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I'm having trouble proving this. If $G$ has a composition series of length 2, then that means there is a maximal normal subgroup $H$ of $G$ such that $H$ is simple, i.e. we would have a composition series $$\{e\}\lhd H\lhd G.$$ Suppose there is another maximal subgroup $H'$ of $G.$ I'm guessing I'd want to show $H'$ is simple as well, but I have no idea how to do it. I tried proving it by way of contradiction. So assume $H'_1$ is a nontrivial maximal subgroup of $H'.$ Then something should go wrong, but I don't know what and how. Any hints would be much appreciated.

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Suppose there is a proper, nontrivial normal subgroup $K$, which is distinct from $H$.

What can you say about the normal subgroup $H K$? What can you say about $H \cap K$?

Hint

$H K/ H$ is a non-trivial normal subgroup of the simple group $G/H$.

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  • $\begingroup$ Well both $HK$ and $H \cap K$ are normal subgroups of $G.$ I'm guessing $HK$ should be equal to $G$? $H$ is normal in $HK$ but $H$ is maximal, so $HK=G.$ Now by the Second Isomorphism Theorem, $G/H \cong HK/H$ is isomorphic to $K/(H \cap K)$ and $G/K \cong HK/H$ is isomorphic to $H/(H\cap K).$ So $H/(H\cap K)$ is simple, but $H$ itself is simple so $H\cap K$ is just $\{e\}.$ That means $K$ is simple, giving us the result. Is this right? $\endgroup$ Jun 3 '16 at 18:21
  • $\begingroup$ Also, I am wondering, is $HK/H$ isomorphic to $K$ when $H$ and $K$ are normal subgroups? $\endgroup$ Jun 3 '16 at 18:24
  • $\begingroup$ I believe your first comment to be correct. AS to the second one, you correctly said $H K / H \cong K/(H \cap K)$ earlier. $\endgroup$ Jun 3 '16 at 20:18
  • $\begingroup$ Great, thank you very much! $\endgroup$ Jun 4 '16 at 3:04

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