Is every homomorphism from $\langle(0,+\infty),\,\cdot\,,{}^{-1}\rangle$ into $\langle\{-1,1\},\,\cdot\,,{}^{-1}\rangle$ a constant valued function?

Question

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 $x$?

score 3 · Accepted Answer · answered Jan 26 '13 at 07:34

3

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

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

answered Jan 26 '13 at 07:34

N. S.

132,525

$f(1)=[f(1)]^2\Rightarrow f(1)=1$ – Myshkin Jan 26 '13 at 07:38
@Panu Well I proved that $f(x)=1$ for ALL $x$ ;) – N. S. Jan 26 '13 at 07:43
hey N.S, how :-o – Myshkin Jan 26 '13 at 07:45
1

@Panu $f(\sqrt(x))$ is $\pm 1$ because that is the range of $f$ ;) – N. S. Jan 26 '13 at 07:47
I understand. Thank you very much. – Popopo Jan 26 '13 at 11:33

Is every homomorphism from $\langle(0,+\infty),\,\cdot\,,{}^{-1}\rangle$ into $\langle\{-1,1\},\,\cdot\,,{}^{-1}\rangle$ a constant valued function?

1 Answers1