This is Exercise 3.1 of Roman's, "Fundamentals of Group Theory: An Advanced Approach".
The Details:
Definition: We call a group $G$ centerless if $Z(G)=\{e\}$, where $Z(G)=\{g\in G\mid \forall h\in G, gh=hg\}$ is the centre of $G$.
The Question:
Let $H\unlhd G$ for a group $G$. If $H$ and $G/H$ are centerless, then $G$ is centerless.
Thoughts:
I had a look at the contrapositive, which states that if $G$ is not centerless, then either $G/H$ is not centerless or $H$ is not centerless. In other words: if $Z(G)$ non-trivial, then if $Z(G/H)$ is trivial, then $Z(H)$ is non-trivial.
Assume there is some $e\neq g\in Z(G)$ and suppose $Z(G/H)=\{e_{G/H}\}=\{H\}$. I aim to show $Z(H)$ is non-trivial.
If $g\in H$, then since $g$ commutes with all of $G$, which contains $H$, we can conclude $g\in Z(H)$.
So suppose $g\notin H$. We know $gH=Hg=H$ since $H\unlhd G$ and $Z(G/H)=\{e_{G/H}\}=\{H\}$. Thus $g\in H$, a contradiction.
I don't think this is quite right. The final paragraph seems a bit suspicious.
This isn't a solution-verification post, so alternative approaches are welcome.
Please help :)
