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!
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3$\begingroup$ Have you tried any examples? There's already a counterexample on a set with three elements ... $\endgroup$– Noah SchweberCommented Sep 2, 2023 at 18:55
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$\begingroup$ @NoahSchweber I tried the three element example but I guess I missed it $\endgroup$– ShawnCommented Sep 2, 2023 at 18:58
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$\begingroup$ You can always take $\eta = \alpha$. $\endgroup$– Qiaochu YuanCommented Sep 2, 2023 at 18:58
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$\begingroup$ @QiaochuYuan Thank you! I missed that too.. $\endgroup$– ShawnCommented Sep 2, 2023 at 19:20
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2 Answers
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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 $\eta=(34)$.
answered Sep 2, 2023 at 18:58
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$\begingroup$ Thank you! I cannot believe I missed these... $\endgroup$– ShawnCommented Sep 2, 2023 at 19:00
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$\begingroup$ @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. $\endgroup$ Commented Sep 2, 2023 at 19:09
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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.
