Structural Equivalence in Graphs and Complete Skeletons

$(\Longrightarrow)$ Suppose $(u\,v)\in\textrm{Aut}(\Gamma)$. Clearly, $N^{1}(u)\setminus\{v\}=N^{1}(v)\setminus\{u\}$ because the transposition $(u\,v)$ must preserve adjacency relations. Next, assume there is some $i\leq\textrm{diam}(\Gamma)$ such that $N^{i}(u)\setminus\{v\}\neq N^{i}(v)\setminus\{u\}$. Thus, we can say without loss of generality that there is some $x\in V(\Gamma)$ such that $d(u,x)>d(v,x)$. This cannot be, though, because $N^{1}(u)\setminus\{v\}=N^{1}(v)\setminus\{u\}$, which is non-empty if $|V(\Gamma)|>2$. This means that a shortest path between $u$ and $x$ can also be a shortest path between $v$ and $x$, simply by replacing the $u$ with the $v$. Hence, $d(u,x)=d(v,x)$ for all $x\in V(\Gamma)\setminus\{u,v\}$, so $N^{i}(u)\setminus\{v\}=N^{i}(v)\setminus\{u\}$. If $|V(\Gamma)|=2$, then the statement holds trivially.

$(\Longleftarrow)$ Suppose $N^{i}(u)\setminus\{v\}=N^{i}(v)\setminus\{u\}$ for all $i\leq\textrm{diam}(\Gamma)$. This means $u$ and $v$ can be swapped without affecting the adjacency relations of $\Gamma$. Therefore, $(u\,v)\in\textrm{Aut}(\Gamma)$, which tells us that $u$ and $v$ are structurally equivalent. ∎

Clearly, $SEP(\Gamma)\subseteq\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$ because there is some $1\leq i\leq s(\Gamma)$ such that $(u\,v)\in S_{\gamma_{i},i}$ for all $(u\,v)\in\mathrm{Aut}(\Gamma)$, which tells us that all the generators of $SEP(\Gamma)$ are included in $\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$. Next, consider some $\sigma\in\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$. Because each of the $S_{\gamma_{i},i}$ groups consist of the permutations on distinct vertex sets, the smallest subgroup containing $\bigcup S_{\gamma_{i},i}$ for all $1\leq i\leq s(\Gamma)$ is the group consisting of all possible compositions of distinct elements from each of the $S_{\gamma_{i},i}$. Because $S_{\gamma_{i},i}\cap S_{\gamma_{j},j}=1$ for $i\neq j$, we see that $\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$ can be expressed as the direct product of the $S_{\gamma_{i},i}$, i.e. $\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}\cong\prod_{i=1}^{s(\Gamma)}S_{\gamma_{i},i}$. We find $\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}\subseteq SP(\Gamma)$ because each of the $S_{\gamma_{i},i}$ can be generated by the transpositions of its vertices, and each element in $\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$, as we saw, is a composition of distinct elements from the $S_{\gamma_{i},i}$, so $\sigma\in\big{\langle}\bigcup S_{\gamma_{i},i}:1\leq i\leq s(\Gamma)\big{\rangle}$ can be generated by the transpositions in Aut$(\Gamma)$ via compositions. The formula for the order of $SEP(\Gamma)$ follows immediately from the fact that $SEP(\Gamma)\cong\prod_{i=1}^{s(\Gamma)}S_{\gamma_{i},i}$. ∎

The theorem is clearly true for the case where the proper sub-cycles are transpositions, given that $SEP(\Gamma)$ is generated by a set of transpositions. Any other proper sub-cycle $\tau$ must also be an element of a unique $S_{\gamma_{i},i}$, so $\tau\in SEP(\Gamma)$ by Theorem 2.3. Similar reasoning can be applied to the disjoint cycles composing $\sigma$. ∎

Suppose first that $\Gamma$ has only one equivalence class. Then $SEP(\Gamma)=S_{\alpha}$, where $\alpha$ is the size of this equivalence class, as expressed in the theorem. First, we observe

because $|SEP(\Gamma)|=\alpha!$, so the primes that divide $|SEP(\Gamma)|$ are precisely the primes less than or equal to $\alpha$. If $p,q\leq\alpha$ and are primes, then the only way to get an element of order $pq$ in $SEP(\Gamma)$ is if we can compose two disjoint cycles of sizes $p$ and $q$, which we can do if and only if $p+q\leq\alpha$.

