Endomorphism Breaking in Graphs
Abstract
We introduce the \symitalicendomorphism distinguishing number of a graph as the least cardinal such that has a vertex coloring with colors that is only preserved by the trivial endomorphism. This generalizes the notion of the distinguishing number of a graph , which is defined for automorphisms instead of endomorphisms.
As the number of endomorphisms can vastly exceed the number of automorphisms, the new concept opens challenging problems, several of which are presented here. In particular, we investigate relationships between and the endomorphism motion of a graph , that is, the least possible number of vertices moved by a nontrivial endomorphism of . Moreover, we extend numerous results about the distinguishing number of finite and infinite graphs to the endomorphism distinguishing number. This is the main concern of the paper.
Keywords: Distinguishing number; Endomorphisms; Infinite graphs; Mathematics Subject Classifications: 05C25, 05C80, 03E10
1 Introduction
Albertson and Collins [1] introduced the \symitalicdistinguishing number of a graph as the least cardinal such that has a labeling with labels that is only preserved by the trivial automorphism.
This concept has spawned numerous papers, mostly on finite graphs. But countable infinite graphs have also been investigated with respect to the distinguishing number; see [9], [15], [16], and [17]. For graphs of higher cardinality compare [10].
The aim of this paper is the presentation of fundamental results for the endomorphism distinguishing number, and of open problems. In particular, we extend the Motion Lemma of Russell and Sundaram [14] to endomorphisms, present endomorphism motion conjectures that generalize the Infinite Motion Conjecture of Tom Tucker [16] and the Motion Conjecture of [4], prove the validity of special cases, and support the conjectures by examples.
2 Definitions and Basic Results
As the distinguishing number has already been defined, let us note that for all asymmetric graphs. This means that almost all finite graphs have distinguishing number one, because almost all graphs are asymmetric, see Erdős and Rényi [5]. Clearly for all other graphs. Again, it is natural to conjecture that almost all of them have distinguishing number two. This is supported by the observations of Conder and Tucker [3].
However, for the complete graph , and the complete bipartite graph we have , and . Furthermore, the distinguishing number of the cycle of length 5 is 3, but cycles of length have distinguishing number 2.
This compares with a more general result of Klavžar, Wong and Zhu [12] and of Collins and Trenk [2], which asserts that , where denotes the maximum degree of . Equality holds if and only if is a , or .
Now to the endomorphism distinguishing number. Before defining it, let us recall that an \symitalicendomorphism of a graph is a mapping such that for every edge its image is an edge, too.
Definition The endomorphism distinguishing number of a graph is the least cardinal such that has a labeling with labels that is preserved only by the identity endomorphism of .
Let us add that we also say colors instead of labels. If a labeling is not preserved by an endomorphism , we say that \symitalicbreaks .
Clearly . For graphs with equality holds. Such graphs are called \symitaliccore graphs. Notice that complete graphs and odd cycles are core graphs, see [7]. Hence , , and for .
Interestingly, almost all graphs are core graphs, as shown by Koubek and Rödl [13]. Because almost all graphs are asymmetric, this implies that almost all graphs have trivial endomorphism monoid, that is, . Graphs with trivial endomorphism monoid are called \symitalicrigid. Clearly for any rigid graph , and thus for almost all graphs .
can be equal to even when . For example, this is the case for even cycles. We formulate this as a lemma.
Lemma 1
The automorphism group of even cycles is properly contained in their endomorphism monoid, but for all .
Proof. It is easily seen that every even cycle admits proper endomorphisms, that is, endomorphisms that are not automorphisms. Furthermore, it is readily verified that and .
Hence, let . Color the vertices and black and all other vertices white, see Figure 1. We wish to show that this coloring is endomorphism distinguishing. Clearly this coloring distinguishes all automorphisms.
Let be a proper endomorphism. It has to map the cycle into a proper connected subgraph of itself. Thus, must be a path, say .
Furthermore, the color of the endpoints of an edge must be preserved under . Hence is mapped into itself. Because is the only edge with two white endpoints that is adjacent to , it must also be mapped into itself. This fixes and . But then and are also fixed.
Now we observe that the path in has only white interior vertices and that it it has to be mapped into a walk in from to that contains only white interior vertices. Clearly this is not possible.
To show that can be smaller than , we consider graphs with trivial automorphism group but nontrivial endomorphisms monoid. For such graphs , but . Easy examples are asymmetric, nontrivial trees . For, every such tree has at least 7 vertices and at least three vertices of degree 1. Let be a vertex of degree 1 and its neighbor. Because has at least 7 vertices and since it is connected, there must be a neighbor of that is different from . Then the mapping
is a nontrivial endomorphism.
