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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.

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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
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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.

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