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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!

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  • $\begingroup$ Not sure if this will help, but apparently, the group is not simple. $\endgroup$ – Noble Mushtak Jan 24 '16 at 15:13
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    $\begingroup$ 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 $G$. $\endgroup$ – Watson Jan 24 '16 at 15:15
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    $\begingroup$ Then let $Q$ be a $q$-Sylow subgroup. Your group must be some semi-direct product of $Q$ and $P$. $\endgroup$ – David Jan 24 '16 at 15:47
  • $\begingroup$ 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? $\endgroup$ – math189925 Jan 24 '16 at 16:56
  • $\begingroup$ 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. $\endgroup$ – gonthalo Jan 24 '16 at 17:03

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