This is part of a bunch of exercises set by my academic supervisor. As such, I'm not sure whether all the hypotheses are needed for the conclusion. Please note that hints are preferred.
The Question:
Suppose $G$ is nonabelian and quasisimple. Show the only proper normal subgroups of $G$ are the subgroups of $Z(G)$.
The Details:
A group $G$ is quasisimple if the derived subgroup $[G,G]$ is $G$ and $G/Z(G)$ is simple.
There are other ways of defining them, but perhaps this way is a slight hint, since my supervisor included it in the email the exercises were set in.
Thoughts:
I have a few of ideas that haven't produced anything yet.
One perspective is to recall that ${\rm Inn}\,(G)\cong G/Z(G).$ However, my ideas are more elementary.
Let $N\lhd G$. Suppose $h\in N\setminus Z(G)$ and try to derive a contradiction. To that end, let $g\in G$. Then $ghg^{-1}\in N.$ In other words, there is some $c_g\in{\rm Inn}\,(G)$ such that $c_g(x)=gxg^{-1}$ for $x\in G$, and $c_g(h)\in N$. Could I use a homomorphism theorem? Also, consider $hZ(G)\subseteq G/Z(G)$. For some reason, simplicity seems like the next thing to use, but I find it hard to articulate my thoughts on that.
If $G$ is nonabelian and simple, then the result is trivial. For suppose $H$ is normal and proper in $G$. Then $H$ is trivial as $G$ is simple, so $H\le Z(G)$.
Some form of contrapositive could be effective. For example, consider showing that if $K$ is not a subgroup of $Z(G)$ but normal in $G$, then $K=G$.
Take $h,k\in N$ with $N$ as above. Try to show $hk=kh$. This is where $G$ being perfect (i.e., $G=[G,G]$) might come into play. Why I think that, I'm not sure.
It is easier than you think. Let $N\unlhd G$ be not in $Z:=Z(G)$. Then $NZ/Z=G/Z$ since $G/Z$ is simple. Hence, $G=NZ$. Now $G/N\cong Z/(Z\cap N)$ is abelian. Therefore, $G=G'\le N$.