$G$ 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.
abstract-algebragroup-theory
$G$ 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.
Best Answer
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: