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Can you find function which satisfies $f(ab)=\frac{f(a)}{f(b)}$? For example $log(x)$ satisfies condition $f(ab)=f(a)+f(b)$ and $x^2$ satisfies $f(ab)=f(a)f(b)$?

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But multiplication is commutative, and division is not. In other words, we would need $\dfrac{f(a)}{f(b)}=f(ab)=f(ba)=\dfrac{f(b)}{f(a)}$. – user641 Apr 12 '13 at 9:38
I'm confused...why the [group-theory] tag? – user1729 Apr 12 '13 at 9:46
@SteveD Why does multiplication have to be commutative? I assume the OP might have more conditions than he's said, but in the absence of such, we don't have to assume it ;) – Luke Mathieson Apr 12 '13 at 9:47
@LukeMathieson: OP mentioned logarithms... – user641 Apr 12 '13 at 9:49
@SteveD, he also labelled it group theory, it might just be an example. I suspect he is thinking about functions in $\mathbb{R}$, but let's have fun. – Luke Mathieson Apr 12 '13 at 9:51
up vote 11 down vote accepted

Assuming the function is defined on non-zero real numbers, and takes all non-zero values (but please do see below for a generalization), one has first $$ f(1) = f(1 \cdot 1) = \frac{f(1)}{f(1)} = 1, $$ and then for all $x$ $$ f(x) = f( 1 \cdot x) = \frac{1}{f(x)}, $$ so that $f(x) \in \{1, -1 \}$. Then $$ f(x y) = \frac{f(x)}{f(y)} = f(x) f(y), $$ so that we get that $$ f(x) = 1, \qquad\text{or}\qquad f(x) = \operatorname{sgn}(x). $$

Addendum One may consider the same problem for $f: G \to H$, where $G, H$ are groups (multiplicatively written, with identity $1$), and then it is possibly clearer. (See also the comments to OP.)

The condition is now $f(ab) = f(a) f(b)^{-1}$. Once more, $f(1) = 1$, and $f(x) = f(1 \cdot x) = f(1) f(x)^{-1} = f(x)^{-1}$, so all values of $f$ are involutions (or the identity) and $f$ is a group homomorphism.

So in this case we have that $f$ is a morphism of $G$ onto a(n abelian) subgroup of $H$ whose non-identity elements are involutions. (Clearly there is a non-trivial such $f$ if and only if $G$ has a non-trivial quotient of exponent $2$.)

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Don't you then have $f(ab) = c$ and $\frac{f(a)}{f(b)} = \frac{c}{c} = 1$? – Luke Mathieson Apr 12 '13 at 9:33
Necessarily $c=1$, so the only solutions are $1$ and $\text{sgn}(x)$. – Sean Eberhard Apr 12 '13 at 9:34
@SeanEberhard Yes, I have fixed that, thanks. – Andreas Caranti Apr 12 '13 at 9:47
@LukeMathieson, indeed, thanks. Should have been fixed. – Andreas Caranti Apr 12 '13 at 9:52

You could take a relatively trivial function like $f(x) = 1$. Or a slightly more general version that takes everything to the identity.

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Let us reformulate the question as classify all maps $f : G \rightarrow H$ which need not be group morphism that satisfies the condition $f(ab)=f(a)f(b)^{-1}$

A simple calculation shows that $f(e)=f(x)f(x^{-1})^{-1}= f(x^{-1})f(x)^{-1}$ or we have $f(x)=f(e)^{-1}f(x^{-1})=f(e)f(x^{-1})$ or $f(e)^{-1}=f(e)$ Now $f(x)=f(ex)=f(e)f(x)^{-1}=f(e)^{-1}f(x)^{-1}=(f(x)f(e))^{-1}=(f(x)f(e)^{-1})^{-1}=f(xe)^{-1}=f(x)^{-1}$ so even if we don't assume a group morphism we have the image involutive. And hence it's a group morphism $f(ab)=f(a)f(b)^{-1}=f(a)f(b)$

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