# Inverse-closed subsets generating simple groups in finitely many steps

Let $G$ be an infinite simple group. Suppose there is a subset $X$ of $G$ with $$X^g = gXg^{-1}=X\qquad (g \in G)$$ which is closed under taking inverses and which even generates $G$ in finitely many steps: $$X^m = G$$ for a suitable $m \ge 2.$

We somehow suspect that $X$ might be equal to $G.$ If so, we have to establish (in the end!) that $X$ is actually closed under the multiplication; this would make $X$ a nonidentity normal subgroup of $G,$ and hence $X=G.$

But suppose this direct approach meets serious difficulties. What are then possible indirect approaches here? Anyone who met this situation in their own research, or elsewhere is kindly requested to share their ideas.

• I don't believe this is true: what about, say, the set of three-cycles in an alternating group? Aug 18, 2010 at 2:13
• It is an example of X with all the properties above, but in your case X isn't equal to A_n. Conjugacy classes generate simple groups (and quite often even in finitely many steps), but I never said X is a conjugacy class.
– Olod
Aug 18, 2010 at 2:29
• Might be I must require G be infinite. The question, however, makes sense for finite groups, either. X is not defined in any exact terms.
– Olod
Aug 18, 2010 at 2:35
• If it helps: X need not be equal to G in either the finite or the infinite case. In other words, your personal X must have some special feature not yet mentioned, so you may also be able to search for the missing hypothesis as a way to solve the problem. Aug 18, 2010 at 3:52
• Yes, something like that. I have a condition \xi(v) that defines X. It might be proven that \xi(v) => \xi(v^{-1}) and that every element of G is a product of at most m elements with \xi. The problem is whether all elements of G satisfy \xi. I just wonder if anyone encountered the similar situation. I hope somebody did.
– Olod
Aug 18, 2010 at 4:05