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For any integer $n$, there is a cyclic group $\Bbb Z(n)$. So for any integer $n$ there is always an abelian group $G$. So why we say the group of order $p^3$ is not always abelian?

K.defaoite
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avijit kundu
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    Well, for a concrete example the Quaternion group of order $8 = 2^3$ is not abelian even though there is a cyclic group of order $8$. Not every group of a given order falls into the same isomorphism class. – Kaj Hansen Jun 06 '15 at 03:33

2 Answers2

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Yes, we can speak about "the" cyclic group of order $p^3$ (up to isomorphism). However, this does not mean that there aren't other, non-cyclic groups of that same order.

Since there's more than one isomorphism class of groups of order $p^3$, there is no single "the" group of order $p^3$ to talk about.

Zev Chonoles
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  • So we see for any integer there always an abelian group. – avijit kundu Jun 06 '15 at 03:48
  • From this we first decompose the order of the group and then use The Fundamental th of finite abelian groups we can find how many non-isomorphic groups (upto isomorphism) for the any integer value n. – avijit kundu Jun 06 '15 at 03:52
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    @avijitkundu: at least one. For four elements there is $\Bbb Z_4$ and the Klein group. They are distinct and both are abelian. For primes $p$ there is only one group, $\Bbb Z_p$, which is abelian. The direct product of abelian groups is abelian, so for composites there are more. There may also be non-abelian groups of the same order. – Ross Millikan Jun 06 '15 at 03:53
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    @avijitkundu: Careful, using FToFAG you only get the number of non-isomoprphic abelian groups of order $n$. But also, this seems unrelated to your original question. – Eric Stucky Jun 06 '15 at 20:54
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EDIT

You can't speak about the group of order $n$, because given $n\in\Bbb N$ there can be more than one group of that order.

If $n=p$ is prime, then (up to isomorphism) there exists exactly one group: $C_p$ (or $\Bbb Z_p$ if you prefer), which is the cyclic group of order $p$ (it's easy to prove).

In other cases it's often difficult to know how many non isomorphic groups of a given order $n$ there exist.

However the case $n=p^3$ is completely solved: there is a classification of finite groups of order $p^3$ (not only of order $8$).

Three of them are abelian: $C_{p^3}$, $C_{p^2}\times C_p$ and $C_p\times C_p\times C_p$.

The remaining two are not abelian (they are expressed as semidirect product).

If $n=8=2^3$ the non abelian groups are the quaternion group $Q_8$ and the dihedral group of the square $D_4$.

Joe
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