Let G be a group of order $p^2q^2$ where $p > q$ are primes. If $|G| \ne 36$, show that G has exactly one Sylow p-subgroup

Question

Let G be a group of order $p^2q^2$ where $p > q$ are primes. If $|G| \ne 36$, show that G has exactly one Sylow p-subgroup.

Ok, I'm not exatly sure where to begin with this one. I know that we are using the Sylow theorems, but what confuses me is being given that $|G| \ne 36$. I am assuming that we will have to prove this for its different cases.

Answer 1

Let $n_p$ denote the number of Sylow p-subgroups of $G$. By the last Sylow theorem, $$ n_p | q^2, $$ so either $n_p = 1$, $n_p = q$, or $n_p = q^2$. If $n_p = q^2$, then by the first Sylow theorem, $$ q^2 \equiv 1 \pmod p \implies p | (q-1)(q+1). $$ Since $p > q$, this forces $p|(q+1)$, so $q=2$, $p=3$. But then $|G| = 36$, a contradiction.

I'll leave the $n_p = q$ case to you.

Answer 2

For completeness' sake, let us show that when $|G| = 36$, the proposition is NOT true. Take $G = S_3 \times \Bbb Z_6$. We have the subgroups:

$\{(e,0),((1\ 2),3),(e,3),((1\ 2),0)\}$, $\{(e,0),((1\ 3),3),(e,3),((1\ 3),0)\}$

and $\{(e,0),((2\ 3),3),(e,3),((2\ 3),0)\}$, which are all of order $4$.