Proof. Let $A:=\{1,a_{1},\ldots ,a_{s}\}$ be a left transversal of H in G. Assume that the number of involutions contained in $a_{i}H$ is $n_{i}$ for $1\leq i\leq s$ . Let $\sigma $ be a permutation on $\{1,\ldots ,s\}$ such that $a_{i}^{-1}H=a_{\sigma (i)}H$ . Since H is normal in G,

$$ \begin{align*} a_{\sigma^{2}(i)}H=a_{\sigma(i)}^{-1}H=Ha_{\sigma(i)}^{-1} =(a_{\sigma(i)}H)^{-1}=(a_{i}^{-1}H)^{-1}=Ha_{i}=a_{i}H. \end{align*} $$

It follows that $\sigma $ is the identity permutation or an involution. Assume that $\sigma $ fixes t integers in $\{1,\ldots ,s\}$ . Then $0\leq t\leq s$ and $s-t$ is even. Set $\ell :={(s-t)}/{2}$ . Without loss of generality, we assume that

$$ \begin{align*} \sigma(i)=\begin{cases} i & \text{if}~i\leq t, \\ i+\ell & \text{if}~t<i\leq t+\ell,\\ i-\ell & \text{if}~t+\ell<i\leq s.\\ \end{cases} \end{align*} $$

Then $a_iH$ is inverse closed if $i\leq t$ and $(a_{t+j}H)^{-1}=a_{t+j+\ell }H$ for every positive integer j not greater than $\ell $ . In particular, $n_{i}=0$ if $i>t$ . For every $i\in \{1,\ldots ,s\}$ , take a subset $X_i$ of $a_iH$ of cardinality $\tau $ by the following rules:

• • if $i\leq t$ and $n_i\geq \tau $ , then $X_i$ consists of exactly $\tau $ involutions;

• • if $i\leq t$ , $n_i<\tau $ and $\tau -n_i$ is even, then $X_i$ consists of $n_i$ involutions and ${(\tau -n_i)}/{2}$ pairs of mutually inverse elements of order greater than $2$ ;

• • if $i\leq t$ , $n_i<\tau $ and $\tau -n_i$ is odd, then $X_i$ consists of $n_i-1$ involutions and ${(\tau +1-n_i)}/{2}$ pairs of mutually inverse elements of order greater than $2$ ;

• • if $t<i\leq t+\ell $ , then $X_i$ consists of exactly $\tau $ elements of order greater than $2$ ;

• • if $i> t+\ell $ , then set $X_i=X_{i-\ell }^{-1}$ .

Note that $X_1,\ldots ,X_s$ are pair-wise disjoint. Set $X=\dot \bigcup _{i=1}^{s}X_{i}$ . Then X is an inverse closed subset of G satisfying $X\cap H=\emptyset $ and $\vert X\cap gH\vert =\tau $ for every $g\in G\setminus H$ . It follows that H is a $(0,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,X)$ and therefore a $(0,\tau )$ -regular set of G.