3 The Endomorphism Motion Lemma
Russel and Sundaram [14] proved that the distinguishing number of a graph is small when every automorphism of moves many elements. We generalize this result to endomorphisms and begin with the definition of motion.
The \symitalicmotion of a nontrivial endomorphism of a graph , is the number of elements it moves:
The \symitalicendomorphism motion of a graph is
For example, .
In the sequel we will prove the following generalization of Theorem 1 of Russell and Sundaram [14].
Lemma 2 (Endomorphism Motion Lemma)
For any graph ,
(1) 
implies
The proof will be an easy consequence of Lemma 3, the Orbit Norm Lemma. We first define orbits of endomorphisms.
Definition An orbit of an endomorphism of a graph is an equivalence class with respect to the equivalence relation on , where if there exist nonnegative integers and such that .
The orbits form a partition , for , of . For finite graphs it can be characterized as the unique partition with the maximal number of sets that are invariant under the preimage . For infinite graphs we characterize it as the finest partition that is invariant under . For automorphisms it coincides with the cycle decomposition.
The \symitalicorbit norm of an endomorphism with the orbits is
and the \symitalicendomorphism orbit norm of a graph is
Notice that may not move all elements of a nontrivial orbit, whereas automorphisms move all elements in a nontrivial cycle of the cycle decomposition. To see this, consider an orbit , where , and . Only one element of the orbit is moved, and the contribution of to the orbit norm of is . Clearly , and thus .
Lemma 3 (Orbit Norm Lemma)
A graph is endomorphism distinguishable if
Proof. We study the behavior of a random coloring of G, the probability distribution given by selecting the color of each vertex independently and uniformly in the set . Fix an endomorphism and consider the event that the random coloring is preserved by , that is, for each vertex of . Then it is easily seen that
Collecting together these events, we have
If this sum is strictly less than one, then there exists a coloring such that for all nontrivial there is a , such that , as desired.
Proof of Lemma 2. From we infer that
(2) 
Hence, if
then the right side of Equation 2 is strictly less than 1, and therefore also less than . Now an application of the Orbit Norm Lemma shows that is distinguishable.
.
Lemma 2 is similar to the \symitalicMotion Lemma of Russell and Sundaram [14, Theorem 1], which asserts that is 2distinguishable if
where
Actually, a short look at the proof of Russell and Sundaram shows that is distinguishable under the weaker assumption
(3) 
Thus, our Endomorphism Motion Lemma is a direct generalization of the Motion Lemma of Russell and Sundaram.
The Motion Lemma allows the computation of the distinguishing number of many classes of finite graphs. We know of no such applications for the Endomorphism Motion Lemma, but will show the applicability of its generalization to infinite graphs.
4 Infinite graphs
Suppose we are given an infinite graph with infinite endomorphism motion and wish to generalize Equation 1 to this case for finite . Notice that
in this situation. Thus the natural generalization would be that
(4) 
implies endomorphism 2distinguishabilty. We formulate this as a conjecture.
Endomorphism Motion Conjecture Let be a connected, infinite graph with endomorphism motion . If , then .
This is a generalization of the Motion Conjecture of [4] for automorphisms of graphs. Notice that we assume connectedness now, which we did not do before. The reason is, that we not only have to break all endomorphisms of every connected component if the graph is disconnected, but that we also have to worry about breaking mappings between possibly infinitely many different connected components, which requires extra effort.
A special case are countable graphs. Let be an infinite, connected countable graph with infinite endomorphism motion . Then and , where c denotes the cardinality of the continuum.
Notice, for countable graphs, . This means that Equation 4 is always satisfied for countably infinite graphs with infinite motion. This motivates the following conjecture:
Endomorphism Motion Conjecture for Countable Graphs Let be a countable connected graph with infinite endomorphism motion. Then is endomorphism 2distinguishable.
In the last section we will verify this conjecture for countable trees with infinite endomorphism motion. Their endomorphism monoids are uncountable and we will show that they have endomorphism distinguishing number 2.
We now prove the conjecture for countable endomorphism monoids. In fact, we show that almost every coloring is distinguishing if the endomorphism monoid is countable.
Theorem 4
Let be a graph with infinite endomorphism motion whose endomorphism monoid is countable. Let be a random coloring where all vertices have been colored independently and assume that there is an such that, for every vertex , the probability that it is assigned a color satisfies
Then is almost surely distinguishing.
Proof. First, let be a fixed, nontrivial endomorphism of . Since the motion of is infinite we can find infinitely many disjoint pairs . Clearly the colorings of these pairs are independent and the probability that preserves the coloring in any of the pairs is bounded from above by some constant . Now
Since there are only countably many endomorphisms we can use subadditivity of the probability measure to conclude that
which completes the proof.
We will usually only use the following Corollary of Theorem 4.
