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.
 A: You are possibly confused here since you write the elements of $\mathbb Z_n$ as $[0],[1],[2],\cdots $. This leads you to think that $\mathbb Z_5=\{[0],[1],[2],[3],[4]\}\subseteq \{[0],[1],[2],[3],[4],[5]\}= \mathbb Z_6$.This however is wrong. To understand why, recall that when you write $[1]\in \mathbb Z_5$ you mean that $[1]$ is the equivalence class of $1$ for the equivalence relation of $1$ modulo $5$. Similarly, when you write $[1]\in \mathbb Z_6$ you refer to the equivalence class of $1$ modulo $6$. These are very different sets and thus the element $[1]\in \mathbb Z_5$ is not in $\mathbb Z_6$. The same goes for all the other elements, so in fact $\mathbb Z_5\cap \mathbb Z_6=\emptyset$. 
A: @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 $n\mathbb{Z}\lhd m\mathbb{Z}$).

A: 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 $\Bbb Z_{6}$. 
