Let $G$ be a simple group, $H\leq G,[G:H]=n$ Prove $G \cong K,K\leq{\rm Sym}(n)$.

Question

Let $G$ be a simple group, $H\leq G,[G:H]=n$. Prove $G \cong K,K\leq{\rm Sym}(n)$.

Denote an action by $G \curvearrowright G/H$.

This action induces the homomorphism $\rho : G\to{\rm Sym}(G/H)\cong{\rm Sym}(n)$.

Using isomorphism first theorem, $\exists \ker\rho \triangleleft G$ such that $\frac{G}{\ker\rho}\cong {\rm Im}(\rho)\leq{\rm Sym}(n)$.

Since $G$ is simple, $\ker\rho=\{1\},G$.

If $\ker\rho=G$ then $\frac{G}{G}=\{1\}\implies {\rm Im}(\rho)\cong{\rm Sym}(1),$ which implies $H$ is a normal subgroup, contradicting the fact that $G$ is simple. (I am not sure about this explanation.)

Also $$\ker\rho=\{1\}\implies G\cong{\rm Im}(\rho)\leq {\rm Sym}(n)$$ as required.

Is my solution correct?

Any feedback is welcome, thanks!

Answer 1

For the action you're considering, the kernel can't be $G$, if $n\gt1$, because it's equal to $\bigcap_{g\in G}gHg^{-1}\subset H$.

$\therefore$ the kernel of $\rho $ is trivial. And so you have an embedding of $G$ into $S_n$.