Proper and normal in quasisimple implies central

Verbal statement
Any proper normal subgroup in a quasisimple group must be a central subgroup, viz it must be contained inside the center.

In other words, a quasisimple group is a group in which every proper normal subgroup is central. This is equivalent to saying that every proper subnormal subgroup is central.

Facts used

 * 1) uses::Second isomorphism theorem
 * 2) uses::Normality satisfies image condition

Proof
Given: A group $$G$$ with $$[G,G] = G$$ and $$G/Z(G)$$ simple. A normal subgroup $$N$$ of $$G$$. $$Z(G)$$ is the center of $$G$$.

To prove: $$N= G$$ or $$N \le Z(G)$$.

Proof:


 * 1) $$NZ(G)/Z(G)$$ is a normal subgroup of $$G/Z(G)$$: Since $$N$$ is normal in $$G$$, fact (2) tells us that the image of $$N$$ under the quotient map $$G \to G/Z(G)$$ is a normal subgroup of $$G/Z(G)$$. This image is precisely $$NZ(G)/Z(G)$$.
 * 2) Either $$N \le Z(G)$$ or $$NZ(G) = G$$: Since $$G/Z(G)$$ is simple, $$NZ(G)/Z(G)$$ is either trivial or $$G/Z(G)$$, so $$NZ(G) = Z(G)$$ (forcing $$N \le Z(G)$$) or $$NZ(G) = G$$.
 * 3) If $$NZ(G) = G$$, then $$G/N$$ is abelian: By fact (1), $$G/N = NZ(G)/N \cong Z(G)/(Z(G) \cap N)$$, which is a quotient of abelian groups, hence abelian.
 * 4) If $$NZ(G) = G$$, then $$G = N$$: Since $$G/N$$ is abelian, $$[G,G] \le N$$, so $$G \le N$$, since $$[G,G] = G$$by assumption.

Steps (2) and (4) complete the proof.