If $G$ is a group let $I(G)$ be the number of involutions (elements of order 2) in $G$. My question is then easily stated: does there exists a constant $C > 1$ such that for every $n \ge 1$ and every subgroup $G \subset S_n$ of the symmetric group $S_n$ we have $$ C^{-n} |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. $$

Some remarks :

*This is true for primitive groups, since those groups are either very small (of size at most $D^n$ for some absolute $D$) or they have to be the full symmetric or alternating groups (for which the number of involutions is precisely known).

*Using iterated wreath products decompositions and the previous remark it is possible to show something like $$ C^{-n\log\log(n)} |G|^{\frac 1 2} \le I(G)+ 1 \le C^{n\log\log(n)} |G|^{\frac 1 2}. $$ (This is actually good enough for the application I have in mind but I was wondering whether a sharper result to which I could refer existed).

*The lower bound would be sharp, since for example a 3-Sylow of $S_n$ is of size roughly $3^{n/2}$ as $n$ goes to infinity and contains no involutions.

*This question: Number of involutions in a finite group seems like it could be relevant but estimating the number of conjugacy classes in this setting seems to have to be rather involved.

Edited to add :

*I am interested only in the exponential aspect of the bound, but one might also ask for optimal $c < 1 < C$ such that $$ c^n |G|^{\frac 1 2} \le I(G)+ 1 \le C^n |G|^{\frac 1 2}. $$ (see Yves' comments below).

*With this notation the third point above (which I edited for clarity) gives an upper bound $< 1$ for $c$; looking at subgroups of exponent 2 in $S_n$ (for example $(\mathbb Z/2\mathbb Z)^{n/2}$) also gives a lower bound $> 1$ for $C$. An upper bound for $C$ is given by Geoff Robinson's answer, now we only lack a lower bound $>0$ for $c$.

  • $\begingroup$ It depends what you call "sharp". Your constant $C$ would be an absolute constant (depending on nothing) and your "sharpness" observation is rather that necessarily $C\ge 3^{1/4}$. Maybe you want to mean that the minimum $\alpha_n$ of $(I(G)+1)/|G|^{1/2}$ over all subgroups $G$ of $S_n$ grows at least as fast as a negative exponential. $\endgroup$
    – YCor
    Commented Feb 8, 2018 at 18:01
  • $\begingroup$ Yes, I mean "sharp" in the exponential growth aspect, not the particular growth rate. $\endgroup$ Commented Feb 8, 2018 at 18:07
  • $\begingroup$ Also the upper bound is (by this argument) sharp in this sense only because you forced a symmetry using the same constant. Are you also interested in the upper bound, i.e., the growth of the maximum $\beta_n$ of $I(G)/|G|^{1/2}$ over all subgroups $G$ of $S_n$? or maybe it has nothing mysterious (achieved by $S_n$ itself??) $\endgroup$
    – YCor
    Commented Feb 8, 2018 at 18:21
  • $\begingroup$ For $G = S_n$ the ratio $I(G) / |G|^{1/2}$ is subexponential in $n$. But I think that with $G$ a maximal abelian 2-subgroup containing only involutions you gan get something like $I(G) / |G|^{1/2} > c^n$ for an absolute $c$ (it is possible to embed $(\mathbb Z/2\mathbb Z)^{n/2}$ int $S_n$). $\endgroup$ Commented Feb 8, 2018 at 18:36
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    $\begingroup$ @GerryMyerson: That's just an example of a relatively large group with only one involution, showing that the lower bound can't be improved. $\endgroup$ Commented Feb 9, 2018 at 0:50

1 Answer 1


I can answer in one direction. L.G. Kovacs and I proved ( around 1993) that any finite subgroup $G$ of $S_{n}$ has at most $5^{n-1}$ conjugacy classes. Later authors showed that it is possible to replace $5$ by a smaller constant, but the existence of such a constant seems to be enough for this question. It follows that any subgroup $G$ of $S_{n}$ has at most $\sqrt{5}^{n-1}|G|^{\frac{1}{2}} -1$ involutions, using properties of the Frobenius-Schur indicator, as in the question you refer to.

Later edit: It might be worth recasting the problem ( this is not so relevant for the direction already proved above): if the group $G$ above has no involutions, ( equivalently, has odd order), then $G$ is certainly solvable, so we have $|G| \leq 24^{\frac{n-1}{3}}$ by a result of J.D. Dixon ( again, the constant can be improved here, but its existence is enough for present purposes). This shows that $c = 24^{\frac{1-n}{6}}$ works above when $|G|$ is odd, and clearly $C = 1$ works here too ( for the other direction).

So we only now need to consider the case when $G$ has even order. It is well known that every $2$-subgroup of $S_{n}$ has order at most $2^{n-1}$, so that a Sylow $2$-subgroup of $G$ has order at most $2^{n-1}$ and, in particular, $G$ certainly has at most $2^{n-1}$ conjugacy classes of involutions.

If there is a positive constant $c$ as asked for in the later version of the question, then $G$ has an involution $t$ such that $[G:C_{G}(t)] > \left(\frac{c}{2} \right)^{n}|G|^{\frac{1}{2}}$ and hence $|C_{G}(t)| < \left(\frac{2}{c} \right)^{n}|G|^{\frac{1}{2}}.$

It follows that the question is equivalent to asking whether it is true that there are positive (finite) constants $d,D$ such that whenever $G$ is an even order subgroup of $S_{n},$ there are involutions $t,u \in G$ such that $|C_{G}(t)| \leq d^{n}|G|^{\frac{1}{2}}$ and $|C_{G}(u)| \geq D^{n}|G|^{\frac{1}{2}}.$

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    $\begingroup$ In fact, the number of conjugacy classes of a subgroup of $S_{n}$ is at most $2^{n-1}$ by a result of Liebeck and Pyber, and this can be improved to $\sqrt{3}^{n-1}$ when $n \geq 3$ by a result of A. Maroti. $\endgroup$ Commented Feb 9, 2018 at 11:15
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    $\begingroup$ Even better: for $n \ge 4$ Garonzi and Maróti prove the sharper upper bound $(5^{1/3})^{n-1}$: users.renyi.hu/~maroti/classes.pdf $\endgroup$ Commented Feb 11, 2018 at 8:56

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