Next, suppose $\Gamma$ has at least two equivalence classes. Then, we can say

It is clear that reasoning from above also applies to this case, so all we have left to show is that for $p,q\in V(\tilde{\Gamma})$, even if $p+q>\alpha$, $pq$ can still be an edge in $\tilde{\Gamma}$, given that $p\leq\alpha$ and $q\leq\beta$ (where $p>q$). This follows from the fact that $S_{\alpha}$ and $S_{\beta}$ are permutation groups on different equivalence classes. If there is a cycle of size $p$ in $S_{\alpha}$ and a cycle of size $q$ in $S_{\beta}$, then the theorem follows. ∎

First, consider the case where $\Gamma$ has at least two equivalence classes, or $SEP(\Gamma)\cong S_{\alpha}\times S_{\beta}\times\prod_{i=1}^{s(\Gamma)-2}S_{\gamma_{i},i}$. By Theorem 3.1, we can deduce that $p_{k}p_{k-1}\in E(\tilde{\Gamma})$ implies that $p_{i}p_{j}\in E(\tilde{\Gamma})$ for all $p_{i}+p_{j}\leq p_{k}+p_{k-1}$. Hence, we just need to find the conditions under which $p_{k}p_{k-1}\in E(\tilde{\Gamma})$ is true in order to prove that $\tilde{\Gamma}$ has a $k$-clique. However, this is also taken care of by Theorem 3.1, so the first case is proved. The second case follows by similar reasoning. ∎

Suppose $\tilde{\Gamma}$ is a complete graph on at least two vertices. Thus, it contains a $n$-clique and $\alpha<p_{n+1}$; otherwise, there would be at least $n+1$ vertices. By Theorem 3.2, $SEP(\Gamma)\cong S_{\alpha}\times S_{\beta}\times\prod_{i=1}^{s(\Gamma)-2}S_{\gamma_{i},i}$ such that $p_{n+1}>\alpha\geq p_{n}+p_{n-1}$ or $p_{n+1}>\alpha\geq p_{n}$ and $\beta\geq p_{n-1}$, or $SEP(\Gamma)$ is just $S_{\alpha}$, where $p_{n+1}>\alpha\geq p_{n}+p_{n-1}$. However, it is known that there is no $n$ such that $p_{n+1}>p_{n}+p_{n-1}$ (This can be seen, for example, in [6]). Thus, we see that the only option for $SEP(\Gamma)$ is $SEP(\Gamma)\cong S_{\alpha}\times S_{\beta}\times\prod_{i=1}^{s(\Gamma)-2}S_{\gamma_{i},i}$ such that $p_{n+1}>\alpha\geq p_{n}$ and $\beta\geq p_{n-1}$. The reverse direction follows immediately from Theorem 3.2 and restriction $p_{n+1}>\alpha$. ∎

First, we will show $V(\tilde{\Gamma}^{(i)})\subsetneq V(\tilde{\Gamma}^{(i-1)})$. Suppose $|V(\tilde{\Gamma}^{(i-1)})|>1$. We note $SEP(\tilde{\Gamma}^{(i-1)})$ is isomorphic to the direct product of symmetric groups on the vertices in the equivalence classes of $\tilde{\Gamma}^{(i-1)}$. Because $|SEP(\tilde{\Gamma}^{(i-1)})|=\prod_{i=1}^{s(\tilde{\Gamma}^{(i-1)})}\alpha_{i}!$ by Theorem 2.3, we note

(Note that the first set inclusion is an equality in the case that $\tilde{\Gamma}^{(i-1)}$ is complete.) Thus, we have found $V(\tilde{\Gamma}^{(i)})\subsetneq V(\tilde{\Gamma}^{(i-1)})$. Note that in the special case where $|V(\tilde{\Gamma}^{(i-1)})|=1$, $\{p:p\,\textrm{is prime and}\,p\leq|V(\tilde{\Gamma}^{(i-1)})|\}=\emptyset$, so $V(\tilde{\Gamma}^{(i)})$ is also empty. If $\tilde{\Gamma}^{(i-1)}=\emptyset$, then it makes no sense to consider $\tilde{\Gamma}^{(i)}$ (although, we could say that, in a trivial sense, $\tilde{\Gamma}^{(i)}=\emptyset$, the empty graph).

