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:
$2^3 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot  \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^3 \cdot 3 \cdot 5$

You have the right idea, but there are 5 abelian groups of order  $3^4$, not 4.  You can have:


*

*$\def\zt{\Bbb Z_3}\Bbb Z_{81}$

*$ \Bbb Z_{27}\times\zt$

*$\Bbb Z_{9}\times\Bbb Z_{9}$

*$\Bbb Z_{9}\times\zt\times\zt$

*$\zt\times\zt\times\zt\times\zt$


These correspond to the 5 ways (not 4) that you can express 4 as a sum of positive integers: $4, 3+1, 2+2, 2+1+1, $ and $1+1+1+1$, respectively. (Similarly, there are not 5 but 7 abelian groups of order $3^5$.)
I would observe this, list the groups of order $2^3$ and $3^4$, and then say that there were $3\cdot5\cdot1 = 15$ abelian groups of order $2^3\cdot3^4\cdot5$, without listing all 15, but on the other hand listing them couldn't hurt. If you do list them, do it methodically, not all mixed up as you did above, so  that you (and the grader) can be certain you didn't omit any or list any twice.
I agree with vadim123 that you should cite the theorem by name, particularly since the whole point of this exercise is to show that you know the fundamental classification theorem of finite abelian groups. Find out what it is called in your text, and call it that.
