# finding groups $G$ such that $G \simeq \operatorname{Aut}(G)$

I proved that the mapping $$\phi: G \to \operatorname{Aut}(G)$$ that sends $$g \longmapsto c_g (x) = gxg^{-1}$$ is a homomorphism, but I need to find a group $$G$$ for which this map establishes an isomorphism. I know that $$S_3$$ is isomorphic to $$\operatorname{Aut}(G)$$, but I don't know off-hand whether this bijection is the correct one. I referenced a similar answer here Proving that ${\rm Aut}(S_3)$ is isomorphic to $S_3$, but it doesn't give an explicit formula for the bijection, and though I understand that $$S_3$$ is generated by its transpositions and so any automorphism is determined completely by how it acts on $$(12)$$, $$(13)$$, and $$(23)$$, I don't understand how a bijection follows.

Updated Attempt:

I claim that $$\phi: S_3 \to \mathrm{Aut}(S_3)$$ sending $$g \longmapsto c_g (x) = gxg^{-1}$$ is an isomorphism. In general, $$\phi: G \to \mathrm{Aut}(G)$$ sending $$g \longmapsto c_g$$ is a homomorphism, so it suffices to show that $$\phi$$ is a bijection. We have: \begin{align*} g \in \mathrm{ker}(\phi) & \iff \phi(g) = c_g (x) = x, \; \forall x \in G \\ & \iff gxg^{-1} = x, \; \forall x \in G \\ & \iff gx = xg, \; \forall x \in G \\ & \iff g \in Z(G) = \{e\}, \end{align*} so the kernel of $$\phi$$ is trivial, so $$\phi$$ is injective. I claim that $$\phi$$ is also surjective. We have $$S_3 = \langle a = (12), b = (13), c = (23) \rangle$$, each of which has order $$2$$, so given $$g \in S_3$$, we have $$g = a^i b^j c^k$$ for $$i,j,k \in \{0,1,2\}$$. Since an automorphism $$f \in Aut(S_3)$$ must preserve the order of elements, $$f$$ must send transpositions to transpositions. Furthermore, upon fixing where $$f$$ send these transpositions, the rest of the map is determined. Since there are $$3!$$ possibilities for where to send the permutations, there are exactly $$6$$ automorphisms, so $$|S_3| = |\mathrm{Aut}(S_3)|$$. But $$\phi(S_3) \leq \text{Aut}(S_3)$$, so that they have the same order immediately implies that $$\phi(S_3) = \text{Aut}(S_3)$$, so $$\phi$$ is surjective, hence bijective.

• General thought: (If I'm not mistaken) If for every element in $G$ (excluding $e$) there is an element that does not commute with it, this question is equivalent to asking if there are no outer automorphisms. Maybe that's one place to look? – Cameron Williams Mar 10 at 2:53
• Hint: What is the kernel of the map $G\to \operatorname{Aut}(G)$? Also show that the fact automorphisms are determined by their actions on $3$ elements means that there is an embedding $\operatorname{Aut}(S_3)\to S_3$. – jgon Mar 10 at 3:05
• The kernel is the set of $g \in G$ such that $\phi$ sends $g$ to the identity map $c_e (x) = xex^{-1} = e$. I'm not sure how to prove in closed-form that the only such $g$ is the identity element, however. It's certainly the case that two-cycles are stable under conjugation by two cycles, but not necessary (I believe) by three cycles. Could you explain a bit more about how there is an implied embedding to $S_3$? – user861776 Mar 10 at 3:19
• Yes, this condition is equivalent to the one that $Z(G)=1$ and $Out(G)=1$. There are many examples of such groups, $S_3$ is one of them. (Actually, every $S_n$ when $n\ne 2, 6$.) See for instance here. – Moishe Kohan Mar 10 at 3:34
• Just to clarify earlier comments: the conditions $Z(G)=1$ and ${\rm Out}(G) = 1$ together imply that $G \cong {\rm Aut}(G)$, and that proof works for $G=S_3$, but the converse is false. The group $D_8$ (dihedral of order 8) satisfies ${\rm Aut}(G) \cong G$, but $Z(G) \ne 1$. – Derek Holt Mar 10 at 8:03

Your map $$\phi:G\to Aut(G)$$ has image $$im(\phi)=Inn(G)=\{c_g| \ g\in G\}$$ and $$Aut(G)/Inn(G)=Out(G)$$ so as mentioned in the comments, $$\phi$$ is onto iff $$Out(G)=1$$. Also $$ker(\phi)=Z(G)$$ hence $$\phi$$ is $$1-1$$ iff $$Z(G)=1$$. To sum up, $$\phi$$ is an isomorphism iff $$Out(G)=Z(G)=1$$.
It could be $$Aut(G)\cong G$$ with the isomorphism between $$G$$ and $$Aut(G)$$ being othen than $$\phi$$.