# 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:

1. $\def\zt{\Bbb Z_3}\Bbb Z_{81}$
2. $\Bbb Z_{27}\times\zt$
3. $\Bbb Z_{9}\times\Bbb Z_{9}$
4. $\Bbb Z_{9}\times\zt\times\zt$
5. $\zt\times\zt\times\zt\times\zt$

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.

• I get it now that there are 5 abelian groups of order $3^4$, 3 abelian groups for $2^3$, and 1 abelian group for 5. Thanks.
– 9959
Apr 25, 2013 at 4:36
• I found out that it is the Fundamental Theorem of Finite Abelian Groups.
– 9959
Apr 25, 2013 at 4:37
• +1 MJD for the tip about partitioning $4$ to find the number of ways to sum to $4$, to determine the number of decompositions corresponding to $3^4$ Apr 25, 2013 at 16:19

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 $3^4$.

• You included more details than I recommended, and there's no harm in that. However you didn't mention the theorem you used, which in my opinion is a significant omission. I wouldn't call it wrong, just skipping an important step. Apr 25, 2013 at 3:51
• The step that I would include that you did not is mentioning the Fundamental Theorem to justify your calculation. Apr 25, 2013 at 4:15
• You omitted $(2^2\cdot 2)(3\cdot3\cdot3\cdot3)(5)$ from your list, but as I noted in my answer elsewhere on this page, there are actually 15 groups, not 12.
– MJD
Apr 25, 2013 at 4:19
• (The preceding comment was addressed to 9959, who had asked which of the 12 was missing from her list in the original question.)
– MJD
Apr 25, 2013 at 4:27
• This site could use threading for comments, to avoid these sorts of confusions when comments get deleted. Apr 25, 2013 at 4:42