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.

The only subgroups of $\Bbb Z_{6}$ are $\Bbb Z_{6}, \{0,2,4\}, \{0,3\}$, and $\{0\}$, corresponding to the divisors $1, 2, 3$, and $6$ of 6. There is no subgroup of order $5$. So there is no way that $\Bbb Z_{5}$ could be a subgroup of $\Bbb Z_{6}$. 
