2 lemmas about Cayley's Group Theorem

hbghlyj

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Lemma1: Let $G$ be simple group and $H$ be a subgroup of index $n$. Then $G\leq A_n$.

Lemma2: Let $G$ be  group and $H$ be a subgroup of index $n$ such that $H$ does not contain any nontrivial normal subgroup of $G$. Then $G\leq S_n$.

How to prove?

Czhang271828

USE GROUP ACTION.

Proof of Lemma1: $\Omega:=\{gH\mid g\in G\}$ is an $n$ -element set. Let $S(\Omega)$ denotes the symmertry group of $\Omega$, thus $|S(\Omega)|=|S_n|=n!$. Consider the group action $f\in \mathrm{Hom}(G,S(\Omega))$:
\[
x:\Omega\to \Omega , aH\mapsto xa H,\quad \forall x\in G.
\]Then the kernel of the such action satisfies $\mathrm{ker}(f)\lhd G$, hence it is either $G$ or $\{e\}$. Obviously $\mathrm{ker}(f)=\{e\}$. As a result, $G\leq S(\Omega)=S_n$. Since $(G\cap A_n)\lhd G$, $G\lhd A_n$.

Proof of Lemma2: Also consider the group action $f\in \mathrm{Hom}(G,S(\Omega))$.
\[
x:\Omega\to \Omega , aH\mapsto xa H,\quad \forall x\in G.
\] $\mathrm{ker}(f)$ is defined as $\{x\in G\mid aH=xaH,\forall a\in G\}$. Since
\[aH=xaH\,(\forall a\in G)\Leftrightarrow aH\ni xa\,(\forall a\in G)\Leftrightarrow x\in \cap _{a\in G}aHa^{-1},\] Then $\mathrm{ker}(f)=\cap _{a\in G}aHa^{-1}\leq H$. We notice that $\mathrm{ker}(f)\lhd G$, hence $\mathrm{ker}(f)=\{e\}$. It yields that $G\leq S_n$.