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I came across the following question:

If $f$ is a group homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Q}-\{0\},.)$ such that $f(2)=\frac{1}{3}$, then what is the value of $f(-8)$?

By property of group homomorphism, we can write - $f(8) = f(2+2+2+2) = f(2)^4 = \frac{1}{81}$.

But we are asked to find $f(-8)$. How does the negative sign change the answer (i.e. if it changes, I am not sure)?

Thank you.

Ritu
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2 Answers2

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Since $$f(x)f(-x)=f(x-x)=f(0)=1$$ by the homomorphism property then $f(-x)=\frac{1}{f(x)}$.

T. Eskin
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    Thank you. I would like to ask another thing. I can see that this property is true here. Is it true in general for all homomorphisms? – Ritu Jun 07 '15 at 03:37
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    @Ritu: Yes. If $f$ is a homomorphism then $f(x^{-1}) = f(x)^{-1}$ (I have used multiplicative notation so that we don't worry about commutativity). The proof is the same as the one Thomas gave, only with the different notation. – Kopper Jun 07 '15 at 03:40
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This question is intentionally confusing.

The fact that it's a group homomorphism means $f(x+y)=f(x)f(y)$, since the domain is an additive group, and the codomain is a multiplicative group. Thus actually,

$f(8)=f(2+2+2+2)=f(2)f(2)f(2)f(2)=\frac{1}{81}$.

Since $1=f(0)=f(8-8)=f(8)f(-8)$, $f(-8)=\frac{1}{f(8)}=81$.

Noah Olander
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