I am trying to solve the problem of the title. I'm not sure if my solution is correct, but maybe there is a more elegant way to solve it? Thanks for any input.
My attempt: Since $6^4$, we know that $6^4 = 3^42^4$ and so by the fundamental theorem of abelian groups, i would just need to count all the possible invariant factors. But, this is just the same as counting the possible ways of writing $4$ as a sum of decreasing numbers, so for instance $4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 +1 + 1 +1$ and so we see that we have 5 such ways to write 4. Since the group would be a direct product of invariant factors of $Z/3^4Z$ and the invariant factors of $Z/2^4Z$, we have that there are 25 such isomorphism classes. Is this a good approach? Anybody knows a better approach? Thanks!
You are correct. I used to write it like this: If an Abelian group has order $p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_t^{\alpha_t}$ where each $p_i$ is prime and $P(n)$ denotes the partition function, or the number of ways to write $n$ as a sum of positive integers, then the number of isomorphism classes is the product $P(\alpha_1)\cdot P(\alpha_2)\cdots P(\alpha_t)$