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I was trying to solve the following problem:

Find number of abelian groups of order $27$ ?

Could someone point me in the right direction?

Thanks in advance for your time.

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Have you learned the Fundamental Theorem of Finitely Generated Abelian Groups? That's all you need. – Graphth Dec 12 '12 at 15:26
up vote 12 down vote accepted

$27 = 3^3\;$ and $\;3\;$ is prime: So....

We know from the Fundamental Theorem of Finitely Generated Abelian Groups, all abelian groups of order $27$ are equivalent to (isomorphic to) one of the following groups:


$\mathbb{Z}_{3}\times \mathbb{Z}_9$

$\mathbb{Z}_{3}\times \mathbb{Z}_3 \times \mathbb{Z}_3$.

These groups are NOT isomorphic. Why not?

There are no other distinct (up to isomorphism) abelian groups of order $27$. Why not?

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Nice questions left for the OP. – Babak S. Dec 12 '12 at 15:47

$27=3^3$ and we have:

$27=1\times 27$ $(1\mid 27)$, $27=3\times 9$ $(3\mid9)$, or $27=3\times 3\times3$ $(3\mid3\mid3)$ so according to abelian fundemental theorem we have thw following groups of order $27$ respectively:

$$\Bbb{Z}_{27},\;\; \mathbb Z_3\times\mathbb Z_9,\;\; \mathbb Z_{3}\times\mathbb Z_3\times\mathbb Z_3$$

Other form of direct product are isomorphic to each others. For example, $$\mathbb Z_3\times\mathbb Z_9\cong\mathbb Z_9\times\mathbb Z_3$$ Note that $A\times B$ in which $A$ and $B$ are groups is abelian iff both $A$ and $B$ are abelian.

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Yes, I love those abelian groups! – amWhy Apr 11 '13 at 1:48

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