If $G^{'}$ is Commutator subgroup of $G$ and $G=G{'} $. Can I show that $Z(G)= \{e \} $?

I think it's not True but I can not find example.


Hint: $SL(2, \mathbb{R})$ has nontrivial center and equals its derived group.

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    $\begingroup$ Or for a finite example, the same works with $\mathbb{R}$ replaced by $\mathbb{F}_5$. $\endgroup$ – Tobias Kildetoft Dec 4 '15 at 9:56
  • $\begingroup$ dang you beat me to it. Although if you've got the time it might be worth it to recall what $SL_2$ is equal to its commutator. $\endgroup$ – BenSmith Dec 4 '15 at 9:58
  • $\begingroup$ @Tobias Kildetoft: Yes, totally. I liked the Lie aspect , but the algebra should work the same-- I guess using elementary matrices, for not too small fields. This got me to read a bit about perfect groups ( good call with the editing!) $\endgroup$ – Orest Bucicovschi Dec 4 '15 at 10:09
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    $\begingroup$ @Ben Smith: Yep, here goes my Lie side. Btw, I have to redo the generation of $SL_2$ in terms of elementary matrices, that should help I suppose. $\endgroup$ – Orest Bucicovschi Dec 4 '15 at 10:13
  • $\begingroup$ @orangeskid is there any way to collaborate on answers? I know I have this proof typed up already from an assignment in a course on linear algebraic groups. $\endgroup$ – BenSmith Dec 4 '15 at 10:15

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