. Therefore

$$n\le m=[A_n:H].$$

This proves part (b) of Theorem 12.1.10.