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Let $f$ be a homomorphism from $\langle(0,+\infty),\,\cdot\,,{}^{-1}\rangle$ into $\langle\{-1,1\},\,\cdot\,,{}^{-1}\rangle$. Must $f$ be a constant valued function, i.e. is $f(x)=1$ for all $x$?

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1 Answer 1

up vote 4 down vote accepted

Yes, the only group homomorphism is $f(x)=1$. The reason why is very simple:

$$f(x)=f(\sqrt{x} \sqrt{x})=[f(\sqrt{x})]^2 \,.$$

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$f(1)=[f(1)]^2\Rightarrow f(1)=1$ –  El Angel Exterminador Jan 26 '13 at 7:38
    
@Panu Well I proved that $f(x)=1$ for ALL $x$ ;) –  N. S. Jan 26 '13 at 7:43
    
hey N.S, how :-o –  El Angel Exterminador Jan 26 '13 at 7:45
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@Panu $f(\sqrt(x))$ is $\pm 1$ because that is the range of $f$ ;) –  N. S. Jan 26 '13 at 7:47
    
I understand. Thank you very much. –  Popopo Jan 26 '13 at 11:33

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