Formula for Product of Subgroups of $\mathbb Z$, Problem What is the product of $\mathbb{Z}_2$ and $\mathbb{Z}_5$ as subgroups of $\mathbb{Z}_6$?
Since $\mathbb{Z}_n$ is abelian, any subgroup should be normal. From my understanding of the subgroup product, this creates the following set: $\{ [0], [1], [2], [3], [4], [5] \}$ which has order 6 and is in fact $\mathbb{Z}_6$. However, the subgroup product formula
$$
|\mathbb{Z}_2\mathbb{Z}_5| = \frac{|\mathbb{Z}_2||\mathbb{Z}_5|}{|\mathbb{Z}_2 \cap \mathbb{Z}_5|} = \frac{2 \cdot 5}{2} = 5
$$
I feel like I'm doing something wrong in the subgroup product, in particular understanding what closure rules to follow when considering the individual product of elements.

@Ittay Weiss, made you a complete illustration,  but for noting a good point about the subgroups of $\mathbb Z$, we memorize:

If $m|n$ then $n\mathbb{Z}\leq m\mathbb{Z}$ (or $n\mathbb{Z}\lhd m\mathbb{Z}$).