Corollary 5
Let be a graph with infinite motion whose endomorphism monoid is countable. Then
The endomorphism motion conjecture for countable graphs generalises the
Infinite Motion Conjecture of Tucker [16] Let be a connected, locally finite infinite graph with infinite motion. Then is 2distinguishable.
It was shown in [4], and follows from Theorem 4, that it is true for countable . There are numerous applications of this result, see [11].
For the Endomorphism Motion Conjecture for Countable Graphs we have the following generalization of [10, Theorem 3.2]:
Theorem 6
Let be a finitely generated infinite group. Then there is a 2coloring of the elements of , such that the identity endomorphism of is the only endomorphism that preserves this coloring. In other words, finitely generated groups are endomorphism 2distinguishable.
Proof. Let be a finite set of generators of that is closed under inversion. Since every element of can be represented as a product of finite length in elements of , we infer that is countable.
Also, if , then
Hence, every endomorphism is determined by the finite set
Because every is a word of finite length in elements of there are only countably many elements in . Hence is countable.
Now, let us consider the motion of the nonidentity elements of . Let be such an element and consider the set
It is easily seen that these elements form a subgroup of . Since does not fix all elements of it is a proper subgroup. Since its smallest index is two, the set is infinite. Thus is infinite. As was arbitrarily chosen, has infinite endomorphism motion.
By Corollary 5 we conclude that is 2distinguishable.
The next theorem shows that the endomorphism motion conjecture is true if , even if is not countable.
Theorem 7
Let be a connected graph with uncountable endomorphism motion. Then implies .
Proof. Set , and let be the smallest ordinal number whose underlying set has cardinality n. Furthermore, choose a well ordering of of order type , and let be the smallest element with respect to . Then the cardinality of the set of all elements of between and any other is smaller than .
Now we color all vertices of white and use transfinite induction to break all endomorphisms by coloring selected vertices black. By the assumptions of the theorem, there exists a vertex that is not fixed by . We color it black. This coloring breaks .
For the induction step, let . Suppose we have already broken all by pairs of vertices , where and have distinct colors. Clearly, the cardinality of the set of all , , is less than . By assumption, moves at least vertices. Since there are still n vertices not in , there must be a vertex that does not meet . If is white, we color black. This coloring breaks .
Corollary 8
Let be a connected graph with uncountable endomorphism motion. If the general continuum hypothesis holds, and if then .
Proof. By the generalized continuum hypothesis is the successor of . Hence, the inequality is equivalent to .
5 Examples and outlook
So far we have only determined the endomorphism distinguishing numbers of core graphs, such as the complete graph and odd cycles, and proved that for . Furthermore, it is easily seen that and if .
In the case of infinite structures we proved Theorem 6, which shows that for finitely generated, infinite groups .
We will now determine the endomorphism distinguishing numbers of finite and infinite paths and we begin with the following lemma.
Lemma 9
Let be an endomorphism of a (possibly infinite) tree such that for two distinct vertices . Then there exist two vertices on the path between and such that and .
Proof. Suppose dist. Hence dist. Let be the path connecting and in , and let be the subgraph induced by the image . Clearly, is a finite tree with at least one edge.
Because every nontrivial finite tree has at least two pendant vertices, there must be a pendant vertex of that is different from . Thus for some internal vertex of . If and are the two neighbors of on , then clearly and dist.
The above lemma implies the following corollary for finite graphs, because any injective endomorphism of a finite graph is an automorphism.
Corollary 10
Let be a finite tree. Then for every there exist two vertices of distance 2 such that .
Lemma 11
The endomorphism distinguishing number of all finite paths of order is two.
Proof. Clearly, since . To see that consider the following labeling
The only nontrivial automorphism of a path (symmetry with respect to the center) does not preserve this labeling. By Corollary 10, any other endomomorphism has to identify two vertices of distance two. Then cannot preserve the coloring, because any two vertices of distance two have distinct labels.
Next let us consider the ray and the double ray which can be viewed as an infinite analogs to finite paths. It turns out that their endomorphism distinguishing number is as well.
Lemma 12
The endomorphism distinguishing number of the infinite ray and of the infinite double ray is two.
Later in this section Theorem 15 will show that every countable tree with at most one pendant vertex has endomorphism distinguishing number two. Clearly Lemma 12 constitutes a special case of this result. It is also worth noting that by the following theorem every double ray has infinite endomorphism motion. Hence we verify the Endomorphism Motion Conjecture for the class of countable trees.
Theorem 13
An infinite tree has infinite endomorphism motion if and only if it has no pendant vertices.
The proof uses the following lemma which may be interesting as such. Note that in the statement of the lemma there is no restriction on the cardinality of the tree or the motion of the endomorphism.
Lemma 14
Let be a tree and let be an endomorphism of . Then the set of fixed points of induces a connected subgraph of .
