# Finding all subgroups of Direct product of groups [duplicate]

I'm having a hard time finding when I left a subgroup out. For example, lets take $G=\mathbb{Z}_2 \times \mathbb{Z}_4.$

Subgroups:

$s_1:$$\langle(0,0)\rangle s_2:$$\langle(0,1)\rangle=\{(0,0),(0,1),(0,2),(0,3)\}$

$s_3:$$\langle(0,2)\rangle=\{(0,0),(0,2)\} s_4:$$\langle(0,3)\rangle=\{(0,0),(0,3),(0,2),(0,1)\}=s_2$

$s_5:$$\langle(1,0)\rangle=\{(0,0),(1,0)\} s_6:$$\langle(1,1)\rangle=\{(0,0),(1,1),(0,2),(1,3)\}$

$s_7:$$\langle(1,2)\rangle=\{(0,0),(1,2)\} s_8:$$\langle(1,3)\rangle=\{(0,0),(1,3),(0,2),(1,1)\}=s_6$

By looking at this I'd say all the subgroups are here, but for this $G$ there is also the subgroup $\mathbb{Z}_2 \times 2\mathbb{Z}_4$. How should I know that?

If I take some other examples like $\mathbb{Z}_3 \times \mathbb{Z}_4$ or $\mathbb{Z}_3 \times \mathbb{Z}_3$, the type of "list" I made is enough to determinate all the subgroups.

## marked as duplicate by Stefan4024, Jim, José Carlos Santos, Antonios-Alexandros Robotis, LeucippusOct 8 '17 at 0:52

$\mathbb{Z}_2 \times \mathbb{Z}_4$ isn't a cyclic group, so there's no reason to expect that all subgroups will be cyclic.
The reason why this method works for $\mathbb{Z}_3 \times \mathbb{Z}_4$ is that it's cyclic, as $\mathbb{Z}_3 \times \mathbb{Z}_4 \cong \mathbb{Z}_{12}$. Why it works for $\mathbb{Z}_3 \times \mathbb{Z}_3$ is because $\mathbb{Z}_3$ is a cyclic simple group.
Goursat's lemma gives a complete classification of the subgroups of $G_1 \times G_2$, where $G_1$ and $G_2$ are arbitrary groups. One statement (a corollary of what's stated in the link, with different notation) is that every subgroup has the form $\{(g_1, g_2): g_1 \in H_1, g_2 \in H_2, \varphi_1(g_1) = \varphi_2(g_2)\}$, where $H_i \leq G_i$ are subgroups and $\varphi_i: H_i \to K$ are surjective homomorphisms to a common quotient group $K$.
For example, your missing subgroup $\mathbb Z_2 \times 2\mathbb Z_4$ is obtained with $H_1 = \mathbb Z_2$, $H_2 = 2\mathbb Z_4$, $K =$ the trivial group, and $\varphi_i: H_i \to K$ the trivial homomorphisms.