If $a$ and $b$ are elements of a group, then both $ab$ and $ba$ have always the same order.

Proof. Let $e$ be the indentity element of the group. For $n > 1$ , we have the equivalent conditions $$e \;=\; (ab)^n \;=\; \underbrace{(ab)(ab)\cdots(ab)}_{n} \;=\; a(ba)^{n-1}b,$$ $$a^{-1}b^{-1} \;=\; (ba)^{n-1},$$ $$(ba)^{-1} \;=\; (ba)^{n-1},$$ $$e \;=\; (ba)^n.$$ As for the infinite order, it makes the conditions false.

Note. More generally, all elements of any conjugacy class have the same order.