Commuting graph of a group on a transversal

By hypothesis, there are isomorphisms $\varphi:G/Z(G)\rightarrow H/Z(H)$ and $\psi:G^{\prime}\rightarrow H^{\prime}$ such that $\psi(\alpha_{G}(xZ(G),yZ(G)))=\alpha_{H}(\varphi(xZ(G)),\varphi(yZ(G)))$, for all $x,y\in G$. Let $T(G)=\{1,t_{1},\ldots,t_{n}\}$ be a transversal of $Z(G)$ in $G$, and, for each $i\in\{1,\ldots,n\}$, let $y_{i}$ be a representative of the coset $\varphi(t_{i}Z(G))$. Therefore $T(H)=\{1,y_{1},\ldots,y_{n}\}$ is a transversal of $Z(H)$ in $H$. If we put $T(G)\setminus\{1\}$ as the vertex set of ${\cal T}(G)$ and $T(H)\setminus\{1\}$ as the vertex set of ${\cal T}(H)$, the mapping $t_{i}\longmapsto y_{i}$ defines a bijection between $V({\cal T}(G))$ and $V({\cal T}(H))$. In addition, if $t_{i}$ e $t_{j}$ are adjacent in ${\cal T}(G)$, then $[y_{i},y_{j}]=\psi([t_{i},t_{j}])=1$, that is, $y_{i}$ and $y_{j}$ are adjacent in ${\cal T}(H)$. On the other hand, if $[y_{i},y_{j}]=1$, we have $\psi([t_{i},t_{j}])=1$. Since $\psi$ is an isomorphism, we obtain $[t_{i},t_{j}]=1$. Therefore ${\cal T}(G)$ and ${\cal T}(H)$ are isomorphic. ∎

Let $x$ be a vertex of ${\cal T}(G)$. If $N(x)=\{x_{1},\ldots,x_{n}\}$ is the neighborhood of $x$ in ${\cal T}(G)$ and if $\mathcal{A}:=\bigcup_{i=1}^{n}x_{i}Z(G)$, it is easy to verify that $C_{G}(x)=Z(G)\cup xZ(G)\cup\mathcal{A}$, so that deg$(x)=n=[C_{G}(x):Z(G)]-2$. This proves (i).

Let $x$ and $y$ be distinct vertices of ${\cal T}(G)$ and put $N=N(x)\cap N(y)$ and $\mathcal{B}:=\bigcup_{a\in N}aZ(G)$. We note that $\{x,y\}\cap N=\emptyset$. If $x$ and $y$ are adjacent in ${\cal T}(G)$, then $\{x,y\}\subseteq C_{G}(x)\cap C_{G}(y)$ and, in this case, we obtain $C_{G}(x)\cap C_{G}(y)=Z(G)\cup xZ(G)\cup yZ(G)\cup\mathcal{B}$. It follows that $|N|=[C_{G}(x)\cap C_{G}(y):Z(G)]-3$, which proves (ii). If $x$ and $y$ are not adjacent in ${\cal T}(G)$, then $x\notin C_{G}(y)$ and $y\notin C_{G}(x)$ and it is easy to see that $C_{G}(x)\cap C_{G}(y)=Z(G)\cup\mathcal{B}$. Therefore $|N|=[C_{G}(x)\cap C_{G}(y):Z(G)]-1$ and part (iii) is proved. ∎

If $G=D_{8}$ or $G$ is the quaternion group of order $8$, the result is clearly true. Consider an extraspecial $2$-group $G$ of order $m$ and suppose that the result is true for all extraspecial $2$-group of order less than $m$. By [15, 5.3.8], $G$ contains subgroups $A$ and $B$ such that $A$ and $B$ are extraspecial $2$-groups of order less than $m$ and $G=AB$ is a central product. Hence, the elements of $A$ commute with the elements of $B$ and $A\cap B=Z(G)=Z(A)=Z(B)$. By induction hypothesis, the following two conditions are satisfied:

(i) $C_{A}(a_{1})\neq C_{A}(a_{2})$, for all $a_{1},a_{2}\in A$ such that $a_{1}\notin a_{2}Z(A)$.

(ii) $C_{B}(b_{1})\neq C_{B}(b_{2})$, for all $b_{1},b_{2}\in B$ such that $b_{1}\notin b_{2}Z(B)$.

