If $S_7$ is the symmetric group on 7 elements and $H \leq S_7$ is a subgroup with $|H| = 15$, then $H$ is a subgroup of order $pq$ where $p \nmid q-1$ for $p=3$ and $q=5$. This would imply that $H$ is cyclic.
I think the remaining portion of this proof relies on arriving at a contradiction when assuming that $S_7$ has a cyclic subgroup of order 15, but I'm struggling with the details. If $H$ is such a cyclic subgroup, then we can write $H = \langle \pi \rangle$ for some $\pi \in S_7$, but $\pi$ can be written as a product of disjoint cycles. So $\pi$ might look like
$$ \pi = (1,2)(3,4,5)(6,7), \quad \text{or}\quad \pi = (1,2,3,4)(5,6,7), $$
etc. For either of the two examples above, the order of the cyclic subgroup generated by $\pi$ should be the least common multiple of all of the cycle lengths, i.e., least common multiple of 2 and 3 which is 6, or the least common multiple of 4 and 3 which is 12.
How can I show that the least common multiple of the disjoint cycles for any possible $\pi$ cannot be equal to 15?
In $S_7$ there can't be elements of the cycle types which would lead to $15$ as order (namely $(a_1a_2a_3a_4a_5)(a_6a_7a_8)$ and $(a_1\dots a_{15})$), because $S_7$ comprises all and only the permutations on $7$ symbols. So, there are elements of order $15$ in $S_n$ only for $n\ge 8$.