TITLE: How many subgroups does $\mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 $ have?
QUESTION [15 upvotes]: How many subgroups does $H = \mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 $ have?

$\mathbb{Z}_6$ has 4 subgroups (including itself), so the answer is at least $4^4$. But, not all subgroups of $H$ are products of subgroups of $\mathbb{Z}_6$, for example the group generated by $(1,1,1,1)$ is not. Any ideas how to count this type of groups in a simple way?

REPLY [10 votes]: Let $G=G_1\times G_2$ be a direct product of finite groups, $|G_1|=n_1$, $|G_2|=n_2$ and $(n_1,n_2)=1$. Then for every subgroup $H$ of $G$ there exists $H_i$ subgroup of $G_i$ such that $H=H_1\times H_2$. 

Since $H\le G$ we have $|H|=k_1k_2$ with $k_i\mid n_i$. Let $H_i=p_i(H)$, where $p_i:G\to G_i$ is the canonical projection. Obviously $H\subseteq H_1\times H_2$. But $|H_i|\mid n_i$ and $|H_i|\mid |H|=k_1k_2$. It follows that $|H_i|\mid k_i$, so $|H_1\times H_2|\le k_1k_2$. On the other side, $|H_1\times H_2|\ge |H|=k_1k_2$, therefore they are equal.
In your case $G_1=\mathbb Z_2^4$ and $G_2=\mathbb Z_3^4$. In general, the number of subgroups of $\mathbb Z_p^n$ ($p$ prime) equals the number of $\mathbb Z_p$-vector subspaces of $\mathbb Z_p^n$ and this is well known.