Digital topological groups

Since the argument for $\nu_{x}$ is similar, we will prove the statement for $\mu_{x}$. Assume that $G$ is an $\textrm{NP}_{i}$-digital topological group for some $i\in\{1,2\}$.

First we show that $\mu_{x}$ is continuous for each $x\in G$. Let $f_{x}:G\to G\times G$ be defined by $f_{x}(y)=(x,y)$ for each $x\in G$. Note that $\mu_{x}=\mu_{G}\circ f_{x}$. Since $f_{x}$ is $(\kappa,\textrm{NP}_{j}(\kappa,\kappa))$-continuous for both $j=1$ and $j=2$, and $\mu_{G}$ is $(\textrm{NP}_{i}(\kappa,\kappa),\kappa)$-continuous, this means that $\mu_{x}$ will be $(\kappa,\kappa)$-continuous as desired.

By the same reason, $\mu_{x^{-1}}$ is continuous. But $\mu_{x^{-1}}$ is the inverse of $\mu_{x}$, and so $\mu_{x}$ is continuous with continuous inverse as desired. ∎

It suffices to show that the inverse is continuous: let $x\leftrightarrow y$, and we will show that $x^{-1}\leftrightarrow y^{-1}$.

Note that in Theorem 3.4, no reference was made to continuity of $\iota_{G}$. Thus we may use Theorem 3.4 even though we have not assumed continuity of the inverse.

Since $\mu_{x^{-1}}$ is an isomorphism by Theorem 3.4, we may apply $\mu_{x^{-1}}$ to $x\leftrightarrow y$ to obtain $e\leftrightarrow x^{-1}y$, where $e\in G$ is the identity element. Then we apply $\nu_{y^{-1}}$ and obtain $y^{-1}\leftrightarrow x^{-1}$ as desired. ∎

By Lemma 3.5, we need only show that $\mu_{G}$ is $(\textrm{NP}_{1},\kappa)$-continuous. We must choose two pairs of $\textrm{NP}_{1}$-adjacent points in $G$. Without loss of generality choose pairs $(x,y)\leftrightarrow_{\textrm{NP}_{1}}(x,z)$, and we will show that $xy\leftrightarrow_{\kappa}xz$. Since $(x,y)\leftrightarrow_{\textrm{NP}_{1}}(x,z)$, we will have $y\leftrightarrow z$, and applying the isomorphism $\mu_{x}$ gives $xy\leftrightarrow_{\kappa}xz$ as desired. ∎

By Theorem 3.6, we need only show that $\mu_{x}$ and $\nu_{x}$ are isomorphisms for every $x$ in $G$. Clearly $\mu_{x}$ and $\mu_{x^{-1}}$ are inverse functions. Since we have assumed that both are continuous, $\mu_{x}$ must be an isomorphism. Similarly $\nu_{x}$ is an isomorphism as desired. ∎

Given a digital topological group $G$ and elements $x,y\in G$, by Theorem 3.4 the map

is a graph automorphism carrying $x$ to $y$ as required. ∎

By Theorem 3.8, $G$ must be a $d$-regular graph for some $d$. Since we are using $c_{1}$-adjacency, the degree $d$ is at most 4. We will consider each possible value for $d\in\{0,1,2,3,4\}$, and conclude that the only allowable situation is when $G$ is a simple closed curve, or a set of 2 or fewer points.

If $d=0$ then $G$ is a single point. If $d=1$, then $G$ is a set of two adjacent points. For $d=2$, any connected $2$-regular graph is a simple cycle, which in our terminology is a simple closed curve. For $d=4$, the only $4$-regular subset of $(\mathbb{Z}^{2},c_{1})$ is all of $\mathbb{Z}^{2}$, and since $G$ is finite it cannot be $4$-regular.

It remains to show that the case $d=3$ is also impossible. To obtain a contradiction, assume that $G\subset(\mathbb{Z}^{2},c_{1})$ is a finite 3-regular graph. Since $G$ is finite we may compose with a translation to assume without loss of generality that $(0,0)\in G$, and that there are no points $(x,y)\in G$ with $x<0$.

