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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 $H$. – user1729 Jun 22 '12 at 10:35
If $H$ is not normal then how is $G/H$ a group, let alone abelian? – anon Jun 22 '12 at 10:37
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. – user1729 Jun 22 '12 at 10:41
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. – user1729 Jun 22 '12 at 11:04
@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. – roy Jun 22 '12 at 11:17

By the look of the comments above, the wording of the question has been changed a few times. But it is true that if a maximal subgroup $H$ of a finite group $G$ is normal, then $[G:H]$ is prime, and $G/H$ is cyclic of prime order. Note that a maximal subgroup which is normal is not the same as a maximal normal subgroup. A maximal normal subgroup can be strictly contained in a proper subgroup, as the $A_{5}$ example given by Derek Holt shows. However, if $H$ is a maximal subgroup which happens to be normal, then the factor group $G/H$ has only one proper subgroup, the trivial group. But, for example by Cauchy's theorem, if $p$ is a prime divisor of $[G:H],$ then $G/H$ has a (necessarily cyclic) subroup of order $p,$ which must then be all of $G/H.$

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As others have remarked, it isn't necessary to assume $H$ has finite index. If $G/H$ has no non-trivial proper subgroups, then $G/H$ is certainly cyclic. But an infinite cyclic group has a proper subgroup of index $2,$ for example, so $G/H$ is necessarily finite. – Geoff Robinson Jun 22 '12 at 21:17

Your result follows almost immediately from this currently active question. The question proves that if $H$ is maximal then $G/H$ is simple.

Now, you are assuming that $H$ is maximal and that $G/H$ is finite and of prime-power order. As $H$ is maximal, $G/H$ is simple and of prime-power order. Every finite $p$-group has a non-trivial centre and the centre is always normal and so $Z(G/H)=G/H$. Thus, $G/H$ must be abelian, and you are done!

(Indeed, $G/H$ is cyclic or prime order. Can you see why this is?)

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Remember the correspondence theorem: for groups $\,N\triangleleft G\,$ , there is a $\,1-1\,$ correspondence between the subgroups of the quotient $\,G/N\,$ and the subgroups of $\,G\,$ that contain $\,N\,$ , and this correspondence respects normality and index.

Thus, if $\,H\triangleleft G\,$ is maximal, there is no proper subgroup of $\,G\,$ containing $\,N\,\Longrightarrow\,G/N\,$ has no non-trivial proper subgroups$\,\Longleftrightarrow \,G/N\,$ is cyclic of prime order and, thus, abelian.

Exercise: Prove that the only groups that don't have non-trivial proper subgroups are the cyclic ones of order a prime number.

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The answer to your question is no. Let $G$ be any finite nonabelian simple group (the smallest such is $A_5$ of order 60). Then the trivial subgroup $H=\{1\}$ is a maximal normal subgroup of $G$, but $G/H \cong G$ is not abelian.

Added later: Re-reading the question after seeing Jyrki Lahtonen's comment, I see that he is correct. If $H$ is a maximal subgroup of $G$ and $H$ is also normal in $G$, then $G/H$ is indeed cyclic of prime order. That follows from the fact that cyclic groups of prime order are the only groups (finite or infinite) with no nontrivial proper subgroups.

Sorry about that!

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Hmm. But that $H$ is not a maximal subgroup. IOW I think in the question $H$ is assumed to be a maximal subgroup that also happens to be normal. It is not assumed to be maximal among just the normal subgroups. – Jyrki Lahtonen Jun 22 '12 at 12:41
I think in the discussion above one is mixing up maximal normal subgroup (maximal w.r.t. normality) and normal maximal subgroup (normal w.r.t. maximality). These are different things as Derek showed. In addition, obeserve that a non-nilpotent group must have a maximal subgroup that is not normal. – Nicky Hekster Jun 22 '12 at 14:16

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