6
$\begingroup$

Let $G$ be a finite non-abelian simple group and $t$ is equal to the number of involutions of $G$. We know that $t<|G|/3$ or $3t+1 \leq |G|$. Is this the best upper bound for the number of involutions of $G$? and if not what is it (if there is any)?

$\endgroup$

1 Answer 1

18
$\begingroup$

If $n$ denotes the number of involutions of the non-Abelian simple group $G,$ then we have $n + 1 = \sum_{\chi} \nu(\chi)\chi(1),$ where $\chi$ runs over the irreducible characters of $G$ and $\nu(\chi) \in \{0,1,-1\}$ denotes the Frobenius-Schur indicator of $\chi.$ Hence, by Cauchy-Schwarz, we have $n < \sqrt{k(G)} \sqrt{|G|},$ where $k(G)$ denotes the number of conjugacy classes of $G.$ By a Theorem of Fulman and Guralnick, which uses the classification of finite simple groups, we have $k(G) < |G|^{0.41},$ so that $n < |G|^{0.705}.$ This is certainly less than $\frac{|G|}{4}$ when $|G| > 256.$ The only non-Abelian simple groups of order less than $256$ are $A_{5}$ and ${\rm PSL}(2,7),$ which have respectively $\frac{|G|}{4}$ and $\frac{|G|}{8}$ involutions. Hence every finite non-Abelian simple group $G$ has at most $\frac{|G|}{4}$ involutions and $4$ can't be replaced by any larger constant. However, the upper bound $|G|^{0.705}$ for the number of involutions in $G$ is asymptotically much stronger ( and I suspect this is far from best possible). In fact, it is possible to prove (without using the classification of finite simple groups- I believe that R. Brauer knew this fact) that for any $\varepsilon > 0,$ there are only finitely many finite simple groups $G$ which have more than $\varepsilon |G|$ involutions. A more careful analysis of the argument at the beginning shows that if $d$ is the smallest degree of a non-trivial complex irreducible character of $G,$ then $G$ has less than $\frac{|G|}{d}$ involutions. But by Jordan's theorem on linear groups, for any integer $d >1,$ only finitely many non-Abelian simple groups have a non-trivial complex irreducible character of degree $d$ or less. In particular only finitely many non-Abelian simple groups have a non-trivial irreducible character of degree at most $\varepsilon^{-1},$ so only finitely many non-Abelian simple groups $G$ have more than $\varepsilon |G|$ involutions.

$\endgroup$
3
  • 1
    $\begingroup$ In fact, instead of counting all conjugacy classes, you could count only the self-inverse (or real) conjugacy classes. Would that give you a better exponent? $\endgroup$
    – Alex B.
    Mar 3, 2014 at 8:51
  • $\begingroup$ Dear Prof. Robinson, the above information are very useful and valuable for me. Thanks a lot indeed. $\endgroup$
    – Ahmadi
    Mar 3, 2014 at 19:55
  • $\begingroup$ Actually, the classification is not needed to prove the $\frac{|G|}{4}$ bound. The only non-Abelian simple groups with a non-trivial irreducible character of degree less than $4$ are $A_{5}$ and ${\rm PSL}(2,7),$ although a triple cover of $A_{6}$ has an irreducible character of degree $3.$ $\endgroup$ Mar 3, 2014 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.