The number, up to isomorphism, or abelian groups of order 40 is:

I got:

$2\times2 \times 10$



So the total number is $3$. However, the answer says $7$, where







I think the answer is dead wrong. Any ideas?

  • 1
    $\begingroup$ You are correct about the number. Several of those in the given answer are in fact isomorphic, as can be checked. $\endgroup$ – Tobias Kildetoft Sep 23 '14 at 20:24

We have $40=2^3\cdot 5$, so that the possible elementary divisors of the group are $\{2, 2, 2, 5\}, \{2, 4, 5\}$ and $\{8, 5\}$. So we have indeed three different abelian groups of order $40$. They are $C_2\times C_2\times C_2\times C_5$, $C_2\times C_4\times C_5$ and $C_8\times C_5$.


To add to Dietrich's answer, more generally if $n=p_1^{a_1}\cdots p_k^{a_k}$, then the number of abelian groups up to isomorphism is $\prod p(a_i)$ where $p$ is the partition function.

3=1+2=1+1+1, and 1 has one partition, so the answer is $3 \times 1=3$ total isomorphisms.

The bijection isn't too hard to establish, but it more an exercise in combinatorics.

  • $\begingroup$ I'm just wanting clarification. When we partition a number n, we are just breaking that number into the different ways it may be summed to? For 3, because it can be written 3, 1+2, and 1+1+1, it has 3 partitions, is that what you are suggesting? Why not 2+1, because it is the same as 1+2 by commutativity? $\endgroup$ – MrStormy83 Dec 9 '17 at 1:12
  • $\begingroup$ Yes, you might want to look at the wiki for partitions. $\endgroup$ – mdave16 Dec 9 '17 at 1:14
  • $\begingroup$ Thanks. I will do just that. $\endgroup$ – MrStormy83 Dec 9 '17 at 1:17

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