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Prove that $\theta$ is a group isomorphism.
Let
$\theta :\mathbb{Q}^* \rightarrow \operatorname{Aut}(\mathbb{Q})$
$w \mapsto f_w$
and
$f_w(x) = wx$ where $w \in \mathbb{Q}^* $ and $x \in \mathbb{Q}$

Is there a way to prove this without resorting to proving that $\theta$ is a homomorphism followed by proving the surjectivity and injectivity?

Maybe using Cayley's theorem since $f_w$ is a permutation.

Thanks for the help

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While $f_w$ is a permutation, it doesn't really lead anywhere in my opinion.

So yes, showing that $\theta$ is injective and surjective is a good way to do that. Together with some linear algebra: you just have to realize that if $f:\mathbb{Q}\to\mathbb{Q}$ is a group homomorphism, meaning it preserves addition, then it also preserves multiplication automatically. This is because in $\mathbb{Q}$ multiplication comes from addition. I encourage you to show that.

This implies that every group homomorphism $f:\mathbb{Q}\to\mathbb{Q}$ is actually a linear map with $\mathbb{Q}$ treated as a vector space over $\mathbb{Q}$. General linear algebra applies and so $f(x)=\lambda x$ for some unique $\lambda\in\mathbb{Q}$. Note that $f$ is invertible if and only if $\lambda\neq 0$.

So surjectivity of $\theta$ is equivalent to existance of $\lambda$. Injectivety of $\theta$ to a simple observation that different linear maps have to have different $\lambda$ coefficient.

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