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.)

#Functions #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup #FunctionalAnalysis

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

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 - 7 years, 1 month ago

Nice!

Eddie The Head - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$