# Subgroup lattice of the direct product of groups

Determine the Hasse diagram of $G=\mathbb{Z}_2 \times \mathbb{Z}_4.$

So I did this so far:

I know $G$ is not cyclic and $G=\lvert\mathbb{Z}_2 \times \mathbb{Z}_4\rvert=8$, therefore the subgroups of $G$ have order $1,2,4,8$ (because they divide $8$). In order $1$ we have $(0,0)$ and for order $8$ we have the group itself.

Finding other 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$

So the diagram should look like this

However I have the feeling it's not right.

Another question I have is: If I'm asked to draw a Hasse diagram of $\mathbb{Z}_m \times \mathbb{Z}_n$ and $m,n$ are prime beetween can I simply draw the Hasse diagram of $\mathbb{Z}_{m n}?$ Because if $m,n$ are prime beetween I know $\mathbb{Z}_m \times \mathbb{Z}_n$ is isomoprh to $\mathbb{Z}_{m n}$

• As the mother group is not cyclic, there MAY be non-cyclic subgroups. – Randall Oct 3 '17 at 16:21
• \langle (0,0) \rangle $\langle (0,0) \rangle$ – Kenny Lau Oct 3 '17 at 16:22

Elements of product $G \times H$ of two groups $G, H$ can be represented by pairs $(g, h)$ with component-wise multiplication. If we restrict to the case of commutative groups, then lattice of a product has pretty simple description. Of course, if $A \leq G, B \leq H$, then $A \times B \leq G \times H$. But there are also other subgroups; if there's a (abstract) surjective homomorphism $\pi: A \to B/B'$ for some $B' \leq B$, then there's a subgroup $A \pi B$ consisting of elements $\{(a, b),\pi a = b \, \mathrm{mod} \, B'\}$. Denote $A \pi B$ as $C$ for brevity. One may note that $A = CH \cap G$, $B = CG \cap H$ and $B' = C \cap H$. Groups of type $A \times B$ correspond to case $B' = B, \pi$ trivial. From above identifications we see that $A_1 \pi_1 B_1 \leq A_2 \pi_2 B_2$ iff $A_1 \leq A_2$ and $\pi_1 A \leq \pi_2 A$.
Proving that we obtain all groups in product this way is straightforward — assume we have $C \leq G \times H$. Let $A := CH \cap G$ and define multivalued function $\tilde \pi (g, h)$ as all $b \in H$ such that $(g, hb) \in C$; it's not a homomorphism yet, as it's even multivalued. It takes values in $CG \cap H$; all indeterminacy is eliminated if we mod out by $C \cap H$ and we have well-defined surjection.
• In fact, your new description also generalizes to non-abelian groups, with subgroups corresponding to $5$-tuples $(A,A',B,B',\varphi)$ consisting of subgroups $A'\unlhd A\leq G$, $B'\unlhd B\leq H$ and an isomorphism $\varphi: A/A'\to B/B'$. – Tobias Kildetoft Oct 3 '17 at 20:15