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 $.
Best Answer
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.