# $\mathrm{Aut}(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)$ for $G$ a non-abelian simple group

For the last couple of days I've been doing a lot of group theory problems, but I found the following particularly difficult, and quite interesting. My level is up to an introductionary course in group theory excluding Sylow Theory.

I've made a big edit to include all the work done this far and exclude non-relevant parts.

Let $$G$$ be a non-abelian group such that all normal subgroups are trivial (i.e. the only normal subgroups are $$\{e\}$$ and $$G$$ itself). Prove the following.

a. $$G\cong\mathrm{Inn}(G)$$

b. If $$\psi\in\mathrm{Aut}(G)$$ and $$\forall\varphi\in\mathrm{Inn}(G), \ \psi\circ\varphi=\varphi\circ\psi$$, then $$\psi=\mathrm{id}_G$$.

c. If $$N\lhd\mathrm{Aut}(G)$$ such that $$N\cap\mathrm{Inn}(G)=\{\mathrm{id}_G\}$$, then $$N=\{\mathrm{id}_G\}$$.

d. $$\mathrm{Inn}(G)\ char \ \mathrm{Aut}(G)$$

e. $$\mathrm{Aut}(\mathrm{Aut}(G))=\mathrm{Inn}(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)$$

a. Since $$Z(G)\lhd G$$, either $$Z(G)=\{e\}$$ or $$Z(G)=G$$. In the latter case $$G$$ is abelian, which is not the case, thus $$Z(G)=\{e\}$$, and it follows that $$\mathrm{Inn}(G)\cong G/Z(G)\cong G$$.

b. Elaborating on the comments by Max and Servaes, we see that the given identity yields $$\psi(g)\psi(a)\psi(g^{-1})=g\psi(a)g^{-1}$$ and from there $$g^{-1}\psi(g)\psi(a)=\psi(a)g^{-1}\psi(g)$$. Since $$\psi\in\mathrm{Aut}(G)$$, $$\psi$$ is bijective (thus surjective), and thus $$\forall g\in G \ g^{-1}\psi(g)\in Z(G)=\{e\}$$, thus $$g=\psi(g)$$ for all $$g\in G$$, so $$\psi=\mathrm{id}_G$$

c. Suppose $$N\lhd\mathrm{Aut}(G)$$ such that $$N\cap\mathrm{Inn}(G)=\{\mathrm{id}_G\}$$. Since $$\mathrm{Inn}(G)\lhd\mathrm{Aut}(G)$$ and the intersection between both normal subgroups is trivial, $$\forall\varphi\in\mathrm{Inn}(G)\ \forall \psi\in N, \ \ \psi\circ\varphi=\varphi\circ\psi$$, thus by applying part b conclude that $$\psi=\mathrm{id_G}$$ if $$\psi\in N$$, thus $$N=\{\mathrm{id}_G\}$$ indeed.

d. Is worked out by Servaes.

e. The last part that remains. Some conjectured that part e) might contain a typo, but I don't think so. I managed to proof the following: we know from part b) that if $$\psi\in\mathrm{Aut}(G)$$ commutes with all $$\chi\in\mathrm{Aut}(G)$$, then certainly $$\psi$$ commutes with all $$\varphi\in\mathrm{Inn}(G)\subset\mathrm{Aut}(G)$$. So $$Z(\mathrm{Aut}(G))=\{\mathrm{id}_G\}$$, and thus $$\mathrm{Inn}(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)/Z(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)$$.

The only thing that remains to be proven is that $$\mathrm{Aut}(\mathrm{Aut}(G))=\mathrm{Inn}(\mathrm{Aut}(G))$$, for a simple non-abelian group the automorphism group is complete. According to Wikipedia, this should be true, but I've no idea how to prove this, but I guess it might follow quite easily. I would appreciate it a lot if someone could write a proof for this very last part.

• Just a note: A group like this is called simple. Also, the group itself is usually not referred to as a trivial subgroup. I am not really sure what text would introduce such an exercise without defining simple groups. Commented Apr 25, 2018 at 9:51
• Thanks for this! I didn't know it was called a simple group, indeed it is not introduced in my (Dutch) syllabus. Using the name of these groups I can search better for any related posts Commented Apr 25, 2018 at 9:53
• For b. you are right that a priori there's no reason why such an $h$ would exist and so your proof is not correct as such. Perhap you could try to see what $\{g\in G\mid \psi(g) = g\}$ looks like ? Commented Apr 25, 2018 at 10:17
• Well your identity yields that if $a$ belongs to it then $gag^{-1}$ does too Commented Apr 25, 2018 at 10:29
• Your identity can also be written out to $$g^{-1}\psi(g)\cdot \psi(a)=\psi(a)\cdot g^{-1}\psi(g).$$ Since $\psi$ is surjective this means $g^{-1}\psi(g)\in Z(G)$ for all $g\in G$. Commented Apr 25, 2018 at 10:53

For part b: Such a map $h$ does not exist in general. For example, the group $A_5$ is simple with $\operatorname{Aut}(A_5)=S_5$, but there is no surjective homomorphism $h:\ S_5\ \longrightarrow\ A_5$.

