Group theory questions

Question

If $aH=bH$, show that $aH{a}^{-1}=bH{b}^{-1}$.

If $aH=bH$, then $b^{-1}a\in H$. Now,

$$H=H{b}^{-1}a⟹H{a}^{-1}=H{b}^{-1}⟹aH{a}^{-1}=aH{b}^{-1}⟹aH{a}^{-1}=bH{b}^{-1}.$$

Question

If $H\le G$ and $K◃G$, show that $H\cap K◃H$.

Consider $g\in H\cap K$ and $h\in H$. Clearly, $hg{h}^{-1}\in H$. Also, $hg{h}^{-1}\in K$ since $K$ is normal in $G$. Thus, $hg{h}^{-1}\in H\cap K$. Since conjugation is an injective map, we have $h(H\cap K)h^{-1}=H\cap K$.

Question

Show that if $H\le S_n$ and $[S_n:H]=n$, then $H\cong S_{n-1}$. $n\ge 5$.

Let $S_n$ act on the set of cosets $S_n/H$ by left multiplication: $g(\sigma H)=g\sigma H$. This induces a homomorphism $\phi :S_n\to S_n$.

$$\begin{array}{rl}\ker \phi & =\{g\in S_n|g(\sigma H)=\sigma H\text{ for all }\sigma H\}\\ & =\{g\in S_n|g\in \sigma H{\sigma }^{-1}\text{ for all }\sigma H\}\\ & =\bigcap _{\sigma H\in S_n/H}\sigma H{\sigma }^{-1}\end{array}$$

Thus, $\ker \phi \le H$.

Now, for $n\ge 5$, $A_n$ is simple. If $K$ is a normal subgroup of $S_n$ for $n\ge 5$, $K\cap A_n$ must be a normal subgroup of $A_n$, from the previous question. Thus, $K\cap A_n=\{e\}$. If all non-identity elements of $K$ are odd, $K$ must have order $2$, since the product of two odd permutations is even. If $K$ does have order $2$, it cannot be normal, since for $n\ge 3$ the center of $S_n$ is trivial. Thus, the only possibility left is $K=\{e\}$. Thus, the action is injective.

Now, restrict the group action $\phi$ to $H$ (note that if $\ast$ is an action of $G$, then $\ast$ restricted to any subgroup of $G$ is also an action!). Note that $\phi {|}_H$ remains injective.

Now, observe that all $h\in H$ fix $1H$. Thus, $\{1H\}$ is an orbit. Thus, $H$ acts on $S_n/H\setminus \{1H\}$ by left multiplication (In general, if you remove an entire orbit from the group being acted upon, the action remains an action). Note that the action still remains injective, too. We get an induced homomorphism $\psi :H\to S_{n-1}$. Since $|H|=|S_{n-1}|=(n-1)!$, $H\cong S_{n-1}$.