A subgroup of :
Def. Let .
Lemma. .
Proof. For any and any , is still an edge.
Lemma. , where is a subgroup of induced from .
Proof. For any , and , . Then and so $$ \check G\lhd\check G:\mathrm{Aut}(G,S).
Note that $\check G$ acts on $G$ regularly and $\mathrm{Aut}(G,S)$ fixes the identity of $V=G$. Thus $\check G\cap \mathrm{Aut}(G,S)=\{1\}$. Therefore the proof is completed.$\blacksquare$ ### Normalizer and centralizer of $\check G$ Recall a lemma in previous section: > **Lemma.** For any group $G$, the following equations hold: > - $N_{\mathrm{Sym}(G)}(\check G)=\check G:\mathrm{Aut}(G)$ > - $\check GC_{\mathrm{Sym}(G)}(\check G)=\check G\circ \check G=\check G:\mathrm{Inn}(G).$ Then we get: **Lemma.** Suppose $\Gamma=\mathrm{Cay}(G,S)$. Then - $N_{\mathrm{Aut}(\Gamma)}(\check G)=\check G:\mathrm{Aut}(G,S)$ - $\check GC_{\mathrm{Aut}\Gamma}(\check G)=\check G:\mathrm{Inn}(G,S).$ **Proof.** Consider the intersection of $\mathrm{Aut}(\Gamma)$ and $N_{Sym(G)}(\check G)$, $C_{Sym(G)}(\check G)$. $\blacksquare$ ### Normal Cayley group **Def.** Let $\Gamma=\mathrm{Cay}(G,S)$. Then - $\Gamma$ is called *normal edge-transitive* if $N=N_{\mathrm{Aut}\Gamma}(\check G)$ is transitive on the edge of $\Gamma$. - $\Gamma$ is called *normal Cayley graph* if $\mathrm{Aut}\Gamma=N$. **Lemma.** Let $\Gamma=\mathrm{Cay}(G,S)$. Then - $\mathrm{Aut}(G,S)$ is transitive on $S$ iff $\Gamma$ is normal arc-transitive. - The Cayley graph $X=\mathrm{Cay}(G,S)$ is normal edge-transitive but not arc-transitive if and only if $\mathrm{Aut}(G,S)$ has two orbits in $S$ in the form of $T$ and $T^{-1}$, where $T$ is a non-empty subset of $S$ and $S=T\cup T^{-1}$. - $N$ is 2-arc transitive on $\Gamma$ iff $\mathrm{Aut}(G,S)$ is 2-transitive on $S$. In addition, all elements of $S$ are involution in this case. - $N$ is not 3-arc transitive on $\Gamma$. **Proof.** i) One direction is easy. Now suppose $\Gamma$ is normal arc-transitive. Then $N=\check G:\mathrm{Aut}(G,S)$ is transitive on arc set. For any arcs $(1,s_1)$ and $(1,s_2)$, there is a $x\in N$ mapping $(1,s_1)$ to $(1,s_2)$. Since $x\in N_1$, $N_1$ is transitive on $S$. Note that $N_1\subset \mathrm{Aut}(G,S)$, thus $\mathrm{Aut}(G,S)$ is transitive on $S$. ii) Assume $N$ acting on $S$ is edge-transitive but not arc-transitive. Then by i) there exist $s_1,s_2\in S$ such that $s_1$ and $s_2$ are in the different orbits. By $N$ is edge-transitive, there is $g\in N$ such that $s_1^g=1,1^g=s_2$. Take any $s\in S$. If $s\notin s_1^{\mathrm{Aut}(G,S)}$, then there is $g_1\in N$ such that $s^{g_1}=1,1^{g_1}=s_1$ by $N$ edge-transitive. Then $1^{g_1g}=1$ yields $g_1g\in\mathrm{Aut}(G,S)$ and $s^{g_1g}=s_2$. Thus $s\in s_2^{\mathrm{Aut}(G,S)}$. Therefore, $\mathrm{Aut}(G,S)$ acting on $S$ has two orbits, denoted as $S=\Delta_1\cup\Delta_2$. Take any $s\in\Delta_1,t\in\Delta_2$. By $N$ edge-transitive on $X$, there exists $\hat gh\in N$ satisfying $\{1,s\}^{\hat gh}=\{1,t\}$. If $g^h=1$, then $g=1$ and $s^h=t$, contradiction. Thus $(sg)^h=1$ and $g=s^{-1}$, which derives $(s^{-1})^h=t$ and so $\Delta_1^{-1}\subset \Delta_2$. As $\Delta_1$ and $\Delta_2$ are symmetric, $\Delta_1^{-1}=\Delta_2$, then $S=\Delta_1\cup \Delta^{-1}_1$. Conversely, if $\mathrm{Aut}(G,S)$ acting on $S$ has two orbits which can be written as $S=T\cup T^{-1}$, then $N$ is not arc-transitive by i). For any $s\in T,s^{-1}\in T^{-1}$, there is $\hat s^{-1}\in\hat G$ such that $\{1,s\}^{\hat s^{-1}}=\{1,s^{-1}\}$. Therefore, $N$ acting on $X$ is edge-transitive. iii) Assume $N$ is 2-arc transitive on $\Gamma$. For any tuples $(s_1,s_2)$, $(s_3,s_4)$ where $s_i\in S$, consider 2-routes $1-s_1-s_1s_2$ and $1-s_3-s_3s_4$. Since there is $n\in \mathrm{Aut}(G,S)\subset N$ mapping one of them to another, $\mathrm{Aut}(G,S)$ is 2-transitive on $S$. The converse direction is similar. If there is $s\in S$ such that $s^2\neq id$, then $N$ acting on subset $\{1-s-ss,1-s_1-s_1s\}$ is not transitive. iv) Consider 3-routes $1-s_1-s_1s_2-s_1s_2s_3$ and $1-s_1-s_1s_2-s_1s_2s_1$. $\blacksquare$ **Exm.** Let $\Gamma=K_n=\mathrm{Cay}(G,S)$. Then $S=G\backslash\{id\}$ and $\mathrm{Aut}(G)$ acting on $G\backslash \{id\}$ is transitive. Then $G=\mathbb Z_p^d$ where $p$ is a prime. See [here.](https://math.stackexchange.com/questions/1404814/action-of-rm-autg-on-g#:~:text=Since%20the%20action%20of%20$mathrm%20%7BAut%7D%20%28G%29$%20is,every%20non-trivial%20element%20of%20$G$%20has%20order%20$p$.) ![[Pasted image 20240209160521.png]]