Cayley graphs of groups

466 Views Asked by At

(Directed & undirected) Cayley graphs of groups have been studied a lot in the literature. I would like to know the answer to the following questions. Please give your valuable suggestions.

  1. Is there any result which characterize the given group by using its Cayley graph ?'.

i.e., Let Cay(G,S) be a Cayley graph of a group G, where S is a Cayley subset of G. If G' is a group, T is a Cayley subset of G' such that Cay(G',T) $\cong$ Cay(G,S), then G' $\cong$ G.

  1. Is there any result which characterize 'some properties' of the given group by using its Cayley graph ?'.
1

There are 1 best solutions below

0
On BEST ANSWER
  1. The answer to the first question is positive, so long as you let your Cayley graph be directed and colored. This later condition means that, supposing your generating set is $S$, then for each $s \in S$ there is a directed edge between $x$ and $sx$ of "color" s. Then the automorphisms of a (connected) Cayley graph that preserve direction and edge color is exactly the original group - the crux of this is the observation that there is a unique colored automorphism taking any one vertex to another, since once that point is chosen the rest of the map can be defined inductively using the coloring and direction, which corresponds to right multiplication by some element of the group.

This is explained in more detail here: https://terrytao.wordpress.com/2010/07/10/cayley-graphs-and-the-geometry-of-groups/

Some answers to your second question are also given geometric interpretation in that very interesting blog post.

But without coloring non-isomorphic groups can have isomorphic Cayley graphs. For example if $G$ and $G'$ are non-isomorphic group of the same order, and the generating sets are taken to be the full groups in both cases, then the resulting Cayley graphs are always the complete graphs on $|G|$ vertices. The coloring would have remembered the multiplication law (follow the edge colored s from edge a to edge sa, then edge colored t from sa to tsa, then the color of the path from a to tsa is colored (ts)), but without coloring that information is lost (how do we know which edge from sa is "t"?).

  1. Yes. Again in the colored Cayley graph case, you can tell a lot more because the group is preserved up to isomorphism. Even with just the Cayley graph Cay(G,T) you can tell some things about the generating set T... for instance in this context people study the rate of growth of metric balls. There are connections to representation theory, because the natural action of $G$ on the graph $Cay(G,T)$ gives rise to an action on the Hilbert space $L^2(Cay(G,T))$ (when the graph is finite at least, this is the vector space consisting of functions from the vertex set to the complex numbers.) Anyway I am sure that this part of your question is very broad.

(Or anyway to copy more from that blog post so that I learn it better, the concept of a normal subgroup appears geometrically when thinking about Cayley graphs. Suppose $(G,S)$ is a group and a a generating set, and a subgroup $G'$ is generated by $S' \subset S$. Then the Cayley graph $Cay(G,S)$ consists of connected copies of $Cay(G',S')$ (corresponding to the cosets $G'a$). When we put back generators from $S \setminus S'$, these connected components start connecting to each other... if adding a single $s \in S \setminus S'$ connects each component to exactly one other component, then this means that $sG'a = G' b$, which means that multiplying on the left by generators in $S$ maps right cosets of $G'$ to right cosets, which is the same as normality ($x G' x^{-1} = G' b x^{-1}$, and the latter is a coset of $G'$ containing 1, so is $G'$... while in the other direction if $G'$ is normal, then $aHb = aHa^{-1} a b = Hab$.))