Let $G$ be a finite 2-generated group. Let $p$ be a prime dividing the order of $G$. Must there exist a generating pair $(g,h)$ of $G$ such that $|g|$ is divisible by $p$?


If not, is this true at least for finite simple groups?



For every prime divisor $p$ of a finite 2-generator group $G$, is there a generating pair containing an element of order divisible by $p$?
The answer is yes for finite simple groups $G$. Every element of $G$ is an element of a generating pair. This is proved in


Guralnick, Robert, Kantor, William, *[Probalistic generation of finite simple groups](http://dx.doi.org/10.1006/jabr.2000.8357)*, J. Algebra 234 (2000), p. 743–792. ([MR1800754](http://www.ams.org/mathscinet-getitem?mr=1800754)) 


*Abstract*: For each finite simple group $G$ there is a conjugacy class $C\_G$ such that each nontrivial element of $G$ generates $G$ together with any of more than $1/10$ of the members of $C\_G$. Precise asymptotic results are obtained for the probability implicit in this assertion. Similar results are obtained for almost simple groups.


But the answer in general is no. It is not true for $2$-generated Frobenius groups with a non-cyclic Frobenius complement, because the two generators have to generate the group modulo the Frobenius kernel, and so elements of the kernel do not arise in generating pairs.


For example, $\mathtt{PrimitiveGroup}(9,3)$ in the GAP/Magma databases has order $72$, with Frobenius kernel elementary abelian of order $9$ and complement $Q\_8$. All generating pairs consist of two elements of order $4$.


