How do I show that $\mathbb{Z}_4\times\mathbb{Z}_6$ has only 3 subgroups of size 12? I am aware of what the 3 subgroups are, and now I am trying to show that these are the only 3 subgroups of size 12.
I am trying to work with their indexes being 2, implying that they are normal, but I don't know where to go from here.
 A: You want to know how many subgroups $\mathbb Z_4 \times \mathbb Z_2 \times \mathbb Z_3$ has of size $12$.
Our subgroup must clearly contain an element of the form $(a_1,a_2,1)$ or $(a_1,a_2,2)$. No matter which case it is adding it four times yields $(0,0,1)$ or $(0,0,2)$, and $1$ and $2$ are generators for $\mathbb Z_3$.
This tells us that if $H$ is a subgroup containing $(a_1,a_2,b)$ then $H$ must contain $(a_1,a_2,0)$ $(a_1,a_2,1)$ $(a_1,a_2,2)$. So a subgroup of size $12$ of $\mathbb Z_4 \times \mathbb Z_2 \times \mathbb Z_3$ is of the form $K\times \mathbb Z_3$ where $K$ is a subgroup of $\mathbb Z_4\times \mathbb Z_2$ of size $4$
So we have to find the subgroups of size $4$ of $\mathbb Z_4 \times \mathbb Z_2$
We can proceed by doing some casework:
If $H$ contains $(1,0)$ or $(3,0)$ it must be $\mathbb Z_4\times \{0\}$
If $H$ contains $(1,1)$ or $(3,1)$ it is the subgroup $\{(0,0)(1,1)(2,0)(3,1)\}$
So the only other case remaining is when $H$ contains only elements of the form $(0,a)$ or $(2,b)$. There are four elements of this kind and we must therefore only check if $\{(0,0),(0,1),(2,0),(2,1)\}$ is a group, which indeed turns out to be true.
So these are the only subgroups of $\mathbb Z_4\times \mathbb Z_2$ and by our earlier claim the subgroups of the group $\mathbb Z_4\times \mathbb Z_2 \times \mathbb Z_3$ are obtained by taking the cross product of each of these three subgroups with $\mathbb Z_3$
A: Let $G\leq\mathbb{Z}_4\times\mathbb{Z}_6$ be a subgroup of order 12.  Then $G$ determines a surjective homomorphism.
$$
\mathbb{Z}_4\times\mathbb{Z}_6\to(\mathbb{Z}_4\times\mathbb{Z}_6)/G\cong\mathbb{Z}_2.
$$
That is, $G$ is the kernel of a homomorphism from $\mathbb{Z}_4\times\mathbb{Z}_6$ ${onto}$ $\mathbb{Z}_2$.  Since there are only 3 distinct homomorphisms of this type (determined by the images of $(0,1)$ and $(1,0)$), we have that there are at most 3 subgroups of order 12.
