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

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 $$\alpha = \eta^{-1}$$?

• Have you tried any examples? There's already a counterexample on a set with three elements ... Commented Sep 2, 2023 at 18:55
• @NoahSchweber I tried the three element example but I guess I missed it Commented Sep 2, 2023 at 18:58
• You can always take $\eta = \alpha$. Commented Sep 2, 2023 at 18:58
• @QiaochuYuan Thank you! I missed that too.. Commented Sep 2, 2023 at 19:20

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 $$\eta=(34)$$.
• @Shawn: in general, fixing $\alpha$, the set of $\eta$ such that $\eta \alpha = \alpha \eta$ is called the centralizer of $\alpha$. It is always a subgroup, and always contains every power of $\alpha$, but typically contains other elements too. Commented Sep 2, 2023 at 19:09
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.