This question already has an answer here:

Consider the following question:

Let $\mathbb{Q}$ be the field of all rational numbers.

Let Aut($\mathbb{Q}$) be the group of all Automorphism on $\mathbb{Q}$ (All Isomorphism from $\mathbb{Q}$ to $\mathbb{Q}$).

Show that $\mathbb{Q^{*}}$ is isomorphic to Aut($\mathbb{Q}$).

I really dont know how to prove this base on the Isomorphism theorem.

Any help will be appreciated.


marked as duplicate by Dietrich Burde, drhab, Morgan Rodgers, user147263, John B Apr 11 '16 at 0:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is $\Bbb X$? Regardless, the group of field automorphisms of $\Bbb Q$ is trivial, so presumably by $\operatorname{Aut}(\Bbb Q)$ you mean the group of automorphisms preserving some weaker structure? $\endgroup$ – Travis Apr 10 '16 at 13:37
  • $\begingroup$ Perhaps the superscript "x" is really a star, indicating some sort of dual? Even so, I can't make sense of this, since $Aut(Q)$ is a singleton, as Travis observes. $\endgroup$ – John Hughes Apr 10 '16 at 13:40
  • $\begingroup$ $\mathbb{Q^x}$ means $\mathbb{Q}$ - {0}. $\endgroup$ – bar Apr 10 '16 at 13:44

Let $f:\mathbb Q\to\mathbb Q$ be a $\mathbb Q$-automorphism. Then $f(q)=f(q\cdot1)=q\ f(1)$, so it is determined by the value of $f(1)$. This gives us a bijection from $\text{Aut}_\mathbb Q(\mathbb Q)$ to $\mathbb Q^\times$ since all values but $0$ give an automorphism. If $f,g$ are automorphisms, then $g(f(q))=g(f(q\cdot1))=g(q\ f(1))=q\ g(f(1))=q\ g(f(1)\cdot1)=$ $q\ f(1)g(1)$, so the composition of automorphisms gets assigned the product of the corresponding numbers, which is what we needed.


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