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Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be a strictly increasing function such that $f(2)=2$ and $f(mn)=f(m)f(n)$ for all positive integers $m,n$ such that $\gcd(m,n)=1$ Prove that $f(n)=n$ for every positive integer $n$.

This is the original question which I have solved by induction, but a comment is given that $f:\mathbb{N}\rightarrow\mathbb{C}$ with $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime is called a multiplication function. We have the general result that for any increasing multiplicative function that is not constant is of the form $f(n)=n^\alpha$ for some $\alpha > 0$

My initial approach to the original problem was to use group theory; is that how this additional, more general, result is derived? I thought that the map is a homomorphism and this preserves order of elements, but I wasn't sure how to go further.

Can anyone give a proof (to either the original problem or the more general result) in terms of this method? Or provide an alternate method?

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    You have a proof of this Erdös result here http://www.jstor.org/discover/10.2307/2322314?uid=3738240&uid=2&uid=4&sid=211008413565812012-06-09

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