What are the three subgroups of $\mathbb{Z}_4\times\mathbb{Z}_6$ of 12 elements? My formatting didn't work in the title, here is the question again:
What are the three subgroups of $\mathbb{Z}_4\times\mathbb{Z}_6$ of 12 elements?
I know that this group does not have order 24 since $\gcd(4, 6) \ne 1$, but I am at a loss as to where to start.
 A: The first two 12-element subgroups are easy to find, by crossing the whole of one of the components with a subgroup of the other that has half the elements.  Thus we have
$$ G_1 = \{0, 1, 2, 3\} \mod 4 \otimes \{0, 2, 4 \} \mod 6 \\
G_2 = \{0, 2\} \mod 4 \otimes \{0, 1, 2, 3, 4, 5 \} \mod 6 
$$
The third is more subtle:
$$ G_3 = \{k \mod 4\} \otimes \{m \mod 6\} | (k+m) \text{ even}
= \\ \{ 0 \otimes 0, 0 \otimes 2, 0 \otimes 4,
1 \otimes 1, 1 \otimes 3, 1 \otimes 5, 
2 \otimes 0, 2 \otimes 2, 2 \otimes 4,
3 \otimes 1, 3 \otimes 3, 3 \otimes 5 \}
$$
Having guessed the form, it is easy to see that $G_3$ is closed under the group multiplication operation, and since it is finite, it thus must form a subgroup of $G$.
As to what would motivate one to guess that there are three and only three subgroups of order 12, that is a tougher issue.
A: Think of homomorphisms $\mathbb Z_4 \times \mathbb Z_6 \to \mathbb Z_2$. If you choose them surjective, the kernel will have order $12$. Conversely, a subgroup of index $2$ is always normal, so corresponds to such an homomorphism. 
A morphism $G \times H \to K$ corresponds to a pair of homomorphisms $G \to K$ and $H \to K$. In your case, the groups are all cyclic, so you only have to check where to map the generator. 
Hope that helps,
