Show that one bijection is the inverse of the other bijection if the two (non-identity) bijections commute

Question

Suppose you have two bijections $\eta, \alpha: S \to S$. Both are not the identity maps on $S$, and that

$$\eta\alpha = \alpha\eta$$

Can we conclude that $\alpha = \eta^{-1}$?

Many thanks in advance!

Answer 1

The following uses cycle notation.

Certainly not: consider $S=\{1,2,3\}$ and take both $\alpha$ and $\eta$ to be the permutation $(123)$. Then $\alpha\eta=\eta\alpha$ but $\alpha\not=\eta^{-1}$.

Even if we require $\alpha\not=\eta$ there are still counterexamples; take $S=\{1,2,3,4\}$, let $\alpha=(12)$, and let $\eta=(34)$.

Answer 2

Consider the maps $f$ and $g$ of the plane defined by $f(x,y) = (-x,y)$ and $g(x,y) = (x,-y)$. These are bijections, they commute, yet are not inverses.