Number of non-isomorphic group of order 8 ?
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I have no clue how to solve this. It's a group of order $p^3$ so is there any general way to calculate the number of non-isomorphic groups of such orders?
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If you can do the abelian case (by the fundamental theorem), then look at this: http://math.stackexchange.com/questions/64406/how-can-you-show-there-are-only-2-nonabelian-groups-of-order-8 – Prahlad Vaidyanathan Nov 02 '15 at 15:34
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For abelian groups use Structure theorem for finitely generated abelian groups groups.
For non abelian Groups:
Hint: As $G$ is a $p$-group so $Z(G)$ can not be trivial and $\frac {G}{Z(G)}$ can not be cyclic because $G$ is nonabelian.Therefore order$(Z(G))=2$ and hence $ \frac {G}{Z(G)} \cong \frac { \mathbb Z}{2\mathbb Z } \times \frac { \mathbb Z}{2\mathbb Z }$.Now proceed...
Conclusion: There are $5$ groups of order $8$ in which $3$ are abelian and $2$ are nonabelian.
Arpit Kansal
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so the answer to the question is $5$ right? As the groups are $\mathbb{Z_8}$, $\mathbb{Z_4\times Z_2}$, $\mathbb{Z_2\times Z_2\times Z_2}$ and $D_4$, quaternions. – Harry Potter Aug 26 '16 at 02:13
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