Let $f,g,h$ be bijections from some set $X$ to itself, i.e. they are permutations of $X$. We say that $f$ and $g$ commute if $f\circ g=g\circ f$.
Is it the case, in general, that if $f$ commutes with $h$, and $g$ commutes with $h$, then $f$ commutes with $g$?
If not, is it always the case if $X$ is finite?
If not, could somebody give a counter example? (It seems like it shouldn't be true, but I can't think of one.)
Thanks to Cayley's theorem, you can realize any group as a permutation group. But there exist noncommutative transitive* groups: take for example a nonabelian group whose center is nontrivial (eg. quaternion group).
*that is a group that is not commutative transitive (nothing to do with transitive groups): a group $G$ is said commutative transitive if two elements of $G$ commuting with a third nontrivial element commute.