# Existence of certain homomorphisms between $\mathbb{Q}$ and $\mathbb{Q}^{\times}$

Letting $\mathbb{Q}^{\times}$ be the group of non-zero rationals under multiplication, and $\mathbb{Q}$ the additive group of rationals, I am seeking to find the following types of homomorphisms (if, of course, they exist):

a) A surjective homomorphism $f: \mathbb{Q}^{\times} \to \mathbb{Q}$

b) An injective homomorphism $f: \mathbb{Q}^{\times} \to \mathbb{Q}$

I know of a homomorphism $\mathbb{Q}^{\times} \to \mathbb{Z}$ but it is not injective (I refrain from describing it, one can see it here: Non-trivial homomorphism between multiplicative group of rationals and integers). Hints, suggestions, solutions will all be much appreciated.

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Yep, I just realised in too. – user44069 Nov 4 '12 at 20:56
For (b), $\mathbb{Q}^{\times}$ has an element of order 2, namely $-1$. But $\mathbb{Q}^+$ does not have an element of order 2. Hence (b) is impossible. – Nicky Hekster Nov 4 '12 at 20:58

If $p_n$ is the $n$th prime, then $$\pm \prod p_k^{e_k}\mapsto \sum \frac{e_k}{k}$$ defines a homomorphism $f\colon\mathbb Q^\times\to \mathbb Q$ that is surjective because for $\frac ab\in\mathbb Q$ you have $f(p_b^a)=\frac ab$ .

Observe that for any homomorphsim $f\colon \mathbb Q^\times\to\mathbb Q$, you have $f(1)=0$ because these are the respective neutral elements and you have $f(-1)f(-1)=f(1)=0$, hence $f(-1)=0$ and thus no injectivity.

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Let's try this ...

$$(-1)^a 2^b 3^c 5^d 7^e 11^f\cdots \mapsto b + \frac{c}{2!}+\frac{d}{3!} + \frac{e}{4!} +\dots$$ On the left is the prime factorization of a nonzero rational: $a$ is $0$ or $1$, and $b,c,d,\dots$ are integera. Allow negative integers to get all rationals. This is surjective, not injective.

[Injective is impossible, since $-1$ and $1$ both have to map to $0$.]

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This seems to work nicely! Thank you very much! – user44069 Nov 4 '12 at 21:01