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I don't understand some steps in the proof by B.S.. Start with some definitions.

http://en.wikipedia.org/wiki/Maximal_subgroup#Maximal_normal_subgroup: $H \unlhd G$ is a maximal normal subgroup (or maximal proper normal subgroup) of G
(i) if $H \unlhd G$
(ii) and there is no $K \lhd G$ such that $H \lhd K \lhd G$ .
(ii ♥) Equivalently, for any $\color{brown}{K} \lhd G$ such that $H \lhd K$, either $K = H$ or $K = G$.

A group is simple if its only normal subgroups are $\{id\}$ and the group itself.

Forward step: $H \text{ maximal } ⊲ G \implies G/H$ simple.

Proof blueprint: By means of (ii ♥) , $G/H$ simple means if $\color{brown}{\frac{A}{H}} ⊲\frac{G}{H}$,
then either $\color{brown}{\frac{A}{H}} = \{id \text{ of } \frac{G}{H} \} = \color{red}{\{eH = H\}} $ or $\color{brown}{\frac{A}{H}} = \frac{G}{H}$.

Proof: Let $\frac{A}{H} ⊲\frac{G}{H}$ wherein $H ⊴A⊴G$. Since H is presupposed a maximal $\unlhd G$, $\begin{cases} H = A \implies \frac{A}{H}=1 \\ \text{ or } A=G \implies \frac{A}{H}=\frac{G}{H} \end{cases}$. ♥

(1.) Where does $H ⊴A⊴G$ spring from? Why presuppose this?
(2.) Shouldn't $\frac{A}{H}= \color{red}{\{eH = H\}}$ ? Not $= id$?

Backward step: Now suppose that $H ⊲G$ and $\frac{G}{H} $ is simple.
If we have $H ⊴A⊴G$ then obviously $\frac{A}{H} ⊲\frac{G}{H}$.

(3.) Could someone please flesh this out? It's not obvious to me.

By reason of the presupposition for this backward step, $\frac{G}{H}$ is simple.
Hence $\frac{A}{H}=\frac{G}{H}$ or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$. ♥

(4.) Where does $\frac{A}{H} =\{H\}$ loom from?
(5.) How do $A=G$ or $H=A$ follow?
(6.) What's the intuition for this theorem?

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  • The intuition is that $G \to \frac G H$ maps normal subgroups to normal subgroups. – Karolis Juodelė Feb 22 '14 at 07:23
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    A natural number $d$ is a maximal divisor of $n$ if and only if $n/d$ is prime. There is a correspondence between the intermediate divisors in between $d$, $n$ and the divisors of $n/d$. Do you understand these basic arithmetic facts? They are actually order-theoretic fact regarding lattices of divisors. The same ideas apply with lattices of (normal) subgroups. Are you aware of the correspondence theorem for groups? – anon Feb 22 '14 at 07:34

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