subgroups of $S_{n}$ which every element is a cycle

Question

We know that the elements of $S_{3}$ are cycles. Since we can put $S_{3}$ into $S_{n}$ for $n>3$ I have the following question: is there subgroups $H$ ($H\neq S_{3}$) of $S_{n}$ which all elements are just cycles?

Answer 1

Any subgroup generated by a $p$-cycle works ($p$ is prime).

In fact, this is essentially the only possible example because if $\sigma$ is an $m$-cycle, then $\sigma^k$ is the product of $d$ disjoint $\frac md$-cycles, where $d=\gcd(m,k)$.

It also indicates why $S_3$ is the only other example: the possible orders of cycles are primes (except for the identity).

Answer 2

It's not too hard to show that such a (finite) group must be either cyclic of prime order, or isomorphic to $S_3$. Here's a sketch of a proof.

First, the order of every element must be prime. (Some power of an element of composite order would have multiple cycles.) Moreover, we cannot have a subgroup of the form $C_p\times C_p$ as such a group must contain an element with multiple cycles. (This is an easy exercise.) It follows that the group has squarefree order. The structure of these groups is well known.

Suppose the group does not have prime order and consider $H$ a Hall $pq$-subgroup. This must be non-abelian, with say $p$ dividing $q-1$. Now the $q$ element consists of a single $q$-cycle, and the $p$-element consists of at least $q-1/p$ cycles of length $p$. So we must have $q-1=p$ and so $p=2$ and $q=3$.