On groups with Cayley graph isomorphic to a cube
We say that a group is a cube group if it is generated by a set of involutions such that the corresponding Cayley graph is isomorphic to a cube. Equivalently, is a cube group if it acts on a cube such that the action is simply-transitive on the vertices and the edge stabilizers are all nontrivial. The action on the cube extends to an orthogonal linear action, which we call the geometric representation. We prove a combinatorial decomposition for cube groups into products of -element subgroup, and show that the geometric representation is always reducible.
Let be a group and let be a subset consisting of involutions. We say that the pair is a cube group (of rank ) if the corresponding Cayley graph is isomorphic to the -skeleton of the -cube. Groups acting on CAT() cube complexes such that the action is simply-transitive on vertices and has nontrivial edge stabilizers were studied in  as a natural generalization of right-angled Coxeter groups. Cube groups are precisely the finite groups in this class.
The purpose of this note is to prove two theorems about cube groups. The first is a product decomposition that implies the existence of a type of “boolean” normal form.
Let be a cube group of rank . For each , let denote the subgroup generated by . Then there exists an ordering on the set such that
In particular, for any , there exist a unique choice of , such that
The action of a cube group on extends canonically to an isometric linear action on the full -cube . We call this linear representation the geometric representation of . The second theorem is a decomposition theorem for this representation.
If is a cube group of rank , then the geometric representation is reducible.
Both of these theorems are consequences of the following general fact about certain actions of -groups, the proof of which is reminiscent of one of the standard combinatorial proofs of Sylow’s theorem.
Let be a set with more than one element, and let be a -group acting on . If has a generating set such that every element fixes some element in , then has at least two orbits.
2. Decorated graphs and group presentations
In this section, we described the relations among the generators of a cube group in terms of a certain graph with involutions. Let be a finite set, and let denote its permutation group. For each , let be an involution satisfying . We can represent this data pictorially as follows. Let denote the complete graph with vertex set . At each vertex , draw an arc between all pairs of edges and whenever . We shall refer to such a choice of involutions (and the resulting picture) as a decorated graph.
For the decorated graph on shown in Figure 1, the involutions , , , and are the transpositions , , , and , respectively. The involution is the product .
Given a decorated graph on and any two distinct elements , one can define a sequence in inductively by . Such a sequence will be called a trajectory. A trajectory is -periodic if for all . We shall say the decorated graph has no holonomy along a trajectory if is the identity permutation in .
A decorated graph on is admissible if every trajectory is -periodic and has no holonomy along it.
The -periodicity condition simply means that the edges of a decorated graph (with their connecting arcs) can be partitioned into subsets of the form
a -cycle with arcs joining consecutive edges,
an angle (two edges meeting at a vertex joined by an arc), or
a single edge (with no connecting arcs touching it).
For example, the decorated graph in Figure 1 satisfies -periodicity since the edges can be partitioned as a single -cycle (), two angles (, ), and two single edges ( and ). The no-holonomy condition is more difficult to verify but also holds for this decorated graph (for example, along a trajectory corresponding to the -cycle, we have ).
Given a decorated graph on , we can define a group by the presentation
The following is a special case of the main result (Theorem 3.2) in .
is a cube group if and only if there exists an admissible decorated graph on and an isomorphism that restricts to the identity on .
We refer the reader to  for the full proof of this theorem, but indicate here how one obtains the decorated graph from the pair .
Suppose is a cube group. Recall that the Cayley graph has vertex set , and two vertices are joined by a directed edge if for some . Each directed edge is labeled by the element , but since , we can unambiguously label the undirected edge by the element as well. We shall therefore regard as an undirected graph on with edges labeled by elements of . Recall that a presentation for can be obtained from the Cayley graph by taking products of labels around all cycles in the graph.
Since the Cayley graph for is isomorphic to a cube, cycles are generated by cycles of length , and any such -cycle can be translated by the (left) -action (preserving labels) so that it starts at the identity vertex . For a presentation, it therefore suffices to consider the -cycles touching the vertex (and the trivial -cycles obtained by repeating an edge). At the vertex , there are precisely incident edges, labeled by the elements of . Any two elements , determine two such edges and hence span a unique -cycle (-dimensional subface of the cube). Reading the labels around this -cycle, we obtain a relation .
