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?

  • $\begingroup$ Do you mean isomorphic, or does the group actually have a metric? $\endgroup$ – Matt Samuel Aug 28 '16 at 23:20
  • $\begingroup$ Yes, I do -- my mistake. $\endgroup$ – SquarerootSquirrel Aug 28 '16 at 23:20
  • $\begingroup$ I'm really stumped here and could use some help! $\endgroup$ – SquarerootSquirrel Aug 29 '16 at 0:17
  • 3
    $\begingroup$ 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. $\endgroup$ – stewbasic Aug 29 '16 at 3:46

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