$G$ has maximal simple subgroup; show that exists $N$ s.t $N$ is a normal subgroup and $G/N$ is simple. $G$ has maximal simple subgroup; show that exists $N$ s.t $N$ is a normal subgroup and $G/N$ is simple.
Can it be solved using the Jordan-Holder theorem?
I’ll appreciate solution/guidance using that theorem if it is possible.
Thanks.
 A: Just an initial remark: this is trivial if $G$ is finite (or more generally finitely generated, or more generally normally finitely generated), since every nontrivial such group has a simple quotient: the question of existence of simple quotients only addresses infinitely generated groups. (An example of a nontrivial group without simple quotient is the group $\mathbf{Q}$ of rationals.)
Anyway in general, let $M$ be a maximal subgroup, which is simple. Then every normal subgroup either contains $M$ or has trivial intersection with $N$. If $M$ is normal, then $G/M$ is simple abelian. Otherwise, a normal subgroup is a proper subgroup iff it has trivial intersection with $N$. Hence Zorn's lemma applies to the family of proper normal subgroups (nonempty since $G\neq 1$ is forced by the assumption), yielding a maximal-normal subgroup, hence a simple quotient group.

Edit: Robert Chamberlain provides a more precise description, and also avoids Zorn's lemma; essentially there are 3 exclusive cases:


*

*$M$ is normal in $G$. Then $G/M$ is simple abelian. (Example: the symmetric group $\mathfrak{S}_n$ for $n\ge 5$.)

*$G$ is simple (which simple groups admit a maximal subgroup that is simple is another question)

*$M$ is not normal and $G$ is not simple: if $N$ is a nontrivial proper normal subgroup, then $M\cap N=\{1\}$ and $MN=G$, so $G=N\ltimes M$, and $N$ is characteristically simple. For finite groups, this can happen with $N$ abelian (e.g., one chooses a prime $p$ and a nontrivial irreducible $\mathbf{F}_pM$-module $N$), or non-abelian, e.g., $G=N\times N$ with $N$ simple non-abelian and $M$ is the diagonal.

A: Not sure I can get there with Jordan-Holder, but I'd do the following:
If $M$ is normal in $G$ and $G/M$ is simple because $M$ is maximal. 
If $G$ is simple then we are done.
If $G$ is not simple and $N$ is a proper non-trivial normal subgroup of $G$, not equal to $M$, then $N\cap M$ is normal in $M$ so trivial (as $M$ is simple), so $G=NM$ and $G/N\cong M$ is simple.
