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Let $H \leq G$, $H$ is a maximal subgroup of $G$. Is it true that if $H \lhd G$ then $G/H$ is abelian? If yes, is it also true when $H$ is not normal?

I know that if $G' \leq H$ then $G/H$ is abelian, but I don't think this is necessary.

(I'm trying to prove that $G/H$ is finite and of a prime order, and the only thing that's missing is the above)

Thank you!

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    No. If $H$ is normal then this is necessary and sufficient. That is, $G^{\prime}\leq H\Leftrightarrow G/H$ is abelian. Think about it...if $abH=baH$ then you need $aba^{-1}b^{-1}\in H$. As this must hold for all $a, b\in G$ you need $G^{\prime}$ to be in $H$.2012-06-22
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    If $H$ is not normal then how is $G/H$ a group, let alone abelian?2012-06-22
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    To prove your result, you might want to try the correspondence theorem (also called...the lattice theorem? Or something.) Basically, the subgroups of $G/H$ and the subgroups of $G$ which contain $H$ are in a one-to-one correspondence with each other, and their homomorphic images will be the same (for corresponding subgroups). So take a maximal subgroup of $G/H$. As $G/H$ is a $p$-group this maximal subgroup contains the derived subgroup of $G/H$ (why?) and so the quotient is abelian, and you're done.2012-06-22
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    Do you mean prime order, or prime-power order? Every group of prime order is cyclic, as such a group cannot have any subgroups, by Lagrange, and as every non-trivial element generates a subgroup it must generate the group itself.2012-06-22
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    @user1729: I mean an order that is equal to a prime. I'm not assuming that $G/H$ is of a prime order, that's what I'm trying to prove. I've proved the next: (1) If $H$ is maximal normal then $G/H$ is simple. (just like the question) (2) An abelian group is simple iff it is finite and of a prime order So all I have left to prove is that $G/H$ is abelian.2012-06-22
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    What is it you are trying to prove, precisely, in the form of "If $X$ then $Y$"?2012-06-22
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    @roy: please consider [registering your account](http://math.stackexchange.com/users/login) that way you will be able to post comments to your own questions.2012-06-22
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    @Willie: Unregistered users can still post comments, but their accounts often fork without notice and then they cannot post comments anymore.2012-06-22

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