What are the Subgroups of $\Bbb{Z}_2 \times \Bbb{Z}_4$? I know that if $Z_n$ is cyclic, then every subgroup of $Z_n$ is also cyclic, and every such subgroup is generated by a single element. 
However, $\Bbb{Z}_2 \times \Bbb{Z}_4$ is clearly not cyclic, because $2$ and $4$ are not co-prime. 
My main concern is that I can list out the elements of $\Bbb{Z}_2 \times \Bbb{Z}_4$, and I can find the subgroups each of these elements generate, but there may be more groups than that. I can't just look at all the subgroups given by the elements, i.e. $<(0,0)>,<(0,1)>,\ldots,<(1,3)>$ but I'd also have to consider groups like $<(0,1)+(1,3)>$. Is that right? 
Can anyone help me out and tell me what the subgroups are, or how to generate them? I need this as a stepping stone in the context of a larger problem. I've looked at this related question but that didn't bring me much further.
 A: By Lagrange theorem, the possible order of the  subgroups are 1, 2, 4,8. Clearly you only need consider the case that 2, 4. 
Case 1. $|H|=2$. Such group is generated by an element of order 2. So you just list all elements of order 2, and you can get them all.
Case 2. $|H|=4$. There are two subcases: $H$ is cyclic or not. If $H$ is cyclic, you need to list all elements of order 4: (0,1), (0,3), (1,1), (1, 3). That gives us two different cyclic subgroups of order 4. Now we need to find the non-cyclic subgroups of order 2, which is generated by two different elements of order 2. As you have find all elements of order 2, (1,0), (1,2), (0,2). Note $(1,0)+(1,2)=(0,2)$. So we get only one such subgroup:$\{ (0,0), (1,0), (1,2), (0,2)\}$.
A: $\mathbb{Z_{2}\times \mathbb{Z}_{4}}$ has two generators $(1,0)$ and $(0,1)$, so are its subgroups. Every subgroup is generated by $(g_{1},0)$ and $(0, g_{2})$ for some $g_{1}$ in $\mathbb{Z}_{2}$ and $g_{2}$ in $\mathbb{Z}_{4}$. Try to find all such pair of generators. Of course you need to check if two pairs generates the same subgroup.