All we have left to prove is that the edge set of $\tilde{\Gamma}^{(i)}$ is contained in the edge set of $\tilde{\Gamma}^{(i-1)}$. Consider the two graphs as labelled prime graphs and assume that there is some edge $pq\in E(\tilde{\Gamma}^{(i)})$ such that $pq\not\in E(\tilde{\Gamma}^{(i-1)})$. Because we have seen $V(\tilde{\Gamma}^{(i)})\subsetneq V(\tilde{\Gamma}^{(i-1)})$, it follows from Theorem 3.1 that the largest equivalence class in $\tilde{\Gamma}^{(i)}$ has fewer vertices than the largest equivalence class in $\tilde{\Gamma}^{(i-1)}$. Let $S_{\alpha^{i}}$ and $S_{\alpha^{(i-1)}}$ be the permutation groups on these largest equivalence classes for $\tilde{\Gamma}^{(i)}$ and $\tilde{\Gamma}^{(i-1)}$, respectively. Define $S_{\beta^{i}}$ and $S_{\beta^{i-1}}$ similarly for the permutation groups on the second largest equivalence classes, if they exist. Hence, we have found $\alpha^{i}<\alpha^{i-1}$. Also, by Theorem 3.1, we know that $V(\tilde{\Gamma}^{(i)})=\{p:p\,\textrm{is prime and}\,p\leq\alpha^{(i-1)}$}. Because equivalence classes partition the vertex set of a graph, we can say $\alpha^{i}+\beta^{i}\leq|\{p:p\,\textrm{is prime and}\,p\leq\alpha^{i-1}\}|$. Next, we can use Theorem 3.1 again to determine that either $p+q\leq\alpha^{i}$ or $p\leq\alpha^{i}$ and $q\leq\beta^{i}$ (where $p>q$) because $pq\in E(\tilde{\Gamma}^{(i)})$. First, suppose $p+q\leq\alpha^{i}$. Then, we find $p+q<\alpha^{i-1}$ because $\alpha^{i}<\alpha^{i-1}$. This is a contradiction, though, because this implies $pq\in E(\tilde{\Gamma}^{(i-1)})$. Next, suppose $p\leq\alpha^{i}$ and $q\leq\beta^{i}$. We find $p+q\leq\alpha^{i}+\beta^{i}\leq|\{p:p\,\textrm{is prime and}\,p\leq\alpha^{i-1}\}|\leq\alpha^{i-1}$, so we have another contradiction. Thus, if $pq\in E(\tilde{\Gamma}^{(i)})$, then $pq\in E(\tilde{\Gamma}^{(i-1)})$, which completes the proof. ∎

By Theorem 3.4 and well-ordering, we know that there is an $i$ such that $\tilde{\Gamma}^{(i)}$ is minimal in the poset of the graphs $\tilde{\Gamma}^{(j)}$ with respect to $\supset$. The theorem follows. ∎

First, we will consider the case where there are no $K_{1}$ vertices in $\Omega(\Gamma)$ that form a disconnected equivalence class. It is apparent that any of the vertices of $\Gamma$ represented by one of the vertices in $\Omega(\Gamma)$ are structurally equivalent, so we just need to show that the vertices of $\Omega(\Gamma)$ are the complete equivalence classes. This follows from the fact that $\Omega(\Gamma)$ is the smallest possible reconfiguration of $\Gamma$ under the reinterpretation of vertices and edges. If there were structurally equivalent vertices in two distinct vertices $K_{a}$ and $K_{b}$ of $\Omega(\Gamma)$, we could conflate the two vertices into a single vertex $K_{a+b}$, where the set of edges incident to $K_{a+b}$ would be the set of edges incident to $K_{a}$ or $K_{b}$ (which are the same because they are structurally equivalent). However, because $\Omega(\Gamma)$ is the smallest possible reconfiguration of $\Gamma$ under the reinterpretation of vertices and edges, no vertices can be conflated in this way, so we deduce that the vertices are the complete equivalence classes.

