Given a permutation on a set, show that its inverse is again a permutation. I started the problem: Let f be a permutation on a set A. Let g be the inverse of f. Show that g is a function, 1-1, and onto. I'm having trouble understanding exactly what it means for g to be a function. Do I need to show it is well defined? If so, then I know since f is 1-1 and onto that each element in A goes to exactly one other element in A. So if I'm "going backwards" when I think about g, then shouldn't g too be well defined?

  • $\begingroup$ Well, a permutation is a bijection from a set to itself. Have you read the wiki page on bijections? Most of what you need should be there. $\endgroup$ – The Chaz 2.0 Mar 27 '12 at 1:25
  • $\begingroup$ Yeah that should help! I think I'm having trouble getting my head around what I need to show to say "g is a function". $\endgroup$ – user23793 Mar 27 '12 at 1:30

Yes, your intuition is completely correct. Whoever posed this problem probably just wants you to be able to write down your reasoning formally.

  1. $g$ is a well-defined function because for every element $x$ the set, $g$ maps it to the single value $y$ in the set that satisfies $f(y) = x$. (This $y$ exists because $f$ is onto; the $y$ is unique because $f$ is one-to-one. Actually, I feel that showing $g$ is a function isn't necessary at all, since the "inverse" of anything has be a function by definition, if you're able to talk about it.)

  2. $g$ is one-to-one: assume not, then there are $x \neq x'$ with $g(x) = g(x') = y$. This means that $f(y) = x = x'$, contradiction.

  3. $g$ is onto: for every $y$ in the set, there is indeed an element $x$, namely $f(y)$, such that $g(x) = y$.

This all may seem like overkill and, yes, it is rather silly. Soon you'll be moving onto more interesting things. Have fun!

  • $\begingroup$ Great! Thanks for the help! $\endgroup$ – user23793 Mar 27 '12 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.