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

Question

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 $.

Answer 1

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.

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.