# Classify all groups of order $p^2q^2$ up to isomorphism

Let $p,q \in \mathbb{N}$ be prime numbers with the properties

$2 < p < q$ and $q - 1 , q + 1 \notin \left\langle p \right\rangle$

Classify all groups of the order $p^2q^2$ up to isomorphism.

This was a question given by my algebra professor and quite frankly I am stumped. My initial thought was that this question is referring to p-Sylow subgroups and one would need to apply the Sylow-theorems. If this is true, how would you apply them? Then what does "up to isomorphism" exactly mean?

I also thought to try and break it down and look at different possible cases. For example something like this:

Since $$q - 1 \notin \left\langle p \right\rangle \Rightarrow p \nmid q - 1$$ $\Rightarrow \exists!$ subgroup of order $p$ $\Rightarrow \exists p - 1$elements of order $p$ and $q-1$ elements of order $q$.

But honestly I am not sure how to answer this question. I would really appreciate if someone could try and explain this to me. Thank you in advance!

• Not sure if this will help, but apparently, the group is not simple. – Noble Mushtak Jan 24 '16 at 15:13
• You can use the following fact : the number $n_p$ of $p$-Sylow subgroups divides $q^2$ and is $\equiv 1$ mod $p$. Under the given hypothesis, you can conclude that $n_p = 1$. Therefore the unique $p$-Sylow $P$ is a normal subgroup of your group $G$. – Watson Jan 24 '16 at 15:15
• Then let $Q$ be a $q$-Sylow subgroup. Your group must be some semi-direct product of $Q$ and $P$. – David Jan 24 '16 at 15:47
• Thank you for your responses. But what does it mean to classify all groups of this order up to isomorphism? How can one show that? – math189925 Jan 24 '16 at 16:56
• For example, the groups S3 and D3 are isomorphic, so if you were asked to classify groups of order 6 up to isomorphism, you should count both of them as just one group. – gonthalo Jan 24 '16 at 17:03