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

Question

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

I got:

$2\times2 \times 10$

$2\times20$

$40$

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

$40$

$10\times4$

$8\times5$

$20\times2$

$10\times2\times2$

$5\times4\times2$

I think the answer is dead wrong. Any ideas?

Answer 1

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 $C_8\times C_5$.

Answer 2

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.