When does $S_n$ have a subgroup with order $p^2$ where $p$ is prime?

Question

I'm attempting this homework problem, and I'm not sure where to start. Here is the problem and how what I've got so far.

Let $p$ be a prime number. What is the least positive integer $n$ such that $S_n$ (The symmetric group on $n$ objects) has a subgroup of order $p^2$.

I think whatever approach is taken, it will probably use Lagrange's theorem. $S_n$ is a finite group with order $n!$, so if it has a subgroup of order $p^2$ then $p^2 | n!$.

But the converse of Langrange's theorem is not true generally. We can't say that if $p^2|n!$ then $S_n$ has a subgroup of order $p^2$. Maybe if I had some insight into WHEN the converse of Lagrange's theorem is true I would know better what to do next.

Answer 1

If $p^2$ divides $n!$ then $n\geq 2p$, and then you can find two disjoint $p$ cycles in $S_n$: they generate a group of order $p^2$.

Answer 2

If $n=2p$, then $p^2|n!$ but $p^3\nmid n!$. Now use Sylow's First Theorem. As @Mariano points out, if $n<2p$, then $p^2\nmid n!$ so no such subgroup exists.