If G is a non-abelian group with non-trivial Center, is it true that the commutator of G is not the whole group G?

Question

Just as the title asks, if we have a non-Abelian group $G$ that has non-trivial center $Z(G)$, i.e. $|Z(G)| \not= 1$, then is it true that $[G,G] \not= G$?

The basis for this question came from showing things about $p$-groups and such but I was wondering if this was true in general. It certainly holds for $G=Q_8$ (Quaternions) but I couldn't see how to generalize to a general group $G$. I looked around for solutions but couldn't find anything, let me know if it is really another question in disguise.

Thanks

Answer 1

Not necessarily. The smallest example is $G = \operatorname{SL}(2,5)$. You can check that $Z(G)$ has order $2$ and that $[G, G] = G$. More generally, $SL(2, q)$ with $q \geq 5$ power of an odd prime is an example.

Also, your example $Q_8$ is a $p$-group. If $G$ is any nontrivial $p$-group, then $G$ has nontrivial center and $[G, G] \neq G$.

Answer 2

$\operatorname{SL}(2,5)$ is a counterexample. It is perfect ($G'=G$) and has a center of order $2$. More generally, any quasisimple group which is not centerless violates this condition.