Let $x=a_{1}b_{1}$ and $y=a_{2}b_{2}$ be non-central elements of $G$ such that $x\notin yZ(G)$, where $a_{1},a_{2}\in A$ and $b_{1},b_{2}\in B$. Thus, in virtue of (i) and (ii), we must have $C_{A}(a_{1})\neq C_{A}(a_{2})$ or $C_{B}(b_{1})\neq C_{B}(b_{2})$. In the first situation, there is $w\in C_{A}(a_{1})\setminus C_{A}(a_{2})$ (or there is $u\in C_{A}(a_{2})\setminus C_{A}(a_{1})$) and it follows that $w\in C_{G}(x)\setminus C_{G}(y)$ (or $u\in C_{G}(y)\setminus C_{G}(x)$), which gives us $C_{G}(x)\neq C_{G}(y)$. Analogously, $C_{B}(b_{1})\neq C_{B}(b_{2})$ implies $C_{G}(x)\neq C_{G}(y)$, as required. ∎

Let $x,y\in G\setminus Z(G)$ be such that $x\notin yZ(G)$. By Lemma 3.1, we have $C_{G}(x)\neq C_{G}(y)$ and, by Theorem 2.7, $G$ is of conjugate type $\{1,2\}$. This shows that $[C_{G}(x):C_{G}(x)\cap C_{G}(y)]\geq 2$. Write $G^{\prime}=\{1,d\}$. Given $g,h\in C_{G}(x)\setminus C_{G}(y)$, we have $gy=ygd$, $h^{-1}y=yh^{-1}d$ and so $(gh^{-1})y=y(gh^{-1})$, that is, $gh^{-1}\in C_{G}(x)\cap C_{G}(y)$. Hence $[C_{G}(x):C_{G}(x)\cap C_{G}(y)]=2$. Applying Proposition 2.5 to calculate the degree of each vertex of ${\cal T}(G)$ and the size of the common neighborhood of two vertices, we conclude that ${\cal T}(G)$ is strongly regular with parameters $(2^{2n}-1,2^{2n-1}-2,2^{2n-2}-3,2^{2n-2}-1)$. Since $2^{2n-2}-1\neq 0$, Lemma 3.2 give us ${\cal T}(G)$ is connected. ∎

Suppose there is a non-abelian group $G$ such that ${\cal T}(G)$ is a connected strongly regular graph with parameters $(v,k,\lambda,\mu)$ and that there are $x,y\in G\setminus Z(G)$ such that $x\notin yZ(G)$ and $C_{G}(x)=C_{G}(y)$. For all $g\in xZ(G)$ and $h\in yZ(G)$ we have $[g,h]=[x,y]=1,$ $C_{G}(g)=C_{G}(x)$ and $C_{G}(h)=C_{G}(y)$. Hence, we may assume that $x,y$ are vertices of ${\cal T}(G)$. From Proposition 2.5, it follows that if $n=[G:Z(G)]$ and $m=[C_{G}(x):Z(G)]$, then $v=n-1$, $k=m-2$ and $\lambda=m-3=k-1$. By Lemma 3.2 the graph ${\cal T}(G)$ is disconnected, a contradiction. ∎

Suppose that ${\cal T}(G)$ is strongly regular. Given $x\in G\setminus Z(G)$, we have $x^{2}\notin xZ(G)$; further, as $C_{G}(x)\subseteq C_{G}(x^{2})$ and deg$(x)$=deg$(x^{2})$ we get $C_{G}(x)=C_{G}(x^{2})$. Hence, by Lemma 3.4, the graph ${\cal T}(G)$ is disconnected, as desired. ∎

From the hypotheses it follows that there exist non-negative integers $m,n,r$ such that $[G:Z(G)]=2^{m+n+r}$, $[C_{G}(x):Z(G)]=2^{m+n}$ and $[C_{G}(x)\cap C_{G}(y):Z(G)]=2^{m}$, for any distinct vertices $x$ and $y$ of ${\cal T}(G)$. Thus, Proposition 2.5 provides us

$$v=2^{m+n+r}-1,\qquad k=2^{m+n}-2,\qquad\lambda=2^{m}-3,\qquad\mu=2^{m}-1.$$

Because $\lambda\geq 0$, we must have $m\geq 2$. From (1) we obtain $(2^{n+r}-2^{n})(2^{m}-1)=(2^{n}-1)(2^{m+n}-2)$ and, consequently, $2^{n}(2^{m+r}-2^{r}+1-2^{m+n}+2)=2.$ Because $\mu\leq k$ we get $n\geq 1$ and from the last equality above we conclude $n=1$. Thus $2^{m+r}-2^{r}+1-2^{m+1}+2=1$ and so $2^{m}(2^{r}-2)=2^{r}-2$, which implies $r=1$. Hence, $[G:C_{G}(x)]=2$, for any $x\in G\setminus Z(G)$, that is, $G$ is of conjugate type $\{1,2\}$. Now, from Theorem 2.7 it follows that $G$ is isoclinic to an extraspecial $2$-group $E$. As $m\geq 2$ and $n=r=1$ we obtain $[E:Z(E)]=[G:Z(G)]\geq 16$ and so $E$ has order at least $32$. ∎

