# Proving isomorphism using Cayley

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

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.