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Okay so say you have a group of order p ( p a prime ) I know that this means that this group is isomorphic to $C_p$ ( the cyclic group of order p) and hence abelian. But what about groups of order $p^2$ are they isomorphic to $C_p\times C_p$ and also abelian because of this ?

Nicky Hekster
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excalibirr
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    Yes, and this is a classic exercise in group theory. :) The usual way is to first show that the center of a group of order $p^2$ is nontrivial using the orbit-stabilizer theorem, and then show that if a group modulo it's center is cyclic, then the group must have been abelian. – Mike Pierce Nov 08 '18 at 18:32
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    You can also have the cyclic group of order $p^2$. There is a theorem that the centre of a group of order $p^n$ is non-trivial for any prime $p$ (it is not hard, and one proof is based on the fact that the sizes of the conjugacy classes have to be factors of the order of the group). This can be used to prove that a group of order $p^2$ is abelian. You might like to see if you can use the fact about the centre to prove this (what happens if the centre is not the whole group) and if a little more ambitious try proving that the centre is not trivial without looking it up. – Mark Bennet Nov 08 '18 at 18:33
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    Every group of order $p^2$ is abelian and thus it is either $C_{p^2}$ or $C_p\times C_p$ up to isomorphism. – freakish Nov 08 '18 at 18:33

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