Since $(0,0)\in G$ but no point of $G$ has negative first coordinate, we have $(-1,0)\not\in G$ and thus since $(0,0)$ is degree 3, the neighbors of $(0,0)$ must be exactly $\{(0,1),(1,0),(0,-1)\}$. In particular, we have $(0,1)\in G$. Inductively applying the same reasoning to $(0,1)\in G$, we see that in fact $(0,y)\in G$ for every $y>0$, and so $G$ is infinite which is a contradiction. ∎

Let $\mu_{G}:G\times G\to G$ and $\mu_{H}:H\times H\to H$ be the multiplication operations for $G$ and $H$, respectively, and let $\iota_{G}:G\to G$ and $\iota_{H}:H\to H$ be the inverses for $G$ and $H$, respectively. Then we define the product and inverse

and

by

and

These operations clearly obey the group axioms, so we need only demonstrate the appropriate continuities, which unfortunately are notationally cumbersome. We will write

for the adjacency relation being used in $G\times H$.

By Lemma 3.5, we need only demonstrate that $\mu_{G\times H}$ is $(\textrm{NP}_{i}(\gamma,\gamma),\gamma)$-continuous. Assume that $((g,h),(g^{\prime},h^{\prime}))$ and $((x,y),(x^{\prime},y^{\prime}))$, elements of $(G\times H)\times(G\times H)$, are $\textrm{NP}_{i}(\gamma,\gamma)$-adjacent, and we must show that $(gg^{\prime},hh^{\prime})$ and $(xx^{\prime},yy^{\prime})$ are $\gamma$-adjacent. We will handle the $i=1$ and $i=2$ cases separately.

For $i=1$, we may assume without loss of generality that $(g,h)=(x,y)$ and $(g^{\prime},h^{\prime})\leftrightarrow_{\gamma}(x^{\prime},y^{\prime})$. Further we can assume without loss of generality that $g^{\prime}=x^{\prime}$ and $h^{\prime}\leftrightarrow_{\lambda}y^{\prime}$. Then when comparing $(gg^{\prime},hh^{\prime})$ with $(xx^{\prime},yy^{\prime})$, we see that the first coordinates are equal. In the second coordinate we have $h=y$ and $h^{\prime}\leftrightarrow_{\lambda}y^{\prime}$, and thus $hh^{\prime}\leftrightarrow yy^{\prime}$ since multiplication by $h=y$ is an isomorphism. Thus $(gg^{\prime},hh^{\prime})$ and $(xx^{\prime},yy^{\prime})$ are $\textrm{NP}_{1}(\gamma,\gamma)$-adjacent as desired.

For $i=2$, we must additionally consider the case where $((g,h),(g^{\prime},h^{\prime}))$ and $((x,y),(x^{\prime},y^{\prime}))$ are $\gamma$-adjacent in each of the 2 coordinates. In this case we will have $g\leftrightarroweq_{\kappa}x$, $h\leftrightarroweq_{\lambda}y$, $g^{\prime}\leftrightarroweq_{\kappa}x^{\prime}$, and $h^{\prime}\leftrightarroweq_{\lambda}y^{\prime}$. By $\textrm{NP}_{2}(\kappa,\kappa)$-continuity of $\mu_{G}$ and $\mu_{H}$ we have $gg^{\prime}\leftrightarroweq xx^{\prime}$ and $hh^{\prime}\leftrightarroweq yy^{\prime}$, and so $(gg^{\prime},hh^{\prime})$ and $(xx^{\prime},hh^{\prime})$ are $\textrm{NP}_{2}(\gamma,\gamma)$-adjacent (or equal) as desired. ∎

The vertices of $X$ are elements of $G$, so there is a natural multiplication $\mu_{X}:X\times X\to X$ and inverse $\iota_{X}:X\to X$ which satisfy the group axioms. Using these operations, $X$ is isomorphic to $G$ as a group. To show that $X$ is an $\textrm{NP}_{1}$-digital topological group, by Lemma 3.5 we must show that $\mu_{X}$ is continuous using $\textrm{NP}_{1}$ in the domain.

Let $(x,y)$ and $(u,v)$ be $\textrm{NP}_{1}$-adjacent, and we will show that $xy\leftrightarrow uv$ in $X$. Without loss of generality, we may assume that $x=u$ and $y\leftrightarrow v$. Since $X$ is a Cayley graph, $y\leftrightarrow v$ means that $v=gy$ for some $g\in S$. So we must show that $xy\leftrightarrow xgy$.

