1
$\begingroup$

I've come across this problem and I've no idea how to handle it: Show that the set of all group isomorphisms from $f:(Q,+)\rightarrow (Q,+)$ is isomorphic with the $(Q^*,\cdot)$ group. Anyone here could lend a hand, maybe an arm?

PS: This question has been market as a possible duplicate of the following: Automorphism group of $\mathbb{Q}$ considered as a group under addition. The person who asked the linked question was wondering if he could prove that $Aut(Q,+)$ is isomorphic with the $(Q^*,+)$ group, whether my question was on $Aut(Q,+)$ and $(Q*,\cdot)$. I hope this is enough of an explanation. Notice me if not.

  • 0
    Possible duplicate of [Automorphism group of $\mathbb{Q}$ considered as a group under addition](http://math.stackexchange.com/questions/1667792/automorphism-group-of-mathbbq-considered-as-a-group-under-addition)2017-01-29

2 Answers 2

2

Hint: Show that any group isomorphism $\varphi: (\mathbb{Q},+) \to (\mathbb{Q},+)$ is of the form $\varphi(x) = ax$ for $a\neq 0$. Clearly, then, composition of such maps is the same thing as multiplication.

To show this, let $a=\varphi(1)$, then show that $\varphi(n) = an$ for $n\in\mathbb{Z}$, then end with extending this to $\mathbb{Q}$.

1

First prove that every isomorphism $(\Bbb Q,+) \to (\Bbb Q,+) $ is of the form $\psi_x(y) = xy$ (where $x\in \Bbb Q^*$) and that every function of this form is an isomorphism

The rest follow immediatly