To get the corresponding decorated graph , we continue the relation to obtain a -periodic sequence . These are the trajectories of our decorated graph . To obtain the involutions , , we simply consider all trajectories of the form and define . One then needs to check that the resulting ’s are well-defined involutions, and that there is no holonomy along trajectories.
Let be the dihedral group of order , represented as the permutation group . Letting where , , and , we obtain the -cube as the Cayley graph (Figure 2, left-hand side), hence is a cube group of rank . Looking at pairs of edges incident to the vertex , we get the trajectories , , , , , and . The coresponding decorated graph on is therefore (Figure 2, right-hand side) where and , and we have a presentation for of the form
3. Product decompositions of cube groups
Given a cube group any subset generates a subgroup of which we denote by . We will call this subgroup a standard subgroup if is also a cube group.
Let be the dihedral group in Example 2.2. The subgroup generated by is a standard subgroup since has Cayley graph isomorphic to a square. On the other hand, if , then is not a standard subgroup. In this case is the entire group () and has only two elements, hence the Cayley graph will be an -cycle, which is not isomorphic to a cube.
In this section we describe how to decompose a cube group into products of standard subgroups. We use the fact that there is a natural action of the group on its generating set .
Let be a cube group with corresponding decorated graph . Then the map given by extends uniquely to a homomorphism .
Proof. One needs only check that the ’s satisfy equations corresponding to the relations in the presentation . This is precisely the requirement that the ’s be involutions and that have no holonomy along trajectories. ∎
We shall call the permutation representation for . It defines an action of on the set by for all and . Invariant subsets (i.e., unions of orbits) of this action give rise to product decompositions of with respect to standard subgroups.
Let be a cube group and let be a -invariant subset of . Then is a standard subgroup and, moreover, we have a product decomposition
(meaning any element can be written uniquely in the form where and ).
Proof. For any subset , let denote the -dimensional subcube of containing the edges incident to that are labeled by . To show that is a standard subgroup, it suffices to show that all of the labels on the edges of are in the subset , since then will coincide with . For this, it is enough to show that the labels around any -cycle in are always in . Since is -invariant, any trajectory starting with will have all terms in , hence if a -cycle in has consecutive edges in , it will have all edges in . It follows that all -cycles in incident to the vertex have edges labeled by elements of , and then by induction on the distance to that all -cycles in are labeled by elements of .
Since will also be -invariant, we have standard subgroups and . The subcubes and will intersect only in the vertex , hence the two subgroups and have trivial intersection. To prove the product decomposition, it is enough to show that any product with and can be rewritten at with and . Consider the trajectory starting with and . Then and . Since is -invariant and , we know , and since is -invariant and , we know . On the other hand, the relation coming from this trajectory is , or (since generators are all involutions) . ∎
It turns out that for any cube group of rank at least , there is always a proper nontrivial invariant subset. This is the key technical result of this paper and uses the same counting argument found in one of the standard proofs of the first Sylow theorem (see ).
Let be a -group acting on a set with . Assume further that there is a generating subset such that for all there exists an such that . Then has at least two -orbits.
Proof. Assume, on the contrary that the action is transitive on . Since any subgroup of is also a -group, the order of any orbit must also be a power of , hence for some . Let denote the set of all subsets of of size . Then
where denotes the -adic valuation of . Since none of the terms in the product on the right have any factors of , we can write for some integer relatively prime to .
Now consider the induced action of on . Since all orbits of this action must have order a power of , and since the sum of these orders must be , there must be at least one orbit of size exactly . Let be such an orbit. Then consists of subsets of of size exactly , and since the action of on is transitive, these subsets form a partition of . Given , we let denote its equivalence class. Now, given , let denote the subgroup it generates, and let denote the elements of fixed by . Since there exists an such that , we know that , hence is not empty. On the other hand, since is a -group, we have
It follows that , so fixes . Since this holds for all in the generating set , the action of on must be trivial. But this contradicts transitivity of the action of on ∎
To apply this theorem (in the case ) to our setting, we suppose is a cube group with and take . The action of on has the property that , hence every generator fixes some element. By Theorem 3.1 it follwos that the action has at least two orbits. Combining this with the previous proposition, we obtain our first theorem from the introduction.