By Lemma 3.6, it is sufficient to show that $|C_{G}(x)\cap C_{G}(y)|=|C_{G}(t)\cap C_{G}(w)|$, for any vertices $x,y,t,w$ of ${\cal T}(G)$, where $x\neq y$ and $t\neq w$. Let us assume the contrary, that is, there are vertices $x,y,t,w$ of ${\cal T}(G)$ such that

$$|C_{G}(x)\cap C_{G}(y)|>|C_{G}(t)\cap C_{G}(w)|.$$

(2)

Taking into account Proposition 2.5 and the fact that ${\cal T}(G)$ is strongly regular, we have two cases to analyze:

Case 1: $[x,y]\neq 1$ and $[t,w]=1$. Due to Proposition 2.5 and (2), we can write

$$v=2^{m+n+r+s}-1\qquad k=2^{m+n+r}-2\qquad\lambda=2^{m}-3,\qquad\mu=2^{m+n}-1,$$

where $m,n,r,s$ are non-negative integers, with $n\geq 1$. From $\mu(v-k-1)=k(k-\lambda-1)$ we obtain

$$(2^{m+n}-1)(2^{n+r+s}-2^{n+r})=(2^{m+n+r}-2)(2^{n+r}-1),$$

which is equivalent to

$$2^{n}(2^{m+n+r+s}-2^{m+n+r}-2^{r+s}+2^{r}-2^{m+n+2r}+2^{m+r}+2^{r+1})=2.$$

Since $n\geq 1$, this forces $n=1$ and

$$2^{m+n+r+s}-2^{m+n+r}-2^{r+s}+2^{r}-2^{m+n+2r}+2^{m+r}+2^{r+1}=1,$$

that is, $2^{r}(2^{m+s+1}-2^{m+1}-2^{s}+1-2^{m+r+1}+2^{m}+2)=1$. For such equality to occur, we must have $r=0$, which give us $k<\mu$, a contradiction.

Case 2: $[x,y]=1$ and $[t,w]\neq 1$. Considering (2), we may write

$$v=2^{m+n+r+s}-1,\qquad k=2^{m+n+r}-2,\qquad\lambda=2^{m+n}-3,\qquad\mu=2^{m}-1,$$

where $m,n,r,s$ are non-negative integers, with $n\geq 1$. Note that $s,m\neq 0$ because $k\neq v-1$ and $\mu\neq 0$. From the identity $\mu(v-k-1)=k(k-\lambda-1)$ we get $(2^{m}-1)(2^{r+s}-2^{r})=(2^{m+n+r}-2)(2^{r}-1)$ and so $2^{r}(2^{m+s}-2^{m}-2^{s}+1-2^{m+n+r}+2^{m+n}+2)=2$, which implies $r\in\{0,1\}$. However, by Lemma 3.2, $\lambda\neq k-1$; thus $r=1$ and, consequently,

$$2^{m+s}-2^{m}-2^{s}+1-2^{m+n+1}+2^{m+n}+2=1.$$

It follows from the last equality that $2^{m}(2^{s}-1-2^{n+1}+2^{n})=2(2^{s-1}-1)$.

If $s=1$, we obtain $1-2^{n+1}+2^{n}=0$, that is, $n=0$, a contradiction.

Suppose now $s\geq 2$. Since $m\geq 1$, we conclude that the number $2^{m-1}(2^{s}-1-2^{n+1}+2^{n})=2^{s-1}-1$ is odd and this implies $m=1$. From the penultimate equality we get $2^{s}-1-2^{n+1}+2^{n}=2^{s-1}-1$, which gives us $2^{s-1}=2^{n}$, that is, $s-1=n$. Hence, in this case, the parameters of ${\cal T}(G)$ are:

$$v=2^{2s+1}-1,\qquad k=2^{s+1}-2,\qquad\lambda=2^{s}-3,\qquad\mu=1.$$

(3)

Putting $\gamma=(\mu-\lambda)^{2}+4(k-\mu)$, it follows from ([5], Theorem 2.16) that the numbers $m_{1}$ and $m_{2}$ given below are non-negative integers:

$$\displaystyle{m_{1}=\frac{1}{2}\left(v-1+\frac{2k+(v-1)(\mu-\lambda)}{\sqrt{\gamma}}\right)},\qquad\displaystyle{m_{2}=\frac{1}{2}\left(v-1-\frac{2k+(v-1)(\mu-\lambda)}{\sqrt{\gamma}}\right)}.$$

Note that $2k+(v-1)(\mu-\lambda)=2^{s+1}(2^{s+2}-2^{2s}+3)-12$ and this number is nonzero for $s\geq 2$.