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:
  
  
*
  
*The Sylow $p$-subgroups or the Sylow $q$-subgroups of $G$ are abelian.
  
*$G/O_{\{p,q\}′} (G)$ = $M$ 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 $q$.
 A: 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 $q$. 

