# What's wrong with my attempt at proving that the automorphisms of a colored Cayley graph of a group are isomorphic to that group?

I'm aware of the existence of this question, but its answer handles the proof differently from how I'm attempting to do so, so I'm posting this one.

I want to show that if $g, g_1 \in G$ where $G$ is a group we have a map $\phi_g: G \mapsto G$ given by $\phi(g_1) = g g_1$, then $\phi$ preserves color when treated as an automorphism of the colored Cayley graph corresponding to $G$ with some set of genereators $S$. To prove that, I'm using that vertices $g_1$ and $g_2$ are adjacent when $g_1 g_2^{-1} \in S$, and the edge is colored $k$ such that $g_1 g_2^{-1} = s_k \in S$. To show $\phi_g$ is color-preserving, I need to show that $\phi_g(g_1)$ $\phi_g(g_2)^{-1} = s_k$ as well. Substituting, and using the group product inverse law, we obtain $g g_1 g_2^{-1} g^{-1} = g s_k g^{-1}$ which only is $s_k$ when the group is Abelian. I know that this result holds for all groups, so where did I go wrong?

• Do you mean isomorphic, or does the group actually have a metric? – Matt Samuel Aug 28 '16 at 23:20
• Yes, I do -- my mistake. – SquarerootSquirrel Aug 28 '16 at 23:20
• I'm really stumped here and could use some help! – SquarerootSquirrel Aug 29 '16 at 0:17
• If the edges in your Cayley graph correspond to left multiplication (ie edges $(g,s_kg)$) then you should construct automorphisms using right multiplication (ie $\phi_g(g_1)=g_1g$). Otherwise $\phi_g$ won't preserve the edges, as you've found. – stewbasic Aug 29 '16 at 3:46