The intended approach seems to be this: If $\psi\circ\varphi=\varphi\circ\psi$ for all $\varphi\in\operatorname{Inn}(G)$, then for all $g,h\in G$ $$\psi(ghg^{-1})=g\psi(h)g^{-1},$$ which can be rearranged to $$g^{-1}\psi(g)\ \psi(h)=\psi(h)\ g^{-1}\psi(g).$$ Because $\psi$ is surjective this implies that $g^{-1}\psi(g)\in Z(G)$ for all $g\in G$. You've already shown that $Z(G)=\{e\}$, and so it follows that $\psi=\operatorname{id}_G$.

For part d: Let $\chi\in\operatorname{Aut}(\operatorname{Aut}(G))$ and let $N:=\chi(G)$. Because $G$ is simple and $\operatorname{Inn}(G)\cong G$, we have either $N\cap\operatorname{Inn}(G)=\operatorname{Inn}(G)$ or $N\cap\operatorname{Inn}(G)=\{\operatorname{id}_G\}$. The latter would, by part c, imply that $N=\{\operatorname{id}_G\}$ which is clearly impossible. Hence $N\cap\operatorname{Inn}(G)=\operatorname{Inn}(G)$ and so $N=\operatorname{Inn}(G)$.

For part e: [This fails for $A_6$, so there must be a typo.] EDIT: I am unsure about this part.

• Yes, and part e follows if you accept my amendment to the question. Commented Apr 25, 2018 at 11:34
• Thank you both! by accepting your amendment @ancientmathematician the relevant part of e becomes $\mathrm{Aut}(\mathrm{Aut}(G))=\mathrm{Aut}(\mathrm{Inn}(G))$ and this follows with a similar approach as approach to d? Commented Apr 25, 2018 at 12:23
• It follows from (d) : any automorphism of $A(G)$ fixes the set $I(G)$. Commented Apr 25, 2018 at 12:35
• I don't see how this follows from d: $\{\mathrm{id}_G\}$ is also a characteristic subgroup of $\mathrm{Aut}(G)$ and this clearly does not apply to $\mathrm{Aut}(\{\mathrm{id}_G\})$ Commented Apr 25, 2018 at 13:45
• I'm not sure what you mean, but do be aware that c does not hold when we replace $\operatorname{Inn}(G)$ by $\{\operatorname{id}_G\}$. I'll expand my answer a bit. Commented Apr 25, 2018 at 14:01

$$\newcommand{\Inn}{\mathrm{Inn}}\newcommand{\Aut}{\mathrm{Aut}}\newcommand{\char}{\quad\mathrm{char}\quad}$$For the last step, considering $$\Inn(G) \char \Aut(G)$$, each automorphism of $$\Aut(G)$$ induces an automorphism of $$\Inn(G)$$, hence we have a homomorphism: $$f:\Aut(\Aut(G))\to\Aut(\Inn(G)),\quad\alpha\mapsto\alpha|_{\Inn(G)}$$ The kernel of this map is a normal subgroup $$K\triangleleft\Aut(\Aut(G))$$. For any $$\alpha\in K$$, $$\alpha$$ induces the trivial map on the subset $$\Inn(G)$$ of $$\Aut(G)$$.

Now consider the normal subgroup $$\Inn(\Aut(G))$$ of $$\Aut(\Aut(G))$$. Using claim (b) above, if an element of $$\Aut(G)$$ commutes with all of $$\Inn(G)$$, it must be trivial. So all non-identity elements in $$\Inn(\Aut(G))$$ must be non-trivial restricted on $$\Inn(G)$$. This means that $$\Inn(\Aut(G))\cap K=1$$. With the same reasoning as (c), replacing $$G$$ with $$\Aut(G)$$, we conclude that $$K=1$$, so $$\Aut(\Aut(G))\cong \Aut(\Inn(G))\cong \Aut(G)$$.

• Addition: We still have to prove that $f$ is surjective. This is straightforward by observing that $f$ is injective and $\mathrm{Aut}(\mathrm{Inn}(G))\cong\mathrm{Inn}(\mathrm{Aut}(G))$, Commented Nov 2, 2021 at 7:39

For (e), based on (d), for any $$\psi\in \text{Aut}(\text{Aut}(G))$$, $$\psi':=\psi|_G\in \text{Aut}(G)$$. Notice that $$(\psi'g\psi'^{-1})(h)=\psi'(g\psi'^{-1}(h)g^{-1})=\psi'(g)h\psi'(g)^{-1}$$ for any $$g,h\in G$$ which suggests $$\psi(g)=\psi'g\psi'^{-1}$$. Want to show that the equation holds on the whole $$\text{Aut}(G)$$. WLOG, we may assume $$\psi|_G=1$$. Then $$xgx^{-1}=\psi(xgx^{-1})=\psi(x)g\psi(x)^{-1}$$ for any $$g\in G$$ and $$x \in \text{Aut}(G)$$. Hence $$x^{-1}\psi(x)$$ commutes with $$g$$. By (b), we get $$x^{-1}\psi(x)=1$$ and finish the proof.