Proof. Denote by Fix the set of fixed points of and assume that it does not induce a connected subgraph. Consider two vertices lying in different components of this graph.
Then maps the unique path in from to to a walk of the same length. But the only such walk is the path connecting and , so this path has to be fixed pointwise.
Proof of Theorem 13. Clearly, if an infinite tree has a pendant vertex, then there is an endomorphism which moves only this vertex and fixes everything else.
So let be a tree without pendant vertices and let be a nontrivial endomorphism of . Assume that the motion of is finite. Then the set of fixed points of contains all but finitely many vertices of . Since has no pendant vertices such a set does not induce a connected subgraph. This contradicts Lemma 14.
Now that we have characterised the trees with infinite endomorphism motion, we would like to show that all of them have endomorphism distinguishing number 2.
Theorem 15
The endomorphism distinguishing number of countable trees with at most one pendant vertex is 2.
Proof. The proof consists of two stages. First we color part of the vertices such that every endomorphism which preserves this partial coloring has to fix all distances from a given vertex . Then we color the other vertices in order to break all remaining endomorphisms.
For the first part of the proof, let be a pendant vertex of , or any vertex if is a tree without pendant vertices. Denote by the set of vertices at distance from , that is the sphere of radius with center . Now color white and all of and black. Periodically color all subsequent spheres according to the pattern outlined in Figure 2. In other words always color two spheres white, then four spheres black, leave two spheres uncolored, color another four spheres black and proceed inductively. Furthermore, we require that adjacent uncolored vertices are assigned different colors in the second step of the proof.
Now we claim that this coloring fixes in every endomorphism. To prove this consider a ray starting at . Clearly holds for every . Assume that there is a color preserving endomorphism of which does not fix and consider the image of the previously chosen ray under , that is, let . Clearly has to lie either in a white sphere or in a sphere which has not yet been colored. We will look at those cases and show that all of them lead to a contradiction. So assume that for some .

If , then must lie in since it must be a black neighbor of . For similar reasons and must hold. Now has to be a white neighbor of but only has black neighbors, a contradiction.

If we get and by the same argument as above. Now would need to be a white neighbor of but only has black neighbors.

If , where , then, for similar reasons as in the previous cases, and . Again has no white neighbors.

If , then lies in one of , and . In the first case clearly has no white neighbors. In the other cases it may have a white neighbor in , but then has no white neighbors, because its neighbor in must have a different color.

If , we can use an argument that is symmetric to the previous case.
Since there are no more cases left we can conclude that has to be fixed by every endomorphism which preserves this coloring.
However, we wish to prove that such endomorphism preserves all distances from , that is, that maps into itself for each .
We first show that any for must have its image in for some . Since is fixed, must be mapped to a vertex at even distance from . Furthermore, this vertex must be black and have a white neighbor, which again must have a white neighbor. It is easy to check that the only vertices for which all of this holds lie in for some .
Now assume that does not map into itself for every and consider the smallest such that . Then there must be some vertex such that . This immediately implies that and that , because otherwise a white vertex would be mapped to a black vertex or vice versa. In order to treat the remaining cases, consider a vertex whose predecessor in is , where is chosen to be minimal with respect to the properties , . The unique path in must be mapped to a walk with length at most . This implies that cannot lie in for . The path does not contain two consecutive white vertices, hence the walk cannot cross the two consecutive white layers and . So cannot lie in for . But this contradicts the fact that must lie in some for .
This completes the proof of the fact that all distances from are fixed by any endomorphism which preserves such a coloring.
For the second part of the proof, consider any enumeration of the vertices of such that, for all , we have for some . It is easy to see that such an enumeration is possible. Now color all vertices in whose predecessor is black and color all other vertices in this sphere white. Color the vertices of whose predecessor is white, and color all other vertices in this sphere black.
We claim that the so obtained coloring is not preserved by any endomorphism but the identity. We already know that every color preserving endomorphism maps every sphere into itself. Assume that there is a vertex which is not fixed by . Then it is easy to see that all vertices in whose predecessor is will be mapped to vertices whose predecessor is . Hence is not color preserving.
We conjecture that this result can be extended to uncountable trees. One does need a lower bound on the minimum degree though, see [10]. As we already noted, the fact that , together with the observations that and , supports the Endomorphism Motion Conjecture. Of course, a proof of the Endomorphism Motion Conjecture is still not in sight, not even for countable structures.
Finally, the computation of seems to be an interesting problem, even for finite cubes. Similarly, the computation of , where denotes the th Cartesian power^{1}^{1}1For the definition of the Cartesian product and Cartesian powers see [8]. of , looks demanding.
Acknowledgement We wish to thank one of the referees for numerous clarifying remarks and helpful comments.
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