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:









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 $\mathbb{Z}_{m n}$

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

(Sorry for writing something utterly meaningless sometime earlier.)

  • $\begingroup$ No, a subgroup need not correspond to a pair of subgroups. Indeed, some of the ones the OP already found are not of that form. $\endgroup$ – Tobias Kildetoft Oct 3 '17 at 19:24
  • $\begingroup$ Well, it's good practice to wake up fully before writing an answer. I think that now it makes sense. $\endgroup$ – xsnl Oct 3 '17 at 20:05
  • $\begingroup$ 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 $\varphi: A/A'\to B/B'$. $\endgroup$ – Tobias Kildetoft Oct 3 '17 at 20:15
  • $\begingroup$ Of course, but in a sense this becomes significantly more complicated — we use not only subgroup lattice, but also normal sublattices for every subgroup and face all problems of determining homs between two prescribed groups. (I guess it'd be better to write "lattice of subgroups has a description which is effectively computable for f. g. abelian groups") $\endgroup$ – xsnl Oct 3 '17 at 20:36

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