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Oct 26, 2016 at 16:16 history edited Stefan Kohl CC BY-SA 3.0
Fixed a typo.
Oct 26, 2016 at 15:54 comment added Derek Holt Actually there is no need for the group $S$ in my example to be generated by elements of order $3$. You can generate the group with a finite number of conjugates of the generators of $P$ in $W$.
Oct 26, 2016 at 15:07 comment added YCor If $G$ is a simple group with an element of order 3, then clearly it's generated by its elements of order 3. Thompson's groups $T$ and $V$ (on the circle and Cantor set) are such groups and are finitely presented.
Oct 26, 2016 at 14:54 comment added CRito Thank you for your answer.I forgot to say that my group is infinite. I think that there is no such simple subgroup. Do you think that the index could be infinite anyway?
Oct 26, 2016 at 14:42 history answered Derek Holt CC BY-SA 3.0