TITLE: Problems with a proof involving graphs and groups
QUESTION [5 upvotes]: I'm studying an article that is the main literature when it comes to non-commuting graph : this article. Originally, a non-commuting graph of a group (denoted by $\Gamma_G$)  is a graph whose vertices are the elements of the group and they're joined by a edge if $[x,y]$ is not trivial, that is, if they don't commute like elements of the group. The article that I'm studying they made a little change: they consider the vertices the elements of $G$ that are not in the center, that is, $G-Z(G)$.
My problem is the proof of 
Preposition 2.4. Let $G$ a non-abelian group and let $S$ a cut set of $\Gamma_G$.  If $x$ and $y$ two vertices of $\Gamma_G$\ $S$ belong to distinct connected components, then $S$ is a union of double cosets of $C_G(x)\cap C_G(y)$. In particular, if $G$ is finite, then $\kappa(\Gamma_G)=t\mid Z(G)\mid$, where $t>1$ is an integer.
Proof: Let $H=C_G(x)\cap C_G(y)$ and $a\in G$ such that $HaH\cap S\neq\emptyset$. Then $HaH\subseteq S$, for if there exist elements $h_1, h_2\in H$ such that $h_1ah_2\notin S$, then $\{x,h_1ah_2\}$ and $\{y,h_1ah_2\}$ are edges of $\Gamma_G$, a contradiction. And continues...
Well, I got the contradiction, but why if $h_1ah_2\notin S$, then $\{x,h_1ah_2\}$ and $\{y,h_1ah_2\}$ are edges? I could see why necessarily $[x,h_1ah_2]$ and $[y,h_1ah_2]$ are not trivial in this case. 
Thank you!

REPLY [2 votes]: I think the first part of this Proposition may be wrong. 
Take for example $G$ to be the dihedral group of order $12$ with presentation $G=\langle r,s\mid r^6=s^2=(rs)^2=1\rangle$. The center is $\{1,r^3\}$ with order $2$ so $\Gamma$ has order $10$. Now take $S$ to be all vertices except $r$ and $r^2$. This is a cut-set since $r$ and $r^2$ commute and thus are not adjacent. So we have $x=r$ and $y=r^2$, say. In this case $H:=C_G(x)\cap C_G(y)=\langle r\rangle$ which is a normal subgroup hence its double cosets are just the usual cosets: $H$ and its complement. But $S$ is not a union of such.
Interestingly, I had a look at the actual paper and they never seem to use Proposition 2.4... (I didn't actually check carefully, I just did a search for "2.4".)
I think it may be possible to fix the proposition by requiring the cut-set to be minimal with respect to inclusion. This makes sense in view of the second part. In fact, I think the second part may still be correct.