# 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

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.

• In an ideal world, you'll argue away happily, using results about congruence mod $q$. But when $p = 2$ and $q = p + 1 = 3$, you'll get into trouble. You'll politely excuse yourself from this case, by saying "No $q = p + 1$ allowed, please." Mar 22, 2015 at 1:39

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.

• I am a little confused on where the q=2 and p=3 comes from. Can you explain that a little further for me please?
– cele
Mar 22, 2015 at 21:30
• Since $q < p$ and $p|(q+1)$, we know $p=q+1$. The only consecutive primes are $2$ and $3$. Mar 22, 2015 at 21:46
• would I be correct in saying that for the $n_p=q$ case, we would have $p=q-1$?? Thats what I am getting but I'm going to assume I am not correct then, since in order for that to be true, p=2,q=3, but p>q
– cele
Mar 22, 2015 at 21:59
• Yes, that is a contradiction, so $n_p = 1$ as you wanted to show. Mar 22, 2015 at 23:20

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 $4$.