Let be a cube group of rank . For each , let denote the subgroup generated by . Then there exists an ordering on the set such that
Proof. The proof is by induction on the rank of . If , then . In general, assume the decomposition holds for all cube groups of rank , and let be a cube group of rank . By Theorem 3.1, there exists a proper, nontrivial -invariant subset . The groups and are cube groups of rank , say and , respectively. Hence by induction, there exists an ordering of the elements in , and an ordering of the elements in , such that
The result then follows from Proposition 3.2. ∎
If is the dihedral group of Example 2.2, then breaks up into -orbits and , and breaks up into -orbits , and . Hence,
and we can take as our ordering on . In other words, the elements of can be listed as . On the other hand, the ordering does not respect the orbit structure and will be a proper subset of (since and ).
For a more complex example, consider the decorated graph in Figure 3 (the missing edges correspond to -periodic trajectories, hence commuting generators).
One can show that this graph is admissible, hence defines a cube group of order . The -orbits are and . The -orbits are , , and the -orbits are , . Finally, all of these orbits break up into singleton orbits, so we can take as our decomposition
In fact any planar representation of the orbit tree shown in Figure 3 will give rise to a different product decomposition by reading off the final nodes from left to right.
4. The geometric representation
In this section we describe the geometric representation of a cube group arising from the left action of on . We use the fact that the -cube is rigid in the sense that any (graph) automorphism of its -skeleton is the restriction of an isometry of the entire cube.
Let be a cube group, and let denote the finite dimensional real Euclidean space with standard basis . There is a natural identification of with the standard cube given as follows. Since is isomorphic to the -cube with , we can index the vertices using subsets of . More precisely, we let be the identity vertex. For any subset , there is a unique minimal subcube containing the vertex and the vertices for all . We let denote the vertex opposite in this subcube. The embedding of into is then given by the mapping
for all . It follows from the way we indexed the elements of that the edges in the Cayley graph will map precisely to the edges in the -skeleton of the cube . For the remainder of the paper, we shall identify with the -skeleton of the cube via this embedding.
Let be a cube group. Any element determines an automorphism of the Cayley graph , hence an isometry of the cube . Any such isometry is the restriction of a unique orthogonal linear transformation . The resulting homomorphism taking to will be called the geometric representation of the pair . Since acts simply-transitively on itself and is identified with the vertices of the cube , the geometric representation is obviously faithful.
The geometric representation can be described explicitly in terms of the permutation representation . Given a product representation for in terms of generators and given any element , we let denote the cardinality of the set
We then obtain the following formula for .
Let be a cube group and let be the corresponding decorated graph. For any , we let denote the image of under the homomorphism . The linear transformation is then given by
(In particular, is independent of the product representation of in terms of generators.)
Proof. First we prove that for any , we have
For each , let (respectively ) denote the facet (codimension- face) of the cube with barycenter (resp., ). The map must take the facets incident to the vertex bijectively to the facets incident to the vertex , permuting those facets incident to both and . The facets incident to are precisely the “positive” ones , and the facets incident to are . So must map to , and hence for we have .
Now suppose , then the facet is the unique facet containing the edges incident to the vertex that are labeled by the set . Equivalently, is the unique facet containing the edges incident to the vertex that are labeled by the set (see Figure 4).
Since restricts to the left action of on the Cayley graph and this action preserves edge labels, it follows that is the unique facet spanned by the edges incident to the vertex that are labeled by the set . This is the same as the facet spanned by the edges incident to the vertex that are labeled by the set , or equivalently, the set . But this is precisely the facet . Restricting to the barycenters of these facets, we then have .
Finally, suppose . Then
as desired. ∎
It follows immediately from this formula for the geometric representation that if is a -invariant subset of , then is an invariant subspace for . By Theorem 3.1, always admits a proper nontrivial -invariant subset, hence we have our second theorem from the introduction.
If is a cube group, then the geometric representation is reducible.
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