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I need to prove that $H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.

My proof to the $(\Rightarrow)$ direction seems too much trivial:

Let us assume there exist $A$ so that $A/H\lhd G/H$. Then by definion, $H$ must be normal in $A$. Because $H$ is maximal, we get $H=A$ and therefore $A/H={1}$

Is it correct?

Update:

Now I see that I need to prove that not only $A\lhd H$ but also $A\lhd G$. Assumig I have proven that, is the proof correct?

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I think it is quite direct to use lattice isomorphism theorem. –  Ben Jun 22 '12 at 9:05
    
I know, but I'm tring no to use it :) –  Roy Jun 22 '12 at 9:08

2 Answers 2

up vote 3 down vote accepted

As you noted let $\frac{A}{H} ⊲\frac{G}{H}$ wherein $H ⊴A⊴G$. Since H is a maximal subgroup, $H=A$ or $A=G$ and so, $\frac{A}{H}=1$ or $\frac{A}{H}=\frac{G}{H} $. This means that $\frac{G}{H} $ is a simple group. Now suppose that $H ⊲G$ and $\frac{G}{H} $ is simple. If we have $H ⊴A⊴G$ then obviously $\frac{A}{H} ⊲\frac{G}{H}$ and that $\frac{G}{H}$ is simple, we get $\frac{A}{H}=\frac{G}{H}$ or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$.

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I flesh out the exquisite answer of B. S.

$\color{darkred}{ \text{ (1.) If I'm not confounded, I think $H ⊴ G$ means $H ⊲ G$ or $H = G$. Is this perfect? } }$

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

Let $\frac{A}{H} ⊲\frac{G}{H}$ wherein $H ⊴A⊴G$. Since H is a maximal subgroup, $\begin{cases} H = A \implies \frac{A}{H}=1 \\ \text{ or } A=G \implies \frac{A}{H}=\frac{G}{H} \end{cases}$. This means that $\frac{G}{H} $ is a simple group. ♥

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

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