Finite Abelian Groups with Toroidal Subgroup Lattices

The graph $G_{n}$ has a clear $K_{3,n}$ minor and $\gamma(K_{3,n})=\lceil\frac{n-2}{4}\rceil$. Hence, $\gamma(G_{n})\geq\lceil\frac{n-2}{4}\rceil$. To demonstrate that $G_{n}$ can be drawn on a surface of genus $\lceil\frac{n-2}{4}\rceil$, we will construct a set of cycles that traverse each edge in both directions (e.g. define the faces of the embedding of the graph on a surface). We will first consider the case when $n\equiv 2\mod 4$. The cycles are as follows:

• •

  $(b,\alpha_{1},\beta_{1})$, $(b,\beta_{2},\alpha_{2})$, $(b,\alpha_{3},\beta_{3})$, $(b,\beta_{4},\alpha_{4})$, $\dots$, $(b,\beta_{n},\alpha_{n})$

• •

  $(b,\beta_{1},c,\beta_{2})$, $(b,\beta_{3},c,\beta_{4})$, $\dots$, $(b,\beta_{n-1},c,\beta_{n})$

• •

  $(b,\alpha_{2},a,\alpha_{3})$, $(b,\alpha_{4},a,\alpha_{5})$, $\dots$, $(b,\alpha_{n},a,\alpha_{1})$

• •

  $(c,\beta_{1},\alpha_{1},a,\alpha_{1+\frac{n}{2}},\beta_{1+\frac{n}{2}})$, $(c,\beta_{3},\alpha_{3},a,\alpha_{3+\frac{n}{2}},\beta_{3+\frac{n}{2}})$, $\dots$, $(c,\beta_{\frac{n}{2}},\alpha_{\frac{n}{2}},a,\alpha_{n},\beta_{n})$,
  $(c,\beta_{\frac{n}{2}+2},\alpha_{\frac{n}{2}+2},a,\alpha_{2},\beta_{2})$, $\dots$, $(c,\beta_{n-1},\alpha_{n-1},a,\alpha_{\frac{n}{2}-1},\beta_{\frac{n}{2}-1})$

Using these cycles, observe that this embedding of the graph $G_{n}$ into some surface has $\frac{5n}{2}$ faces, $2n+3$ vertices, and $5n$ edges. Using Euler’s formula, we deduce that $2-2g=\frac{5n}{2}-5n+2n+3=-\frac{n}{2}+3.$ This equation yields $g=\frac{n-2}{4}=\lceil\frac{n-2}{4}\rceil$ since $n\equiv 2\mod 4$. Observe that the fact that $n\equiv 2\mod 4$ is used inherently in the definition of the $6$-cycles in the fourth bullet point.

Thus we have proved our result in the case when $n\equiv 2\mod 4$. The other cases follow from the fact that $G_{n}$ is a subgraph of $G_{n+1}$, $G_{n+2}$ and $G_{n+2}$. If $n\not\equiv 2\mod 4$, there exists a unique integer $m\in\{n+1,n+2,n+3\}$ satisfying $m\equiv 2\mod 4$ with $\lceil\frac{n-2}{4}\rceil=\lceil\frac{m-2}{4}\rceil$. ∎

The subgraph lattice graph of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}}$ is pictured below in Figure 2.

The red vertices represent the $p+1$ subgroups of size $p$. The blue vertices represent the $p^{2}+p$ cyclic subgroups of size $p^{2}$ and the orange vertex represents the unique non-cyclic subgroup of size $p^{2}$. The green vertices represent the $p+1$ subgroups of size $p^{3}$.

It is easy to see that $\Gamma(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}})$ has a $G_{p+1}$ minor by merging all the blue vertices with the respective red vertex they connect to. Hence, $\gamma(\Gamma(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}}))\geq\lceil\frac{p-1}{4}\rceil$. It is also relatively easy to see that since $G_{p+1}$ embeds onto a surface of genus $\lceil\frac{p-1}{4}\rceil$, we can also embed $\Gamma(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}})$ on the same surface. This is because one can cut out a small neighborhood of the edges $\alpha_{1}\beta_{1},\dots\alpha_{p+1}\beta_{p+1}$ and replace it by the corresponding red vertex, green vertex, and $p$ blue vertices along with the edges that connect them. For example, see Figure 3.

Hence, we have proven that $\gamma(\Gamma(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}}))=\lceil\frac{p-1}{4}\rceil$. ∎

Observe that $H_{n}$ contains two copies of $G_{n}$. By deleting four of the red edges and using Theorem 2, one deduces that $\gamma(H_{n})\geq 2\lceil\frac{n-2}{4}\rceil$.

