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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?

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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. – The Chaz 2.0 Mar 27 '12 at 1:25
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". – user23793 Mar 27 '12 at 1:30
up vote 2 down vote accepted

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!

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Great! Thanks for the help! – user23793 Mar 27 '12 at 1:46

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