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I am looking for a textbook or a paper which include the classification of groups of order $p^4$ ($p$ is prime) using generators and relations. In particular I like to understand which group $G$ "exist" for the odd $p$'s and do not exist for $p=2$.

Olexandr Konovalov
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Ofir Schnabel
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$(\mathbb{Z_p}\times \mathbb{Z_p} \times \mathbb{Z_p})\rtimes \mathbb{Z_p}$ with presentation $$ <a,b,c,d : a^p=b^p=c^p=d^p=e : d^{-1}cd=cb, d^{-1}bd=ba,d^{-1}ad=a,c^{-1}bc=b,c^{-1}ac=a,b^{-1}ab=a> $$is that group which you are finding . It exists for all primes $p>2$, but not for $p=2$. In the latter case, this presentation defines the dihedral group of order $2^3$.

Derek Holt
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HIMANSHU
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    To define a semidirect product you need to specify an action, which you have not done. So you have not defined a group. – Derek Holt Jun 25 '20 at 10:41
  • @Derek I am sorry, what I mean is that you can always define action as shown in Burnside classification of groups of order $p^4$ (xv) which represents the group I have mentioned for $p>2$. But not for $p=2$. – HIMANSHU Jun 25 '20 at 10:45
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    But that's really not useful unless you tell us what the action is. Until you do that, you have not mentioned a group at all. – Derek Holt Jun 25 '20 at 11:00
  • @Derek Is it fine now ? – HIMANSHU Jun 25 '20 at 11:12
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    Not exactly. When $p=2$, that presentation does not define a group of order $p^4$. The group it defines is the dihedral group of order 8 (not elementary abelian of order 16). – Derek Holt Jun 25 '20 at 11:30
  • @DerekHolt I am sorry. It was my mistake.Is it okay now ? or am I missing something ? – HIMANSHU Jun 25 '20 at 12:56
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    I made a small further edit, but it is correct now. – Derek Holt Jun 25 '20 at 13:20