Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Gamma=Cay(G,S)$ be a Cayley graph over a group $G$, $H$ be a proper subgroup of $G$ and $\Sigma=Cay(H,T)$ where $S$ and $T$ are inversed-closed subsets of $G$ and $H$ not containing idendity, respectively. Is there any relation between $Aut(\Gamma)$ and $Aut(\Sigma)$ in general? When $T=H\cap S$, what can we say?

share|improve this question
    
I want a relation according to decomposition of groups (direct product, ...) –  majid arezoomand Feb 28 '13 at 15:22
    
For example, If $G=\Bbb Z_4=<a>$, $H=<a^2>$, $S=\{a,a^3}$ and $T=H\cap S=\emptyset$, then $Aut(\Gamma)=S_2\wr S_2$ and $Aut(\Sigma)=S_2$ where $\wr$ denotes the wreath product of groups and $S_2$ is the symmetric group of 2 letters. –  majid arezoomand Feb 28 '13 at 15:26

2 Answers 2

up vote 0 down vote accepted

I am reasonably confident that the real answer to your question is "No". For example, if $T=H\setminus 1$ then the automorphism group of $\mathrm{Cay}(H,T)$ is the full symmetric group. For another, take $S=T$; then $\Gamma$ will have a large automorphism graph, a wreath product. (If you want your Cayley graphs connected, take complements.) So adding generators may reduce symmetry, or increase it.

Note that questions phrased as you have are a bit difficult to answer, since the answer depends on what you mean by "any relation between".

share|improve this answer
    
Dear Prof. Godsil, thanks for your answer. I guess that when $T=H\cap S$, then one can give a decomposition of $Aut(\Gamma)$ (such as semidirect product, wreath product,...) which one of its factors is $Aut(\Sigma)$. Is this guess true? –  majid arezoomand Feb 28 '13 at 20:17
    
It is not true. As an extreme case, take a Cayley graph for a simple group where the connection set contains an involution and the vertex stabilizer of the graph is trivial. Let $T$ be the involution and $H$ be the subgroup it generates; then there is no decomposition of the automorphism group. –  Chris Godsil Feb 28 '13 at 21:22
    
Thanks so much. –  majid arezoomand Mar 1 '13 at 5:09

I don't know how interested you are in the kinds of things that happen in infinite groups, but this question does have some general interest in that setting.

It seems to be quite common, for example, that $G$ has finite index in the automorphism group of its Cayley graph. This is true, for example, for many examples of lattices in Lie groups, as Alex Furman has noted. For a very specific example, this is true for the fundamental group of a closed surface of genus $g$ with the standard presentation $$\langle a_1,b_1,\ldots,a_g b_g \quad | \quad [a_1,b_1] \ldots [a_g,b_g] \rangle $$

At a different extreme, for free groups with their standard generating set, the automorphism group of the Cayley graph is a locally compact topological group locally homeomorphic to the Cantor set, and hence the free group has uncountable index.

share|improve this answer
    
Thanks a lot for your responsibility. I am interest to finite groups. –  majid arezoomand Mar 1 '13 at 5:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.