# Finite groups with no elements of order $p^2q$

Let $G$ be a finite group. What can be said if $G$ has the following Property P: $G$ has no element of order $p^2q$ for any two distinct primes $p,q$?

In particular, which finite simple groups satisfy Property P?

For instance, the alternating group $\text{Alt}_n$ has Property P if and only if $n\le 8$. The question for $\text{PSL}_2$ of finite fields seems not obvious.

As a side remark, a result of Malle, Moreto and Navarro states the following:

Suppose that $p$ and $q$ are distinct primes. If $G$ does not have any elements of order $pq$, then one of the following holds:

1. The Sylow $p$-subgroups or the Sylow $q$-subgroups of $G$ are abelian.
2. $G/O_{\{p,q\}′} (G)$ = $M$ and $\{p, q\} = \{5, 13\}$ or $\{7, 13\}$.

Here $M$ is the Monster sporadic group, and $O_{\{p,q\}′} (G)$ is the largest normal subgroup of $G$ that is divisible by neither $p$ nor $q$.

• What do you want to know? – user6976 Jun 29 '13 at 17:14
• @Mousavi: this is better but it is also asked to clarify other things. First you seem to mean "finite group" instead of "group". And also "What can be said" is too broad. – YCor Jun 30 '13 at 8:25
• To follow Yves's comment: make your own conjecture, and test it. – Yemon Choi Jun 30 '13 at 17:06
• "Most" finite simple groups will have an element of order 12, and it should not be too hard to use the classification to list exactly which simple groups do not. So the problem is approachable for simple groups. For general finite groups, I doubt whether you can say anything very interesting. – Derek Holt Jun 30 '13 at 17:06
• I had a look. If $p\ge 5$ is prime then both $PSL_3(F_p)$ and $PSL_2(F_{p^2})$ have an element of order 12. There are some other cases I skip. Finally for odd prime $PSL_2(F_p)$ has orders precisely the divisors of $p$, $(p-1)/2$ and $(p+1)/2$. So for infinitely many $p$ this has a element of order $4q$ for some prime $p$, but probably (conjecturally?) for infinitely many $p$ (with positive density), both $(p-1)/2$ and $(p+1)/2$ are square-free and thus $PSL_2(F_p)$ has no element of the required order. – YCor Jun 30 '13 at 22:15

## 1 Answer

It seems hard to give a complete answer to this, but at the very least one should be able to specify what the composition factors of $G$ might be. To do this it is enough to classify which simple groups $G$ have property $P$, in which case:

• if $G$ is alternating, then @Yves asserted that $n\leq 8$;
• if $G$ is sporadic, then we can consult the ATLAS and do each group one by one. For instance, of the Mathieu groups, $M_{11}, M_{12}, M_{22}$ and $M_{23}$ have property P, but $M_{24}$ does not;
• suppose next that $G$ is a quasisimple group of type $A_4$ over a field of characteristic $p$, i.e. is isomorphic to a quasisimple cover of $PSL_5(p^a)$, and suppose that $p^a>3$. Since $p^a>3$, there is a non-trivial element of a split torus of order $q$, that has a centralizer equal to a quasisimple cover of $PSL_4(p^a)$. Since the Sylow $p$-subgroup of $PSL_4(p^a)$ has exponent $>p$ and we obtain an element of order $p^2q$. We can consult the ATLAS to see that $PSL_5(2)$ contains an element of order 12. Thus any quasisimple group of type $A_4$ contains an element of order $p^2q$ for some primes $p$ and $q$, i.e. it does not satisfy property P.
• Now any simple group of Lie type which contains a subgroup of type $A_4$ will also fail to satisfy P. This includes, in particular, $E_6, {^2E_6}, E_7, E_8$ plus all of the classical groups of dimension at least 10.
• We've dealt with all groups of Lie type or large rank ($\geq 5$), and we should be able to deal with `of medium rank' groups (say rank 3 or 4) with ad hoc methods. For instance the ATLAS tells us that the Tits group ${^2F_4}(2)'$ contains an element of order $12$ and so we can immediately rule out $G={^2F_4}(q)$ and $G=F_4(q)$ and that deals with all of the exceptional groups of (twisted) rank at least $3$.
• We are left with the situation when $G$ is a low rank group of Lie type (say rank 1 or 2). In this case (as the comments indicate) the question is subtle and somewhat number-theoretic. For instance for $PSL_2(q)$ and for ${^2B_2}(q)$ one needs to determine when the order of a maximal torus of $G$ is divisible by $p^2q$ for some primes $p$ and $q$.