4
$\begingroup$

Question. Is there a nontrivial group $G$ such that $[G,G]=G$ and $g^3=1$ for every $g\in G$?

All I could think of so far is the following.

  • If such a group exists it must be infinitely generated due to the local finiteness of groups of exponent 3.

  • Perfect groups of large enough prime exponent do exist, e.g. a Tarski Monster.

  • 0
    Why exponen$t$ three? Why not arbitrary exponent?2012-11-15

1 Answers 1

5

Suppose $\exp\,G=3$. Then the nilpotency class of $G$ is at most $3$. Hence $[[[G,G],G],G]=1$. But if $G$ is a perfect group, then $G=[G,G]=[[G,G],G]=[[[G,G],G],G]=1$, an impossibility.

  • 0
    Easy and straightforward approach +12012-11-15