By the construction of the Cayley graph we have $e\leftrightarrow g$ since $g\in S$. Then applying the isomorphisms $\mu_{x}$ and $\nu_{y}$ gives $xy\leftrightarrow xgy$ as desired. ∎

Since $x\in S$ we will have $e\leftrightarrow x$, and so $(e,e)$ and $(x,x)$ are $\textrm{NP}_{2}$-adjacent. But $\mu_{X}(e,e)=e$ and $\mu_{X}(x,x)=x^{2}$ are not equal or adjacent because $x^{2}\not\in S$ and $x^{2}\neq e$ was assumed. Thus $\mu_{X}$ is not an $\textrm{NP}_{2}$-continuous function. ∎

Since the statement for $\nu_{x}$ is similar, we will prove the statement for $\mu_{x}$. Note that

Thus $\mu_{x}$ maps $y$ into $G_{xy}$, and since the continuous image of a digital connected subset is connected, we must have

To see that these sets are equal, note that $\mu_{x^{-1}}$ maps $xy$ into $G_{y}$, and so by the same reasons as above we have

Applying $\mu_{x}$ gives $G_{xy}\subseteq\mu_{x}(G_{y})$, which combines with the above to give

as required.∎

To show that $G_{e}$ is a subgroup, take $x,y\in G_{e}$, and we will show that $x^{-1}y\in G_{e}$. Since $x$ and $y$ are in the same component, we can construct a path from $x$ to $y$ in $G_{e}$. Applying the isomorphism $\mu_{x^{-1}}$ to this path gives a path from $e$ to $x^{-1}y$, which means that $x^{-1}y\in G_{e}$ as desired.

To show that $G_{e}$ is normal, take $x\in G$ and $y\in G_{e}$, and we must show that $xyx^{-1}\in G_{e}$. Since $y\in G_{e}$, by Lemma 3.18 we have $\mu_{x}(y)\in G_{x}$, and thus $xyx^{-1}=\nu_{x^{-1}}(\mu_{x}(y))\in G_{e}$ as desired. ∎

By the group properties of $G$, these operations on $C_{G}$ will satisfy the group axioms. It suffices only to show that these operations are well-defined.

To show that the inverse is well defined, take $x,x^{\prime}$ in $G$ with $G_{x}=G_{x^{\prime}}$, and we will show that $G_{x^{-1}}=G_{x^{\prime-1}}$. Since $G_{x}=G_{x^{\prime}}$, there is a path from $x$ to $x^{\prime}$ in $G$. Since the inverse map

is continuous, we may apply $\iota_{G}$ to this path to obtain a path from $x^{-1}$ to $x^{\prime-1}$, which shows that $x^{-1}$ and $x^{\prime-1}$ are in the same component, and thus that $G_{x^{-1}}=G_{x^{\prime-1}}$ as desired.

Now we show that the product $\cdot$ is well-defined. To do this, take $x,y,x^{\prime},y^{\prime}\in G$ with $G_{x}=G_{x^{\prime}}$ and $G_{y}=G_{y^{\prime}}$, and we must show that $G_{xy}=G_{x^{\prime}y^{\prime}}$. That is, we must show that: when $x$ and $x^{\prime}$ are in the same component, and when $y$ and $y^{\prime}$ are in the same component, then $xy$ and $x^{\prime}y^{\prime}$ are in the same component.

Since $x$ and $x^{\prime}$ are in the same component, there is a path

from $x$ to $x^{\prime}$ in $G$. Similarly there is a path

from $y$ to $y^{\prime}$ in $G$. By perhaps repeating elements several times to lengthen the paths, we may assume that $m=n$ so that these paths have the same length. We will also assume that points are repeated in these paths in the following way:

• •

  for even $i$ we assume that $x_{i}=x_{i+1}$, and

• •

  or odd $i$ we assume that $y_{i}=y_{i+1}$.

Now we may multiply these paths pointwise to obtain:

and it suffices to show that this sequence of points forms a path from $xy$ to $x^{\prime}y^{\prime}$. We must demonstrate that

for each $i<n$. Because of the structure of repeated values in even and odd positions, we see that $(x_{i},y_{i})$ and $(x_{i+1},y_{i+1})$ are $\textrm{NP}_{1}$-adjacent for any $i<n$. Thus their products $x_{i}y_{i}$ and $x_{i+1}y_{i+1}$ are adjacent as desired. ∎

For $x\in G$, let $\bar{x}\in G/G_{e}$ be the representative of $x$ in the quotient. Let

be given by

We will show that $f$ is an isomorphism.

First we show that $f$ is well-defined. Let $\bar{x}=\bar{y}$, and we must show that $G_{x}=G_{y}$, that is, $x$ and $y$ are in the same component. The fact that $\bar{x}=\bar{y}$ means that $x=gy$ for some $g\in G_{e}$. Since $g\in G_{e}$ and $y\in G_{y}$, we have $gy\in G_{y}$ by Lemma 3.18. Thus $x\in G_{y}$, and so $x$ and $y$ are in the same component as desired.

Clearly $f$ is a group homomorphism because

and