Now, we consider the case when $n\equiv 1\mod 4$. To demonstrate that $H_{n}$ can be drawn on a surface of genus $2\lceil\frac{n-2}{4}\rceil$, we will construct a set of cycles that traverse each edge in both directions exactly once. The cycles are as follows:

1. (a)

   $(b_{0},\alpha_{1},\beta_{1})$, $(b_{0},\beta_{2},\alpha_{2})$, …, $(b_{0},\beta_{n-1},\alpha_{n-1})$, $(b_{0},\alpha_{n},\beta_{n})$

2. (b)

   $(b_{1},\delta_{1},\gamma_{1})$, $(b_{1},\gamma_{2},\delta_{2})$, …, $(b_{1},\gamma_{n-1},\delta_{n-1})$, $(b_{1},\delta_{n},\gamma_{n})$

3. (c)

   $(b_{0},\beta_{1},c_{0},\beta_{2})$, $(b_{0},\alpha_{2},a_{0},\alpha_{3})$, …, $(b_{0},\beta_{n-2},c_{0},\beta_{n-1})$, $(b_{0},\alpha_{n-1},a_{0},\alpha_{n})$

4. (d)

   $(b_{1},\delta_{2},c_{1},\delta_{1})$, $(b_{1},\gamma_{3},a_{1},\gamma_{2})$, …, $(b_{1},\delta_{n-2},c_{1},\delta_{n-1})$, $(b_{1},\gamma_{n},a_{1},\gamma_{n-1})$

5. (e)

   $(c_{0},\beta_{1},\alpha_{1},a_{0},\alpha_{(n+3)/2},\beta_{(n+3)/2})$, $(a_{0},\alpha_{2},\beta_{2},c_{0},\beta_{(n+5)/2},\alpha_{(n+5)/2})$, …,
   $(c_{0},\beta_{(n-3)/2},\alpha_{(n-3)/2},a_{0},\alpha_{n-1},\beta_{n-1})$, $(a_{0},\alpha_{(n-1)/2},\beta_{(n-1)/2},c_{0},\beta_{n},\alpha_{n})$

6. (f)

   $(a_{1},\gamma_{1},\delta_{1},c_{1},\delta_{(n+3)/2},\gamma_{(n+3)/2})$, $(c_{1},\delta_{2},\gamma_{2},a_{1},\gamma_{(n+5)/2},\delta_{(n+5)/2})$, …,
   $(a_{1},\gamma_{(n-3)/2},\delta_{(n-3)/2},c_{1},\delta_{n-1},\gamma_{n-1})$, $(c_{1},\delta_{(n-1)/2},\gamma_{(n-1)/2},a_{1},\gamma_{n},\delta_{n})$

7. (g)

   $(b_{0},a_{a},a_{0},\alpha_{1})$, $(b_{0},\beta_{n},c_{0},b_{1})$, $(b_{0},b_{1},\gamma_{1},a_{1})$, $(c_{1},\delta_{n},b_{1},c_{0})$

8. (h)

   $(c_{0},\beta_{(n+1)/2},\alpha_{(n+1)/2},a_{0},a_{1},\gamma_{(n+1)/2},\delta_{(n+1)/2},c_{1})$

Using these cycles, observe that this embedding of the graph $H_{n}$ into some surface has $5n+2$ faces, $4n+6$ vertices, and $10n+5$ edges. Using Euler’s formula, we deduce that $2-2g=5n+2-(10n+5)+4n+6=-n+3.$ This equation yields $g=\frac{n-1}{2}=2\lceil\frac{n-2}{4}\rceil$ since $n\equiv 1\mod 4$. Observe that we inherently used the fact that $n\equiv 1\mod 4$ when we constructed the $6$-cycles in bullets (e) and (f) since we need $\frac{n-1}{2}$ to be even to have an equal number of cycles beginning with $c_{0}$ versus $a_{0}$ (or $a_{1}$ versus $c_{1}$) in our list of $6$-cycles.

Thus we have proved our result in the case when $n\equiv 1\mod 4$. The case when $n\equiv 3\mod 4$ follows from the fact that $H_{n}$ is a subgraph of $H_{n+2}$. ∎

The subgroup lattice of $\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}}$ is pictured below in Figure 4.

The red vertices represent the $p+1$ subgroups of size $p$, and the red vertex with a black outline is the subgroup generated by $(p^{2},0)$. The black vertices in the third row from the bottom represent the $p(p+1)$ cyclic subgroups of size $p^{2}$, and the white vertex represents the unique non-cyclic subgroup of size $p^{2}$. The blue vertices without a black outline represent the $p^{2}$ cyclic subgroups of size $p^{3}$ and the $p+1$ blue vertices with a black outline represent the $p+1$ non-cyclic subgroups of size $p^{3}$. Lastly, the green vertices are the $p+1$ subgroups of size $p^{4}$, and the green vertex outlined in black unique subgroup of size $p^{4}$ that is entirely $p^{2}$ torsion.

