Quotient group by a maximal subgroup is abelian 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!
 A: 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 $G/H.$
A: 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.
A: 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?)
A: 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!
A: First you need this Proposition 1:

If G is a group which has no nontrivial proper subgroups, then either G is the trivial group or G is cyclic of prime order.

The proof is here.
By Correspondence Theorem, if $H \triangleleft G$, and there is $A < G$ that contains $H$, we would also have $A/H < G/H$. As $H$ is maximal, there is no such an $A$, so there is no such an $A/H$. Thus $G/H$ has no nontrivial proper subgroup. (For a more fundamental arguments, see this.)
By Proposition 1, $G/H$ is cyclic of prime order, thus abelian.
