7 From graph to normal edge-transitive Cayley graph

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Find group s.t. Cayley graph is $K_{m[b]}$

Problem. Express $K_{m[b]}=K_{b,\cdots ,b}$ as a normal edge-transitive Cayley graph.

An easier problem

First, try to express $K_n$ as a normal edge-transitive Cayley graph. Suppose $\Gamma =K_n=\mathrm{Cay}(G,G\backslash \{1\})$ and $\Gamma$ is normal edge-transitive. Then $\mathrm{Aut}(G,G\backslash \{1\})$ is transitive on $G\backslash \{1\}$.

Recall:

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.

And we know $G$ is elementary abelian.

Back to the problem

Now suppose $\Gamma =\mathrm{Gay}(G,S)=K_{m[b]}$. Define $N$ as $N=G\backslash S$. Note that the complement graph of $\Gamma$ is also a Cayley graph, i.e. $\overline{\Gamma }={\cup }_{i=1}^m{K}_b^{(i)}$. Thus $N$ is a subgroup with $b=|N|$. Assume $\Gamma$ is normal arc-transitive, then $\mathrm{Aut}(G,S)$ is transitive on $S=G\backslash N$. Hence elements of $G\backslash N$ have the same order and are conjugate in a subgroup of $\mathrm{Aut}(G)$. Simply, we say $G$ is a REA-group relative to $N$.

Def. A finite group $G$is a relative elementary abelian group if $G$ has a subgroup $N$ such that all elements of $G\backslash N$ are conjugate in a subgroup of $\mathrm{Aut}(G)$.

Properties of REA-group

Problem. Characterize REA-groups. Study $G$ and $N$ such that $\mathrm{Aut}(G)$ is transitive on $G\backslash N$ and study $\mathrm{Aut}(G)$ and its subgroups which are transitive on $G\backslash N$.

Lemma. Let $G$ be a REA-group relative to $N$. Then:

• each element of $G\backslash N$ is of order $p^e$ for some prime $p$ and integer $e$, which is independent of the choice of elements;
• an element of $G\backslash N$ does not centralize an element of $G$ of order coprime to $p$.

Proof. i) Let $x\in G\backslash N$. If $|x|=kl$ with $(k,l)=1$, then $x^k,x^l$ and $x$ do not have the same order and so $x^k,x^l\in N$. Then $(k,l)=1$ yields $x\in N$, which is impossible. Hence $|x|=p^e$ for some prime $p$. Since all elements in $G\backslash N$ have the same order, $|y|=|x|=p^e$ for all $y\in G\backslash N$.

ii) If $x$ centralize $z\in G$ such that $(|x|,|z|)=1$, then $|x||z|=|xz|>|x|$ and $xz\notin N$, not possible. $■$

Cor. If $G$ is abelian and REA, then $G$ is a $p$-group.

Proof. By the lemma above, if order of elements of $G\backslash N$ is $p^e$ and order of an element $h$ of $N$ is coprime to $p$, then $h^g\ne h$ for all $g\in G\backslash N$, contradiction.

nilpotent + REA ⇒ p-group

Lemma. Let $M\mathrm{char}G$ such that $M\le N$. Then $G/M$ is a REA-group relative to $N/M$.

Proof. Easy. Only need to show elements in $(G/M)\backslash (N/M)$ are conjugate in $\mathrm{Aut}(G/M)$.

Lemma. If $G$ is a REA-group relative to $N$, then $N⊲G$.

Proof. Let $G$ be a minimal counter-example. Then $N$ is not normal in $G$. Since all elements of $G\backslash N$ have order $p^e$, define $T=\{g\in G:r||g|,r\ne p\mathrm{is}\mathrm{a}\mathrm{prime}\}.$Then $⟨T⟩\le N$ and $⟨T⟩\mathrm{char}G$. Let $\overline{G}=G/⟨T⟩$. Then $\overline{G}$ is a REA-group relative to $N/⟨T⟩$.

If $G$ is not a $p$-group, then $⟨T⟩$ is non-trivial. By $G$ minial counter-example, $N/⟨T⟩⊲G/⟨T⟩$. Then $N⊲G$, contradiction. Thus $G$ is a $p$-group.

Let $Z(G)$ be the center of $G$. Then $1\ne Z(G)\mathrm{char}G$. Suppose $(G\backslash N)\cap Z(G)\ne \varnothing$. Then elements of $G/N$ are conjugate to some elements of center and so $G\backslash N\subset Z(G)$. Thus $G=Z(G)\cup N$, which is impossible.

Fact: A finite group cannot be written as a union of two proper group. For example, suppose $G=H\cup K$ and for any $h\in H,k\in K$, consider $hk$.

Hence $G\backslash N\cap Z(G)=\varnothing$ and $Z(G)\subset N$. Now $G/Z(G)$ is a REA-group relative to $N/Z(G)$ and $N/Z(G)⊲G/Z(G)$, which is also impossible. $■$

Rmk. $N$ is not necessarily a characteristic group. For example:

Lemma. $G/N=Z_p^d$.

Proof. For any $g\in G\backslash N$, if $|g|=p^e$, then $g^p\in N$. Thus each non-identity element of $G/N$ is of order $p$.

Assume first that $G$ is a $p$-group. Then the commutator subgroup $G^{\prime }$ is nontrivial (proof later). We claim that $G^{\prime }\le N$. Otherwise $G\backslash N\cap G^{\prime }\ne \varnothing$. Since $G^{\prime }$ is a characteristic subgroup, every element of $G\backslash N$ lies in $G^{\prime }$, namely, $G\backslash N\subset G^{\prime }$. Thus, $G=N\cup (G\backslash N)=N\cup G_0$, which is impossible. Hence, $G^{\prime }\le N$ and $G/N$ is abelian. Therefore, $G/N=Z_p^d$ for some positive integer $d$.

Fact: $p$-group has non-trivial commutator is non-trivial.

Proof 1. Suppose $G$ is a $p$-group, then $Z(G)$ is non-trivial. Then $G/Z(G)$ has definitely an order smaller than that of $G$. By induction, $(G/Z(G){)}^{\prime }=G^{\prime }Z(G)/Z(G)⊊G/Z(G).$Hence $G^{\prime }Z(G)⊊G$, so $G^{\prime }⊊G$.

Proof 2. Using the following proposition:

• every $p$-group is nilpotent
• a finite group is nilpotent iff all its maximal proper subgroup is normal
• a proper maximal subgroup of a finite group has prime index

and we get a proper maximal subgroup of a $p$-group is normal and has index $p$. Suppose $G$ is a $p$-group and $H$ is a proper maximal subgroup. Then $G/H$ is of order $p$ and so it is cyclic. Then $G^{\prime }\le H$ is a non-trivial subgroup of $G$.

Suppose now that $G$ is not a $p$-group. Since the elements in $G\backslash N$ have the same order $p^e$, each element of $G$ of order not equal to $p^e$ lies in the normal subgroup $N$. Thus, the set $T=\{g\in G:|g|\ne p^e\}$is a subset of the subgroup $N$, and $M:=⟨T⟩\le N$. Clearly, $T$ generates a characteristic subgroup of $G$. By Lemma, $G/⟨T⟩$ is an REA group relative to $N/⟨T⟩$. Since $T$ contains all $p^{\prime }$-elements of $G$, so does $N\ge ⟨T⟩$. So, the factor group $G/N$ is a $p$-group. Therefore, the factor group $(G/M)/(N/M)$ is an elementary abelian $p$-group by the previous paragraph. So is $G/N$, because $G/N\cong (G/M)/(N/M)$ is elementary abelian, completing the proof. $■$