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.
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.
$27 = 3^3\;$ 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:
$\mathbb{Z}_{27}$
$\mathbb{Z}_{3}\times \mathbb{Z}_9$
$\mathbb{Z}_{3}\times \mathbb{Z}_3 \times \mathbb{Z}_3$.
These groups are NOT isomorphic. Why not?
There are no other distinct (up to isomorphism) abelian groups of order $27$. Why not?
$27=3^3$ 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.