# Finite simple groups with the same numbers of elements of orders p and q

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.

• 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 $\frac{q}{p}=(n-p)(n-p-1)\ldots(n-q+1)$. Jul 12 at 11:28
• @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. Jul 12 at 11:58
• @Neil Strickland yes, I totally agree with you Jul 12 at 15:32
• This is a duplicate: mathoverflow.net/questions/347291/… Jul 21 at 15:59

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).
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.