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I know that the map $\varphi: G \to \mathrm{Aut}(G)$ sending an element $g$ to the map conjugation by $g$, which I will denote by $f_g$, is a homomorphism. I'm trying to find a $G$ such that the map is also an isomorphism.

I know from my reading on the subject that $S_3$ works, but I'm trying to fully understand how I would come up with this on my own with being prompted to consider $S_3$. The first important thing is this map $\varphi$ would have to be injective (in order to be an isomorphis), so that would imply it would have to have trivial kernel: for $g \in G$, \begin{align*} a \in \mathrm{ker}(\varphi) & \iff \varphi(a) = \mathrm{id}_G \\ & \iff \forall a \in G, \; c_a = \mathrm{id}_G \\ & \iff \forall a \in G, \forall x \in G, \; c_a (x) = x \\ & \iff \forall a \in G, \; axa^{-1} = x \\ & \iff \forall x \in G, \; ax = xa \\ & \iff a \in Z(G). \end{align*} So if $\phi$ is an isomorphism, it is injective, so its kernel is trivial. But we just showed that $\mathrm{ker}(\varphi) = Z(G)$, so $Z(G)$ is also trivial. So it seems that this is one requirement for $G$ to be an isomorphism.

For $\phi$ to be an isomorphism, it must also be the case that $|G| = |\mathrm{Aut}(G)|$. I think this implies that the entire automorphism group of $G$ must be inner automorphisms. Is that correct? For every $g \in G$, $c_g$ is an automorphism and an "inner" automorphism. At least in the case of finite groups, I only have $|G| = |\mathrm{Aut}(G)|$ if there are no outer automorphisms. Is is true in general that $\mathrm{Aut}(G) =\mathrm{Inn}(G)$ if and only if the center of $G$ is trivial?

If not, how else would I come up with a group like $S_3$?

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    $\begingroup$ Such groups $G$ are called complete groups. $\endgroup$ Commented Sep 6, 2022 at 2:42
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    $\begingroup$ That $|G|=|\mathrm{Aut}(G)|$ implies that the automorphism group equals the inner automorphism group only holds if you are also assuming (i) $Z(G)$ is trivial; and more importantly, that (ii) $G$ is finite. $\endgroup$ Commented Sep 6, 2022 at 21:13

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It is not true in general that $Z(G) = 1$ implies ${\rm Aut}(G) = {\rm Inn}(G)$.

There are many examples, for example $G = A_4$ has $Z(G) = 1$, but $[{\rm Aut}(G) :{\rm Inn}(G)] = 2$. In fact ${\rm Aut}(G) \cong S_4$ in this case. (For an example of an outer automorphism of $G$, take $f: G \rightarrow G$ defined by $f(x) = gxg^{-1}$ where $g = (12)$ is a transposition from $S_4$.)

In general describing automorphism groups is not always so easy.

For the question of how you would come up with $G = S_3$ as an example, as you note you should look for groups with $Z(G) = 1$. A good start for any problem is to look at some small examples, and it turns out in this case the smallest example works.

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Yes, because the kernel of the homomorphism from $G$ to $\rm {Aut}(G)$ given by $\varphi (g)=i_g$ (the inner automorphism $x\mapsto gxg^{-1}$) is the center $Z(G)$. So by the first isomorphism theorem, $G/Z(G)\cong \rm{Inn}(G)$. So when the center is trivial, $G\cong\rm{Inn}(G)\le\rm{Aut}(G)$.

For $n\gt2$, $S_n$ is centerless (fairly easy).


Also, other than $n=2,6$, all the automorphisms of $S_n$ are inner. Putting it together, we get that $S_n\cong\rm{Inn}(S_n)\cong\rm{Aut}(S_n)$. Such groups are called complete.

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  • $\begingroup$ @JasonDeVito I understand that $S_6$ is an exception, but the proof above from the first isomorphism theorem seems to me to work. Am I missing something? $\endgroup$ Commented Sep 6, 2022 at 5:11

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