# Show that $G$ is isomorphic to $\operatorname{Aut}_{G-Set}(G)$

A problem from Aluffi's Algebra Chapter 0.

Let $G$ be a group. Consider $G$ as a $G$-set, by acting with left-multiplication. Prove that $\operatorname{Aut}_{G-Set}(G)\cong G$.

What I have so far:

We are looking for all equivariant bijections from $G$ to itself, i.e. all bijections $\varphi\colon G\to G$ such that

$$(\forall g,a\in G)\quad\varphi(ga)=g\varphi(a)$$ It is easy to see that the map $$\varphi_x\colon G\to G,\quad g\mapsto gx^{-1}$$ is an equivariant bijection for all $x\in G$. So, we have a map $$\psi\colon G\to \operatorname{Aut}_{G-Set}(G),\quad x\mapsto\varphi_x$$ It is clear that this map is an injective homomorphism. It is left to show that $\psi$ is surjective, and this is what I can't do.

Thanks.

• So $G$ is a group? Is there anything else about it that we should know? May 4, 2018 at 0:36
• Yes, it is a group. And no, there is no further information. I edited the question to make it clearer. May 4, 2018 at 0:42

Given $\varphi_x$ you can recover the value of $x$ by evaluating it at the identity and then taking the inverse. In other words the map $\varphi\mapsto \varphi(e)^{-1}$ is a candidate inverse for $\Psi$. All you have to do is prove that for any $\varphi\in \operatorname{Aut}_{G-Set}(G)$, the following holds: $$\varphi=\varphi_{\varphi(e)^{-1}}$$ which is straightforward.
• This is the inverse map $im(\psi)\to G$. But the main question is: why does any equivariant bijection $G\to G$ have the form of $\varphi_x$ for some $x\in G$? May 4, 2018 at 0:53