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$G$ is a group of finite order with center of odd cardinality. $G$ has a non-trivial subgroup $H$ that is simple and $[G:H]=2$. I want to prove that $H$ is the only non-trivial proper normal subgroup of $G$.

I think we should use this fact: if $N$ is normal in $G$ of cardinality 2 then $N<Z(G)$.

This is what I have done: $H$ is normal because of index 2. Take a subgroup $K$ normal, proper and non trivial. $K$ cannot be contained properly in $H$ because $H$ is simple. If I prove that $H\subseteq K$ then $|G/K||K/H|=|G/H|=2$ and so $K=H$. So I want to prove that $H\subseteq K$. Now, $H\cap K\trianglelefteq H$ and $H$ is simple then we have $H\cap K=\{e\}$ or $H\cap K=H$ and so $H\subseteq K$. So I suppose that $H\cap K=\{e\}$ and I want to reach a contradiction.

I don't know if this is the right approach, but I don't know how to continue.

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$|G| = |KH| = |K||H|/|K \cap H|$, right? So $|K| = 2$, and your fact then applies. – Dylan Moreland Jan 2 '12 at 5:41
ah thank you , I was almost there – Alex M Jan 2 '12 at 5:52
@DylanMoreland You could post the comment as an answer, even if the question has been asked long time ago. – Davide Giraudo Apr 23 '12 at 16:52

I'm posting Dylan Moreland's comment as a community wiki answer, since he didn't follow up that suggestion almost a year ago:

$|G|=|KH|=|K||H|/|K\cap H|$, right? So $|K|=2$, and your fact then applies.

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