# Does every transitive permutation group contain a permutation whose cycle lengths have a common divisor?

Let $$H$$ be a transitive subgroup of $$\mathfrak{S}_n$$, $$n \geq 2$$. Using Jordan's lemma ($$H$$ is not a union of conjugate proper subgroups), we see that $$H$$ contains a permutation without fixed points. I'm interested in whether $$H$$ necessarily contains a permutation whose cycle lengths have a common divisor. It's tricky to either prove or disprove this: on the one hand, my attempts at proving this have all been vain (some cases are easy: if $$H$$ embeds via its action on itself, if $$H$$ is a $$p$$-group ; if $$n$$ is a prime power we can reduce to the $$p$$-group case by noticing that any $$p$$-Sylow of $$H$$ acts itself transitively); on the other hand, a colleague of mine has checked that this is true for $$2 \leq n \leq 27$$ using Julia (assuming correctness of his code). A key observation based on his experimentations is that you cannot choose which of the prime divisors of $$n$$ divides the cycle lengths of an element (there is a transitive subgroup of $$\mathfrak{S}_6$$ with no elements having only even-sized cycles, and a transitive subgroup of $$\mathfrak{S}_{12}$$ with no elements having only cycles of length a multiple of three, so neither the smallest nor the largest prime dividing $$n$$ are always candidates for example).

• Note: this apparently stronger statement is actually equivalent to the required statement. Indeed, let $f$ be a permutation in the group, all of whose cycles have size divisible by some prime $p$, and among them, chosen with a maximal number of cycles. If by contradiction $q$ is another prime dividing some cycle length, then $f^q$ has the same property and has more cycles, a contradiction. So $f$ has $p$-power order.