Cayley graph with opposite action: is the group abelian?

Question

Let $G$ be a group, let $S$ be a set of generators and let $\Gamma=\Gamma(G,S)$ be the Cayley graph, where there is an edge between $g$ and $h$ if and only if $h=gs$ for some $s\in S$.

• We know that the action of $G$ on $\Gamma$ via left multiplication acts on $\Gamma$ via graph automorphisms: if $h=gs$, then for each $a\in G$, $$ ah = a(gs) = (ag)s , $$ and so $ag$ ahd $ah$ are connected by an edge whenever $g$ and $h$ are.
• Conversely, the action of $G$ on $\Gamma$ via right multiplication does not in general induce graph automorphisms: $$ ha = gsa \ne ags , $$ unless $G$ is abelian.

Now, is the converse true? If the action of $G$ via right multiplication induces graph automorphisms, can we conclude that $G$ is abelian?

Answer 1

To add a bit to kabenyuk's answer, it's actually easy to show that this is equivalent to the connection set being a union of conjugacy classes. You want $g\sim h\Longleftrightarrow ga\sim ha$ for all $g,h,a\in G$. But this is equivalent to $g^{-1}h\in S\Longleftrightarrow a^{-1}g^{-1}ha\in S$, which is equivalent to $S$ being closed under conjugation.

(Such Cayley graphs are sometimes called normal Cayley graphs, although that terminology can also mean something else.)

Answer 2

No, this is wrong. Let us take the union of several conjugacy classes in the non-Abelian group G as a generating set: $S=a_1^G\cup\ldots\cup a_k^G$. It is easy to see that the action of $G$ via right multiplication induces graph automorphisms. For instance, $G=S_3$, $S=\{(12),(12),(23)\}$.