It is relatively easy to see that $\Gamma(\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}})$ has a $H_{p}$ minor. Hence, $\gamma(\Gamma(\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}}))\geq 2\lceil\frac{p-2}{4}\rceil$. It is also relatively easy to see that since $H_{n}$ embeds onto a surface of genus $2\lceil\frac{p-2}{4}\rceil$, we can embed $\Gamma(\mathbb{Z}_{p^{3}}\times\mathbb{Z}_{p^{2}})$ on the same surface. This is because one can cut out a small neighborhood of the edges $\gamma_{1}\delta_{1},\dots\gamma_{p+1}\delta_{p+1}$ and replace it by the corresponding black vertex, green vertex, and $p$ blue vertices along with the edges that connect them. For example, see Figure 5 below.

A similar operation allows one to cut out a small neighborhood of the edges $\alpha_{1}\beta_{1},\dots\alpha_{p+1}\beta_{p+1}$ and replace it by the corresponding red vertex, blue vertex, and $p$ black vertices along with the edges that connect them.

Now consider the case when $p=2$. We know that $\gamma(\Gamma(\mathbb{Z}_{8}\times\mathbb{Z}_{4}))\geq 1$ by Theorem 4. To demonstrate that $\gamma(\Gamma(\mathbb{Z}_{8}\times\mathbb{Z}_{4}))=1$, we simply need to show that we can draw $\Gamma(\mathbb{Z}_{8}\times\mathbb{Z}_{4})$ on the 1-torus which we do here in Figure 6.

∎

The subgroup lattice of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}\times\mathbb{Z}_{q}$ is pictured below in Figure 7.

The blue vertices represent the $p+1$ subgroups of size $p$, the $a$ vertex represents the trivial subgroup, the $b$ vertex represents the unique subgroup of size $p^{2}$, the $c$ vertex represents the unique subgroup of size $q$, the green vertices represent the $p+1$ subgroups of size $pq$ and the vertex $d$ is the full group.

If one contracts the edges $0_{a}0_{c},\dots,p_{a}p_{c}$ and deletes edges $bd$ and $ac$, then one obtains a $K_{4,p+1}$ graph with genus $\lceil\frac{(4-2)(p+1-2)}{4}\rceil=\lceil\frac{p-1}{2}\rceil$. Hence, $\gamma(\Gamma(\mathbb{Z}_{p}\times\mathbb{Z}_{p}\times\mathbb{Z}_{q}))\geq\lceil\frac{p-1}{2}\rceil$.

First, we will prove the case when $p$ is an $odd$ prime. To show that we can embed the subgroup lattice on a surface of genus $\lceil\frac{p-1}{2}\rceil$, we will exhibit a set of cycles that traverse each edge in both directions exactly once. The cycles are as follows:

• •

  $(a,0_{a},b,1_{a})$, $(a,2_{a},b,3_{a})$, $\dots$, $(a,(p-1)_{a},b,p_{a})$

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  $(c,1_{c},d,0_{c})$, $(c,3_{c},d,2_{c})$, $\dots$, $(c,p_{c},d,(p-1)_{c})$

• •

  $(a,c,0_{c},0_{a})$, $(a,p_{a},p_{c},c)$

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  $(b,d,1_{c},1_{a})$, $(b,2_{a},2_{c},d)$

• •

  $(a,1_{a},1_{c},c,2_{c},2_{a})$, $(a,3_{a},3_{c},c,4_{c},4_{a})$, $\dots$, $(a,(p-2)_{a},(p-2)_{c},c,(p-1)_{c},(p-1)_{a})$

• •

  $(b,0_{a},0_{c},d,p_{c},p_{a})$, $(b,4_{a},4_{c},d,3_{c},3_{a})$, $\dots$, $(b,(p-1)_{a},(p-1)_{c},d,(p-2)_{c},(p-2))a)$

Using these cycles, observe that this embedding of $\Gamma(\mathbb{Z}_{p}\times\mathbb{Z}_{p}\times\mathbb{Z}_{q})$ into some surface has $2p+4$ faces, $2(p+3)$ vertices, and $5p+7$ edges. Using Euler’s formula, we deduce that $2-2g=2p+4-(5p+7)+2p+6=-p+3$. This equation yields $g=\frac{p-1}{2}=\lceil\frac{p-1}{2}\rceil$ since $p$ is an odd prime. Hence, we have shown that $\gamma(\Gamma(\mathbb{Z}_{p}\times\mathbb{Z}_{p}\times\mathbb{Z}_{q}))=\lceil\frac{p-1}{2}\rceil$ when $p$ is an odd prime.

When $p=2$, we know that $\gamma(\Gamma(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{q}))>0$ by Theorem 4. Since, $\Gamma(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{q})$ is a subgraph of $\Gamma(\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{q})$, we deduce that $\gamma(\Gamma(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{q}))=1=\lceil\frac{2-1}{2}\rceil$. ∎