# Show that the probability that two randomly choosen elements from $R$ commute is at most $(5/8)$.

Let $R$ be a finite non-commutative ring. Show that the probability that two randomly choosen elements from $R$ commute is at most $(5/8)$.

i know there is an answer in https://www.whitman.edu/mathematics/SeniorProjectArchive/2010/SeniorProject_CodyClifton.pdf

But I can't get it. Can anyone please give me any new answer or explain what is written here explicitly.

• Maybe a look here can help. – drhab Oct 2 '14 at 9:34

the key idea is that if $a,b$ are chosen at random from $G$ then they commute iff $b$ belongs to $Z_G(a)$, the centralizer of $a$. thus since conjugates share the same centralizer, the probability that $b$ commutes with $gag^{-1}$ is also $$p(a,b) = \frac{Z_G(a)}{|G|}=[G:Z_G(a)]^{-1}$$ since each $a$ is equally likely to be chosen with probability $|G|^{-1}$ we find the probability of two elements commuting is $$|G|^{-1}\sum_{a \in G} [G:Z_G(a)]^{-1}$$ since the sum can be done over conjugacy classes this shows that the probability we require is in fact simply $$\frac{C_G}{|G|}$$
where $C_G$ is the number of conjugacy classes of $G$
since every element of the center is a class, and any other class must have at least two elements in it, we have: $$C_G \le |Z(G)| + \frac12(|G|-|Z|) = \frac12(|G|+|Z(G)|)$$ an easy argument shows that the index of the center of a non-abelian group must be at least $4$, from which the result follows.