Now, we will consider the case with disconnected structurally equivalent vertices separately. Clearly, if we have two or more $K_{1}$ vertices in $\Omega(\Gamma)$ that share the same neighbors, these vertices are not their own complete equivalence classes because we can form a larger one containing them. The paragraph above applies to the rest of the graph, though– it is easy to observe that any of the remaining vertices of $\Omega(\Gamma)$ are still their own complete equivalence classes. ∎

This follows from reasoning similar to what was used to prove Theorem 4.2. If two vertices are connected and are structurally equivalent in $\Omega(\Gamma)$, then we can conflate them as we saw above, which contradicts the minimality of $V(\Omega(\Gamma))$. ∎

The forward direction was established by Corollary 4.3. Now, we just need to show that any graph without connected structurally equivalent vertices can be the structure of a complete skeleton. Let $G$ be such a graph. Reinterpret the vertices and edges of $G$ such that the vertices represent complete induced subgraphs of some other graph $\Gamma$ and two of these vertices are connected if and only if the two complete induced subgraphs are completely connected. Because $G$ contains no connected structurally equivalent vertices, there are no connected vertices $K_{a}$ and $K_{b}$ in $R\Gamma$ such that $N^{1}(K_{a})\setminus\{K_{b}\}\neq N^{1}(K_{b})\setminus\{K_{a}\}$. Thus, no vertices in $R\Gamma$ can be conflated, so $R\Gamma=\Omega(\Gamma)$, which means $G=S\Omega$. ∎

Consider the matrix $I+A(\Gamma)$. This is precisely the adjacency matrix of $\Gamma$ where a loop is attached to each vertex. Let this new graph be $\hat{\Gamma}$. If two vertices $u$ and $v$ were connected and structurally equivalent in $\Gamma$, then we see $N^{1}(u)=N^{1}(v)$ in $\hat{\Gamma}$, so connected structurally equivalent vertices have identical columns in rows in $A(\hat{\Gamma})=A(I+A(\Gamma))$, so they must be linearly dependent. If $u$ and $v$ were disconnected structurally equivalent vertices in $\Gamma$, then $N^{1}(u)\cup\{v\}=N^{1}(v)\cup\{u\}$ in $\hat{\Gamma}$. This means that for disconnected structurally equivalent vertices, their columns and rows differ in two places (corresponding to the coordinates of the two vertices) in $A(\hat{\Gamma})$. Hence, the rows and columns for any disconnected structurally equivalent vertices must be linearly independent with respect to each other, so we cannot a priori establish a lower upper bound for rank($I+A(\Gamma))$. Thus, we can conclude rank$(I+A(\Gamma))$ is bounded above by the sum of the number of connected equivalence classes in $\Gamma$ with the number of the individual disconnected structurally equivalent vertices. Equivalently, rank$(I+A(\Gamma))\leq|V(S\Omega(\Gamma))|$. ∎

It is well known in spectral graph theory that, for a graph on $n$ vertices, the multiplicity of an eigenvalue $\lambda$ is given by $n-\textrm{rank}(I\lambda-A(\Gamma))$, so the theorem follows from this and Theorem 4.5. ∎

This follows by the same reasoning used in the proof of Theorem 4.6 and by Equation 3 ∎

The graphs in Figure 4 are all the possible complete skeletons of four vertices. This can be verified by checking all graphs of four vertices and finding the ones that have no connected structurally equivalent vertices. By Theorem 4.4, these graphs can be complete skeleton structures of graphs. Additionally, by Theorem 4.5, we know that for any graph $\Gamma$ that has one of these five graphs as its complete skeleton, rank$(I+A(\Gamma))\leq 4$. It can easily be shown that $\Lambda=|V(S\Omega(\Gamma))|-\textrm{rank}(I+A(S\Omega(\Gamma)))=0$ for all of these graphs, so rank$(I+A(\Gamma))=|V(S\Omega(\Gamma))|=4$. ∎

There are 34 total graphs with five vertices, which can all be found in [7]. As we saw in the proof of Theorem 5.4, we can find the possible complete skeletons of five vertices by determining all graphs with five vertices that contain no two connected structurally equivalent vertices; these graphs are the structures of the complete skeletons. The reader is encouraged to verify that the fifteen graphs in Figure 7 are all such graphs. Further, it can easily be verified with a CAS that $\Lambda=|V(S\Omega(\Gamma))|-\textrm{rank}(I+A(S\Omega(\Gamma)))=0$ for all of the graphs in Figure 7, excluding the second one in the second row. In other words, rank$(I+A(\Gamma))=5$ unless $S\Omega(\Gamma)=P_{5}$. In this case, there is linear dependence among the columns of $I+A(S\Omega(\Gamma))$ and, in particular, rank$(I+A(S\Omega(\Gamma)))=4$, so $\Lambda=1$ and rank$(I+A(\Gamma))=4$ by Equation 3. ∎