Let R be the relation defined on the set of bijective functions from the set (1 to n) to the set (1 to n) by $fRg$ whenever $f = g^{-1}$

Question

I am not too sure as to what the relation is but I think $R = \{(1, 2), (2, 3), (3, 4), ..., (n - 1, n)\} $.

Any guidance would be appreciated.

Answer 1

To show that $R$ is reflexive you would need to show that $f R f$ for every $f.$ Is it the case that $f=f^{-1}$ for every $f?$

To show that $R$ is symmetric you would need to show that $f R g$ implies $g R f$ for every $f, g.$ Is it the case that $f=g^{-1}$ implies $g=f^{-1}$ for every $f, g?$

To show that $R$ is antisymmetric you would need to show that $f R g$ implies it is not the case that $g R f$ for every $f, g.$ Is it the case that $f=g^{-1}$ implies $g\neq f^{-1}$ for every $f, g?$

To show that $R$ is transitive you would need to show that $f R g$ and $g R h$ implies $f R h$ for every $f, g, h.$ Is it the case that $f=g^{-1}$ and $g=h^{-1}$ implies $f=h^{-1}$ for every $f, g, h?$

Answer 2

For example, if $f,g : \{1,\dots,n\} \to \{1,\dots,n\}$ are defined by $$f(i) = \begin{cases} i+1 & \textrm{if } i < n \\ 1 & \textrm{if } i=n \end{cases} \qquad g(i) = \begin{cases} n & \textrm{if } i = 1 \\ i-1 & \textrm{if } i>1 \end{cases}$$ it is easy to see that $f$ and $g$ are inverses to each other, and then the pair $(f,g)$ is an element of $R$.