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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.

user861776
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  • 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 '21 at 02:53
  • 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 '21 at 03:05
  • 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 '21 at 03:19
  • 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 '21 at 03:34
  • You are confused about the identity map: $c_e(x)=exe^{-1} = x$, not $e$ (you gave the zero map, which is almost never an automorphism). – Arturo Magidin Mar 10 '21 at 04:41
  • Thank you for pointing out that error. Based on these comments, I've been able to understand the following facts. First, because the transpositions, of which there are three, generate the group, and automorphisms have to send transpositions to transpositions, there are exactly $3! = 6$ automorphisms, so $|S_3| = |Aut(G)|$. To show the map is injective, I need to show that the kernel is trivial, which follows from the fact that if $g \in \text{ker} \phi$ if and only if $g \in Z(G) = {e}$. Assuming this is fine, how do I show that the map is onto? – user861776 Mar 10 '21 at 05:55
  • I edited my post with an updated attempt. Is this correct? Is there a better/different way to explain surjectivity? I don't know if I used the fact that an automorphism is uniquely determined by its values on transpositions completely. – user861776 Mar 10 '21 at 06:47
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    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 '21 at 08:03

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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 $\phi$.

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  • Thanks for the answer. I haven't studied Out (G) yet, so I didn't allude to it in my solution. Could you tell me if my attempt is incorrect? – user861776 Mar 10 '21 at 17:40