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.
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.