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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.

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

1 Answer 1

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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.

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  • $\begingroup$ 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 $x\in G$? $\endgroup$
    – lanskey
    May 4, 2018 at 0:53
  • $\begingroup$ @lanskey I edited to make it clearer. $\endgroup$
    – Arnaud Mortier
    May 4, 2018 at 0:54
  • $\begingroup$ I see it now. Thanks! $\endgroup$
    – lanskey
    May 4, 2018 at 0:55
  • $\begingroup$ @lanskey you're welcome! $\endgroup$
    – Arnaud Mortier
    May 4, 2018 at 0:55

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