Finding all group homomorphisms $(\mathbb{Q},+)\to (\mathbb{Q}-\lbrace 0\rbrace,\cdot)$

Question

If $f$ is such a homomorphism, then $f(x+y)=f(x)f(y)$. I know of examples of $f$ which satisfy this property, such as $f(x)=e^x$, but how do I find all of them?

Answer 1

There is exactly one such homomorphism, namely the constant map $x\mapsto 1$, because $1$ is the only number whose $k$th roots are all rational. In more detail:

Let $x$ be arbitrary. Since $f(x)=f(\frac x2+\frac x2)=f(\frac x2)^2$ it must be positive; take $f(x)=n/m$ in lowest terms.

Suppose that $n/m\ne 1$; then there is some prime $p$ that divides either $n$ or $m$. Suppose it is $p\mid n$; the other case is similar. Let $k$ be some integer such that $p^k>n$. Then what can $f(\frac{x}{k})$ be? \We know that $f(x) = f(k\frac{x}{k}) = f(\frac{x}{k})^k$. But if $f(\frac{x}{k})=a/b$ in lowest terms, then we must have $a^k=n$. Since $p$ divides $n$, $p$ must divide $a$ too, but then $p^k$ divides $a^k=n$, which contradicts $p^k$ being larger than $n$.

Answer 2

Consider some such homomorphism $f$. Consider an arbitrary element $a/b \in \mathbb{Q}$. If $f(a/b)=p/q$, we can write $p/q$ in lowest terms and get a unique prime factorization $p/q = \frac{p_1^{r_1} \dots p_n^{r_n}}{{q_1^{s_1} \dots q_m^{s_m}}}$. the $p_1 \dots p_n, q_1 \dots q_m$ are distinct prime numbers, and the $r_1 \dots r_n, s_1 \dots s_m$ are positive integers. If $p/q$ is an integer, then the denominator contains no factors, and if $p/q=1$, the numerator contains no factors either. Now, we suppose $p/q \neq 1$. Then consider $k=\min\{r_1, \dots , r_n, s_1, \dots, s_m\}$. Then $f\left(\frac{a}{b(k+1)}\right) = \frac{p_1^{r_1/{k+1}} \dots p_n^{r_n/{k+1}}}{q_1^{s_1/{k+1}} \dots q_m^{s_n/{k+1}}}$. If this value were equal to a rational number $p'/q'$, then we would have $p'^{k+1}/{q'}^{k+1} =p/q$. But any prime factor in $p'^{k+1}/{q'}^{k+1}$ must have power at minimum $k+1$, and since $p/q$ contains a prime of power $k$ and prime factorization is unique by the fundamental theorem of arithmetic, we have a contradiction.

As the only assumption we made was that $\{p_1, \dots , p_n, q_1, \dots q_m\}$ was nonempty, this means that $f(a/b)=1$. But since $a/b$ was arbitrary, $f(x)= 1 \forall x \in \mathbb{Q}$.

Answer 3

There is only the trivial morphism $x \mapsto 1$.

Indeed, we can check that for any $n \in \mathbb{Z}$ and $x\in \mathbb{Q}$, we have $f(nx)=f(x)^n$.

In particular, since $f\left(\frac{1}{n}\right)^n = f(1)$, the number $f(1)$ has a rational $n^{\textrm{th}}$ root for all $n \ge 1$. So $f(1) = 1$ and $f\left(\frac{1}{n}\right) = 1$ for all $n \ge 1$. And from that we deduce $f\left(\frac{a}{b}\right) = 1^a = 1$.

EDIT : Here's an explanation of my statement that $1$ is the only rational number with a rational $n^{\textrm{th}}$ root for all $n \ge 1$.

Any non zero rational number can be uniquely written as a product of powers of primes (with some negatives exponents, for example $\frac{4}{15} = 2^2 . 3^{-1} . 5^{-1}$). Let $p$ be a prime number, for $x \in \mathbb{Q}^*$ denote $v_p(x) \in \mathbb{Z}$ the exponent of $p$ in this decomposition (called the $p$-adic valuation). We see that $v_p(xy) = v_p(x) + v_p(y)$ for $x, y \in \mathbb{Q}^*$, so in particular $v_p(x^n) = n v_p(x)$. This means the $p$-adic valuation of an $n^{\textrm{th}}$ power is a multiple of $n$.

So if $x \in \mathbb{Q}^*$ is an $n^{\textrm{th}}$ power for all $n \ge 1$, then $v_p(x)$ is a multiple of $n$ for all $n \ge 1$, which means $v_p(x) = 0$. And since this is true for all prime numbers $p$, we get $x = 1$.

Answer 4

Two proofs that there are no nontrivial homomorphisms.

This is kind of the same thing as what others have suggested, but in more abstract terms: ${\bf Q}$ is a divisible group. The image by a homomorphism of a divisible group is always divisible (this is easy to show), but ${\bf Q}^*$ has no nontrivial divisible subgroups (every nonzero rational is not an $n$-th power for a sufficiently large $n$), so the only homomorphism is the trivial one.

For a slightly convoluted, but perhaps interesting proof, it is not very hard to see that ${\bf Q}^*$ is isomorphic (as a group) to the product of the two-element group (corresponding to the sign) and a countably infinitely generated free abelian group (corresponding to the prime decomposition). By noticing that $f(x)=f(x/2)^2$, we see that the image of the homomorphism is a subgroup of the positive rationals, which form a free abelian group under multiplication.

But a subgroup of a free abelian group is always free, and moreover a free generator cannot be divisible (this is a simple consequence of the definition), so the image must be trivial (because it is free and divisible), and so must the homomorphism itself.

This shows that no divisible group (even divisible by $2$, say, or any other prime) admits a nontrivial homomorphism into ${\bf Q}^*$.