# Does a perfect group always have trivial center?

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.

## 1 Answer

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

• Or for a finite example, the same works with $\mathbb{R}$ replaced by $\mathbb{F}_5$. – Tobias Kildetoft Dec 4 '15 at 9:56
• 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. – BenSmith Dec 4 '15 at 9:58
• @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!) – Orest Bucicovschi Dec 4 '15 at 10:09
• @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. – Orest Bucicovschi Dec 4 '15 at 10:13
• @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. – BenSmith Dec 4 '15 at 10:15