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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)?

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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.

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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? –  Alex B. Mar 3 at 8:51
    
Dear Prof. Robinson, the above information are very useful and valuable for me. Thanks a lot indeed. –  Ahmadi Mar 3 at 19:55
    
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 $3.$ –  Geoff Robinson Mar 3 at 21:22

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