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We have the canonical projection $\pi:\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$, which is a group homomorphism. The most natural way I can think of to try extending this to a group homomorphism $\phi:\mathbb{Q} \to \mathbb{Z}/2\mathbb{Z}$ would be to define $$ \phi \left( \frac ab \right) = \pi(a) $$ where $\gcd(a,b) = 1$. Then $\phi$ agrees with $\pi$ on the integers, but $\phi$ isn't a group homomorphism, because $$ 1 = \phi \left( 1 \right) = \phi \left( \frac 12 + \frac 12\right) \neq \phi \left( \frac 12 \right) + \phi\left( \frac 12 \right) = \pi(1) + \pi(1) = 2 = 0 $$ Is is just true that $\operatorname{Hom}(\mathbb{Q},\mathbb{Z}/2\mathbb{Z}) = 0$, in which case such an extension is hopeless?

Joshua Ruiter
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    Yes it is true that $\hom(\Bbb Q,\Bbb Z/2\Bbb Z)=0$. If $\phi$ were any homomorphism and $\phi(x)$ were ever nonzero, what could $\phi(x/2)$ be, since $\phi(x/2)+\phi(x/2)=0$? – anon Sep 28 '17 at 13:12

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You want to consider additive group homomorphisms, where the second map is the canonical projection: $$\Bbb Q\to \Bbb Z \to \Bbb Z/2\Bbb Z$$ First we have to deal with mapping $(\Bbb Q,+)\to (\Bbb Z,+)$.

Consider $\varphi: \mathbb Q \rightarrow \mathbb Z$, and assume that $\varphi$ is a nontrivial map. Now, let $n$ be the smallest positive integer in the image of $\varphi$, and pick $\frac{a}{b} \in \varphi^{-1}(n)$. Then $$n=\varphi\left(\frac{a}{b}\right)=\varphi(a/2b+a/2b)=\varphi(a/2b)+\varphi(a/2b),$$ and hence $\varphi(a/2b)=\frac{n}{2}$. But we assumed $n$ was the smallest positive integer in the image, and hence no nontrivial map exists.

Horizon
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