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\big||g|,r\neq p\mathrm{;is;a;prime}}.

Then $\lang T\rang\leq N$ and $\lang T\rang\space\mathrm{char}\space G$. Let $\overline G=G/\lang T\rang$. Then $\overline G$ is a REA-group relative to $N/\lang T\rang$. If $G$ is not a $p$-group, then $\lang T\rang$ is non-trivial. By $G$ minial counter-example, $N/\lang T\rang\lhd G/\lang T\rang$. Then $N\lhd G$, contradiction. Thus $G$ is a $p$-group. Let $Z(G)$ be the center of $G$. Then $1\neq Z(G)\space\mathrm{char}\space G$. Suppose $(G\backslash N)\cap Z(G)\neq\empty$. 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 $G\backslash N\cap Z(G)=\empty$ and $Z(G)\subset N$. Now $G/Z(G)$ is a REA-group relative to $N/Z(G)$ and $N/Z(G)\lhd G/Z(G)$, which is also impossible. $\blacksquare$ **Rmk.** $N$ is not necessarily a characteristic group. For example: ![[Pasted image 20240430191356.png]] **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 $G'$ is nontrivial (proof later). We claim that $G'\leq N$. Otherwise $G\backslash N\cap G'\neq\empty$. Since $G'$ is a characteristic subgroup, every element of $G\backslash N$ lies in $G'$, namely, $G\backslash N\subset G'$. Thus, $G=N\cup (G\backslash N)=N\cup G_0$, which is impossible. Hence, $G'\leq N$ 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))‘=G’Z(G)/Z(G)\subsetneq G/Z(G).

Hence $G'Z(G)\subsetneq G$, so $G'\subsetneq G$. **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 $G'\leq H$ 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|\neq p^e}

is a subset of the subgroup $N$, and $M:=\lang T\rang\leq N$. Clearly, $T$ generates a characteristic subgroup of $G$. By Lemma, $G/\lang T\rang$ is an REA group relative to $N/\lang T\rang$. Since $T$ contains all $p'$-elements of $G$, so does $N\geq \lang T\rang$. 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. $\blacksquare$