# Homomorphisms from $(\Bbb Q,+)$ to $(\Bbb Q,+)$ and from $(\Bbb Q^*,\cdot)$ to $(\Bbb Q^*,\cdot)$

Hope this isn't a duplicate.

Considering homomorphisms from $(\Bbb Q,+)$ to itself, I notice that all the homomorphisms are of the form $\phi(x) = ax$ , where $a=\phi(1)$ . So there are exactly $\aleph_0$ number of homomorphisms from $(\Bbb Q,+)$ to itself and except the trivial 0-homomorphism, all others are onto-homomorphism . Is it correct?

On the other hand, considering homomorphisms from $(\Bbb Q^*,\cdot)$ to itself, I get that $\phi(1) =1$ and also get some additional restrictions like if $x \in \Bbb Q^*$ and $x$ is a square number then $\phi(x) \neq y$ , where y is either (negative) or (positive but does not admit a rational square root) . Are there some other restrictions on a general homomorphism from $(\Bbb Q^*,\cdot)$ to itself ?

The following maps,(i) $x \mapsto x$ and (ii) $x \mapsto x^{-1}$ (since, $(\Bbb Q^*,\cdot)$ is abelian) are automorphisms of $(\Bbb Q^*,\cdot)$ . How can one find the complete list of $Aut(\Bbb Q^*,\cdot)$ ?

There are lot more automorphisms of $\Bbb Q^*$. By the fundamental theorem of arithmetic, $\Bbb Q^*\cong (\Bbb Z/ 2\Bbb Z)\times\bigoplus_p\Bbb Z$ where the $p$ range over all primes. The automorphism group of $\Bbb Q^*$ is uncountable, as for instance we can consider the maps fixing $-1$ and taking $p$ to $\epsilon_p p$ with $\epsilon_p=\pm1$.

• And of course you can permute the factors, too, and it’s even the case that there are uncountably many permutations of the index set. – Lubin Feb 6 '18 at 5:32
• So $Aut(\Bbb Q^*)$ is uncountable while $Aut(\Bbb Q , +)$ is countable. – reflexive Feb 6 '18 at 5:32

The endomorphism ring of $(\mathbb{Q},+)$ is the field $\mathbb{Q}$, the isomorphism being $f\mapsto f(1)$, for $f$ an endomorphism of $(\mathbb{Q},+)$.

The group $(\mathbb{Q}^*,\cdot)$ is isomorphic to $C_2\times\mathbb{F}$, where $F$ is a free abelian group on a countable basis and $C_2$ is the two-element group (I'll write both additively). If we represent elements of this group as columns $\left[\begin{smallmatrix}a\\x\end{smallmatrix}\right]$, for $a\in C_2$ and $x\in F$, endomorphisms can be represented as matrices $$\begin{bmatrix} \alpha & \beta \\ 0 & \gamma \end{bmatrix}$$ where $\alpha$ is an endomorphism of $C_2$, $\beta\colon F\to C_2$ and $\gamma$ is an endomorphism of $F$. The composition of maps can be written as formal matrix multiplication, so in order to have an automorphism, we need $\alpha$ to be invertible (hence be the identity on $C_2$) and $\gamma$ to be an automorphism of $F$; there's no restriction on $\beta$. Then $$\begin{bmatrix} 1 & \beta \\ 0 & \gamma \end{bmatrix}^{-1}= \begin{bmatrix} 1 & -\beta\gamma^{-1} \\ 0 & \gamma^{-1} \end{bmatrix}$$ Thus you can get the cardinality of the set of all automorphisms:

• it is at least the size of the automorphism group of $F$, which is $2^{\aleph_0}$, because it contains all automorphisms defined by permutations of the basis;

• it can't be greater than $2^{\aleph_0}$, because the cardinality of all maps $\mathbb{Q}^*$ to itself is $2^{\aleph_0}$.

$$\mathbb{Q}^*\cong gr(-1)\times\prod_{p\in P} gr(p).$$ Here, P is the set of primes. From this it follows that $$Aut(\mathbb{Q}^*)=\prod_{p\in P} gr(-1).$$