# How to prove or disprove transitivity of relation

I have a relation $R$ on a set $G$ defined as $(a,b) \in R$ if $ab = ba$. It is pretty clear that this is reflexive and symmetric, but how do I prove or disprove transitivity?

Edit: As mentioned in the comments, I forgot to specify that $G$ is a group.

• Is $G$ actually a group? If so, please say so in your question. Commented Oct 16, 2016 at 4:44
• @AlexG. Edited! Commented Oct 16, 2016 at 4:51

## 3 Answers

To be more explicit than the other answers, take the group $GL_2(\Bbb R)$ of invertible $2\times 2$ matrices. Let $a = \left(\begin{array}{cc} 1 & 1\\ 1 & 0\end{array}\right), b = \left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right),$ and $c = \left(\begin{array}{cc} 0 & 1\\ 1 & 1\end{array}\right)$. Then $ab = ba = a$ and $bc = cb = c$, but $ac \neq ca$. Thus, this relation is not transitive.

Well, transitivity means that $aRb \land bRc \implies aRc$.

In this case, $aRb$ means $ab = ba$. $bRc$ means $bc = cb$.

So, does $(ab = ba) \land (bc = cb) \implies ac = ca$?

• That's my question... Commented Oct 16, 2016 at 4:54

To show transitivity you need to show that if $(a,b) \in R$ and $(b,c) \in R$ then $(a,c) \in R$.

However, depending on the set G, this condition may or may not be satisfied.

For example:

If $G = \mathbb{R}$ then the relation is transitive since $\mathbb{R}$ is abelian. If G is the set of 2x2 matrices then it is not transitive.