Why this graph is connected? Let $R$ be a non-commutative finite ring with center $Z(R)$ and consider the graph $\Gamma_R$ whose vertix set is $R\backslash Z(R)$ and two vertices $a$ and $b$ are adjacent iff $ab \neq ba$. In an article I am reading, [1],it is claimed that $\Gamma_R$ is connected (their reasoning: just because the degree of any vertices is non zero) but I don’t get it. So please let me know why the graph is connected.
[1]  On non-commuting graph of a finite ring,
J. Dutta, D. K. Basnet. 2017 submitted. 
 A: I think the argument in the paper is incomplete, but here is how one can justify that the graph is indeed connected. Take two elements $a,b\not\in Z(R)$. Consider the centralizers $C(a),C(b)$ (the set of elements commuting with $a$ and $b$, respectively). It's easy to see those are additive subgroups of $R$, and since $a,b\not\in Z(R)$, they are proper subgroups, meaning $|C(a)|,|C(b)|\leq |R|/2$. Since $C(a)\cap C(b)$ is nonempty (it contains $0$), it follows $|C(a)\cup C(b)|<|R|$. Take an element $c\in R\setminus(C(a)\cup C(b))$. Then $a,c$ don't commute and neither do $b,c$, so $a,c$ and $b,c$ are edges in $\Gamma_R$, showing $a,b$ are connected by a path (of length $2$).
In fact, this works even if $R$ is infinite, but we need a slightly different argument -- we need to know that a group, even an infinite one, can't be a union of two of its proper subgroups. Let $G_1,G_2\subseteq G$ be two proper subgroups. If $G_1\subseteq G_2$ or $G_2\subseteq G_1$, it's clear their union is not $G$. Otherwise take $g_1\in G_1\setminus G_2,g_2\in G_2\setminus G_1$. Then $g_1+g_2\not\in G_1\cup G_2$.
