<s>[INST] Show abelian groups of order 3240? Show how to get all abelian groups of order $2^3 \cdot 3^4 \cdot 5$.
I just started learning this and was wondering how you would do this?
Is this correct?
$2^3 \cdot 3^4 \cdot 5 = 3240$. Therefore the number of abelian groups of order $3240$ is $3 \cdot 4 = 12$.
Is this the entire proof or do we need to do the table showing divisors like this?
Divisors:
$2^3 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot  \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^3 \cdot 3 \cdot 5$
 [/INST] You are implicitly using the Fundamental Theorem of Finite Abelian Groups.  You should cite this by name.
Of order $8=2^3$, you can have $\mathbb{Z}_8$ or $\mathbb{Z}_2\times \mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$.  Similar options exist for the $3^4$.  
To fully explain, I would write out each one of the above (3 versions for $2^3$, 5 versions for $3^4$), to justify the $3\times 5$ calculation.
Edit: Corrected number of subgroups of order $3^4$.
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