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Proving a nice functional property

A function g from a set \(X\) to itself satisfies \(g^{m} = g^{n}\) for positive integers m and n with \(m > n\). Here \(g^{n}\) stands for g ◦ g ◦ · · · ◦ g (n times). Show that g is one-to-one if and only if g is onto. (Some of you may have seen the term “one-one function” instead of “one-to-one function”. Both mean the same.)

Note by Eddie The Head
3 years, 1 month ago

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Only If part. Assume \(g\) is \(1-1\). We will show that \(g^{-1}\) exists and this will prove the claim.

Since \(g\) is \(1-1\), so is \(g^{j}\), for any positive integer \(j\). Now, we have for all \(x\in X\), and the given \(m\) and \(n\), \(g^m(x)=g^n(x)\), i.e. \(g^{n}\odot g^{m-n}(x)= g^{n}(x)\). From the \(1-1\) property of \(g^{n}\), we get \(g^{m-n}(x)=x\), for all \(x\in X\). Hence, \(g^{-1}\) exists and is given by \(g^{m-n-1}\) (if \(m=n+1\), \(g\) is then the identity mapping). This proves the claim.

If part can be solved by similar ideas. Abhishek Sinha · 3 years, 1 month ago

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@Abhishek Sinha Nice! Eddie The Head · 3 years, 1 month ago

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