Proof. The sufficiency follows from [Reference Wang, Xia and Zhou9, Theorem 1.2]. Now we prove the necessity. Let H be a $(0,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,X)$ . Then $X=X^{-1}$ , $X\cap H=\emptyset $ and $\vert X\cap gH\vert =\tau $ for every $g\in G\setminus H$ . Let $A:=\{1,a_{1},\ldots ,a_{s}\}$ be a left transversal of H in G and set $X_i=X\cap a_iH$ for every $i\in \{1,2,\ldots ,s\}$ . Then $X=\dot \bigcup _{i=1}^{s}X_{i}$ . If $X_{i}$ contains an involution for each $i\in \{1,\ldots ,s\}$ , then H is a perfect code of G with respect to $\{1,y_1,\ldots ,y_s\}$ , where $y_i$ is an involution in $X_i$ , $i=1,\ldots ,s$ . Now we assume that there exists at least one integer $k\in \{1,\ldots ,s\}$ such that $X_{k}$ contains no involution. Then $x^{-1}\neq x$ for every element $x\in X_{k}$ . It follows that $|X_{k}\cup X_{k}^{-1}|$ is even. Since $|X_{k}|=\tau $ and $\tau $ is odd, we get $X_k\neq X_k^{-1}$ . Since H is normal in G, we obtain $(a_kH)^{-1}=(Ha_k)^{-1}=a_k^{-1}H$ . Assume that $a_k^{-1}H=a_{j}H$ for some $j\in \{1,\ldots ,s\}$ . Then $X_{k}^{-1}\subseteq a_{j}H$ . Since $X=\dot \bigcup _{i=1}^{s}X_{i}$ and $X^{-1}=X$ , we conclude that $X_k^{-1}=X_{j}$ . Therefore, without loss of generality, we can assume that $X_i^{-1}=X_{i+\ell }$ if $1\leq i\leq \ell $ and $X_i^{-1}=X_{i}$ if $2\ell < i\leq s$ , where $\ell $ is a positive integer not greater than ${s}/{2}$ . Note that $X_{i}$ contains at least one involution if $X_i^{-1}=X_{i}$ (as it is of odd cardinality). For every $i\in \{1,\ldots ,s\}$ , take an element $y_i\in X_i$ by the following rules:

• • $y_i$ is an arbitrary element in $X_i$ if $i\leq \ell $ ;

• • $y_i=y_{i-\ell }^{-1}$ if $\ell <i\leq 2\ell $ ;

• • $y_i$ is an involution if $i>2\ell $ .

Then H is a perfect code of G with respect to $\{1,y_1,\ldots ,y_s\}$ .

Proof. ( $\Rightarrow $ ) Let H be a $(\kappa ,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,X)$ . Then $|H\cap X|=\kappa $ and $|gH\cap X|=\tau $ for every $g\in G\setminus H$ . Set $Y=X\setminus H$ . Then $|H\cap Y|=0$ and $|gH\cap Y|=\tau $ for every $g\in G\setminus H$ . Since $X^{-1}=X$ and $H^{-1}=H$ , we get $Y^{-1}=Y$ . It follows that H is a $(0,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,Y)$ and therefore a $(0,\tau )$ -regular set of G.

( $\Leftarrow $ ) Let H be a $(0,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,Y)$ . Then $|H\cap Y|=0$ and $|gH\cap Y|=\tau $ for every $g\in G\setminus H$ . Let m be the number of elements contained in H of order greater than $2$ . Then m is even and the number of involutions contained in H is $|H|-1-m$ . Recall that $0\leq \kappa \leq |H|-1$ and $\gcd (2,|H|-1)\mid \kappa $ . If $\kappa $ is odd, then $|H|$ is even and therefore contains at least one involution. Take an inverse closed subset Z of H of cardinality $\kappa $ by the following rules:

• • if $m\geq \kappa $ and $\kappa $ is even, then Z consists of exactly ${\kappa }/{2}$ pairs of mutually inverse elements of order greater than $2$ ;

• • if $m\geq \kappa $ and $\kappa $ is odd, then Z consists of ${(\kappa -1)}/{2}$ pairs of mutually inverse elements of order greater than $2$ and one involution;

• • if $m<\kappa $ , then Z consists of ${m}/{2}$ pairs of mutually inverse elements of order greater than $2$ and $\kappa -m$ involutions.

Set $X=Y\cup Z$ . Then $|H\cap X|=\kappa $ and $|gH\cap X|=\tau $ for every $g\in G\setminus H$ . Therefore, H is a $(\kappa ,\tau )$ -regular set of the Cayley graph $\mathrm {Cay}(G,X)$ and therefore a $(\kappa ,\tau )$ -regular set of G.

Proof of Theorem 1.1.

Lemmas 2.1 and 2.3 imply that H is a $(\kappa ,\tau )$ -regular set of G if $\tau $ is even. Now assume $\tau $ is odd. By Lemmas 2.2 and 2.3, H is a $(\kappa ,\tau )$ -regular set of G if and only if it is a perfect code of G.