# Finding a group such that the map $\varphi: G \to \mathrm{Aut}(G)$, $g \mapsto f_g$ (conjugation by $g$) is an isomorphism

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 $$S_3$$?

• Such groups $G$ are called complete groups. Sep 6, 2022 at 2:42
• 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. Sep 6, 2022 at 21:13

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.

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.

• @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? Sep 6, 2022 at 5:11