# 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?

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

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You don't really need the condition that $G$ is non-abelian, since it is trivially true that $[G,G]\neq G$ when $G$ is abelian and $|G|>1$. – Thomas Andrews Feb 21 '13 at 20:20

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 $[G, G] \neq G$.
Right, the result that this question came from was showing just that, if you have a $p$-group then $[G,G] \not=G$ and I was curious if it was true in general, Thanks for your help – slippery44 Feb 21 '13 at 20:22
Yes, in general, if $G$ is nilpotent, then certainly $Z(G)$ is non-trivial and $[G,G] \lneqq G$. Of course, SL(2,5) is not nilpotent. – Nicky Hekster Feb 22 '13 at 11:11
$\operatorname{SL}(2,5)$ 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.