5
$\begingroup$

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 $.

$\endgroup$
9
  • 2
    $\begingroup$ Hint: Use simplicity of $G$ and Feit–Thompson theorem. $\endgroup$
    – Moishe Kohan
    Mar 31 at 13:52
  • 2
    $\begingroup$ The answer is no, but I suspect you meant finite simple groups. In that case they are all generated by three involutions. $\endgroup$
    – Derek Holt
    Mar 31 at 13:56
  • $\begingroup$ @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. $\endgroup$
    – Ian Gershon Teixeira
    Mar 31 at 14:01
  • 2
    $\begingroup$ @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 $. $\endgroup$
    – Ian Gershon Teixeira
    Mar 31 at 14:09
  • 1
    $\begingroup$ ${\rm PSL}(2,{\mathbb C})$ 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. $\endgroup$
    – Derek Holt
    Mar 31 at 16:53

1 Answer 1

Reset to default
6
$\begingroup$

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.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .