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Let $G$ be a nonabelian finite simple group, and let $p$ and $q$ be distinct prime divisors of the order of $G$. Is it true that the number of elements of $G$ of order $p$ never equals the number of elements of $G$ of order $q$?

Remark: My husband ran a GAP computation checking all nonabelian finite simple groups of order less than 400000000, and did not find a counterexample.

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  • $\begingroup$ This is so for any group of $A_n$. Indeed, the number of elements of order $p$ for odd prime $p$ is $({n \atop p})(p-1)!$. If $p$ and $q$ are odd prime numbers and $n\geq p>q$, then from equality $\left({n \atop p}\right)(p-1)!=\left({n \atop q}\right)(q-1)!$ follows that $\frac{q}{p}=(n-p)(n-p-1)\ldots(n-q+1)$. $\endgroup$
    – kabenyuk
    Jul 12 at 11:28
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    $\begingroup$ @kabenyuk The formula $\binom{n}{p}(p-1)!$ gives the number of $p$-cycles, but if $n\geq 2p$ then you also need to count products of two disjoint $p$-cycles, and so on. $\endgroup$ Jul 12 at 11:58
  • $\begingroup$ @Neil Strickland yes, I totally agree with you $\endgroup$
    – kabenyuk
    Jul 12 at 15:32
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    $\begingroup$ This is a duplicate: mathoverflow.net/questions/347291/… $\endgroup$
    – Nick Gill
    Jul 21 at 15:59
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Here is a partial (though not complete) answer which shows that the number of such elements is often different for pairs $\{p,q\}.$ As long as the prime $p$ divides $|G|$, the number of elements of order $p$ in $G$ is not divisible by $p$ (by virtue of Frobenius's theorem that the number of solutions of $x^{p} = 1$ in $G$ is a multiple of $p$, and the identity is present). Also, the number of elements of order $p$ in $G$ is a multiple of $p-1$, since distinct subgroups of order $p$ have only the identity in common, and each such subgroup contains $p-1$ elements of order $p$.

Hence if $p$ and $q$ are different primes which divide $|G|$ , and either $q|p-1$ or $p|q-1$, then the number of elements of order $p$ in $G$ is different from the number of elements of order $q$ in $G$ (consider the case $q|p-1$. Then the number of elements of order $p$ in $G$ is divisible by $p-1$, so by $q$, but the number of elements of order $q$ in $G$ is not divisible by $q$). Note that this eliminates the case $q = 2$ from all consideration in this problem.

Another (easy) reduction is when a Sylow $p$-subgroup $P$ of $G$ centralizes no element of order $q$ in $G$, but the prime $q$ divides $|G|$. Then the number of elements of order $q$ in $G$ is divisible by $p$, but the number of elements of order $p$ in $G$ is not. Hence we only need to consider cases where $q$ divides $|O_{p^{\prime}}(C_{G}(P))|$ and $p$ divides $|O_{q^{\prime}}(C_{G}(Q))|,$ where $Q$ is a Sylow $q$-subgroup of $G$.

Such considerations eliminate quite a number of pairs of primes for any given simple group $G$ (and similar arguments hold, to show for example that that if $P$ and $Q$ are both Abelian and commute, then $PQ$ can't be contained in a TI Hall subgroup).

But I imagine that a definitive answer to the question requires both use of the classification of finite simple groups, and very detailed examination of properties of simple groups of Lie type.

Later edit: If $G$ is a finite simple group of Lie type in characteristic $p$, then $C_{G}(U) = Z(U)$ for $U$ a Sylow $p$-subgroup of $G$, so that for any prime $r \neq p$, the number of elements of order $r$ in $G$ is divisible by $p$, whereas the number of elements of order $p$ in $G$ is not divisible by $p$ ( the latter as remarked above in general). Hence for such a group $G$ we only need to consider pairs of primes $r,s$ such that the elements of order $r$ and the elements of order $s$ are semisimple.

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