7
$\begingroup$

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 $Aut(\Bbb Q^*,\cdot)$ ?

Thanks in advance for help.

$\endgroup$
4
$\begingroup$

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 $\epsilon_p=\pm1$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lubin Feb 6 '18 at 5:32
  • 1
    $\begingroup$ So $Aut(\Bbb Q^*)$ is uncountable while $Aut(\Bbb Q , +)$ is countable. $\endgroup$ – reflexive Feb 6 '18 at 5:32
1
$\begingroup$

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 $2^{\aleph_0}$.

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.