# Is every finite non-abelian simple group generated by involutions?

Let $$G$$ be a finite non-abelian simple group. Is it true that the set of involutions $$\{ g: g\in G, g^2=1 \}$$ generates $$G$$?

For example consider the group $$G=A_5$$ of order $$60$$. The involutions are the $$15$$ permutations of cycle type $$(23)(45)$$. These indeed generate all of $$A_5$$ since they can immediately be seen to generate all $$20$$ of the 3 cycles e.g. $$(123)=(12)(45)(23)(45)$$. And $$20+15=35$$ is greater than any proper divisor of $$60$$.

• Hint: Use simplicity of $G$ and Feit–Thompson theorem. Mar 31 at 13:52
• The answer is no, but I suspect you meant finite simple groups. In that case they are all generated by three involutions. Mar 31 at 13:56
• @DerekHolt Yes let me put in finite. I guess $PSL(2,\mathbb{C})$ is an example of an infinite non-abelian simple group not generated by involutions. Mar 31 at 14:01
• @MoisheKohan Oh I see the conjugate of an involution is another involution so the set of all involutions is fixed under conjugation and thus the group it generates is normal. So by simplicity $G$ is either generated by involutions or has no involutions. In some infinite cases like $PSL(2,\mathbb{C})$ there are no involutions. But if $|G|$ is finite then we can apply Feit-Thompson and order $|G|$ must be even so by Cayley's theorem there is an involution so involutions generate all of $G$. Mar 31 at 14:09
• ${\rm PSL}(2,{\mathbb C})$ contins involutions, so it is generated by involutions. But there are torsion-free infinite simple groups, which are clearly no generated by involutions. You seem to have answered your question about almost simple groups yourself. $A_6.3$ cannot be generated by involutions. Mar 31 at 16:53

So by simplicity $$G$$ is either generated by involutions or has no involutions.
Since $$|G|$$ is finite non-abelian and simple then it is finite and nonsolvable thus we can apply Feit-Thompson to conclude that $$|G|$$ is even. Thus by Cayley's theorem $$G$$ contains an element of order $$2$$.
So the subgroup generated by involutions is not trivial and by the reasoning above $$G$$ is generated by involutions.