If $H \triangleleft G$, and $H$ is maximal, then $G/H$ is simple

Question

I would appreciate it very much if you could take a look at my proof and tell whether or not it's sufficiently good.

Proof:

Suppose there exists $N \triangleleft G/H$, then $(aH)(nH)(aH)^{-1}=nH=(ana^{-1})H$ for all $a \in G$ and $nH \in N$. This implies that $ana^{-1} = n$, thus $Z(G)=\{n\in G:nH\in G/N\}$, which implies that $H \subset N$, but since $H$ is a maximal normal group of $G$, $N$ cannot be a proper normal subgroup of $H$. Therefore, $H$ is the only normal subgroup of $G/H$. Since $H$ is trivial in $G/H$, $G/H$ is simple.

Answer 1

Take $H$ a proper normal subgroup ofer normal subgroup of $G$ which is maximal among normal subgroups of $G$, that is if we have $H\leq K\leq G$ with $K\triangleleft G$ then either $H=K$ or $K=G$.

Let us denote $Q:=G/H$ the quotient group of $G$ by $H$. It comes with a natural surjective group morphism $\pi :G\rightarrow Q$ defined by $\pi(g):=gH$.

Now you just need to use (and demonstrate) the following lemma :

For any $f:G_1\rightarrow G_2$ group morphism with $N_2\triangleleft G_2$. Then $N_1:=f^{-1}(N_2)\triangleleft G_1$. If $f$ happens to be surjective then $N_1\triangleleft G_1$ if and only if $N_2\triangleleft G_2$.

The proof is actually straightforward.

Now if we take $N\triangleleft Q$ a normal subgroup of $Q$ then by this lemma $K:=\pi^{-1}(N)$ is a normal subgroup of $G$. Furthermore $\pi(H)=\{1_Q\}\subseteq N$ so that $H\leq K$. Hence $K$ is a normal subgroup of $G$ containing $H$, by maximality of $H$ either :

$$K=H\Rightarrow N=\pi(K)=\pi(H)=\{1_Q\} $$

$$K=G\Rightarrow N=\pi(K)=\pi(G)=Q $$

Hence, either $N$ is trivial or the whole group, it means that $Q$ is necessarily simple.

Answer 2

Let $H \lhd G$ be a maximal subgroup. Then $G/H$ cannot have any non-trivial subgroup. Hence $G/H$ must be cyclic of prime order. In particular $G/H$ is simple.