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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.

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    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Apr 25 '18 at 9:51
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    $\begingroup$ 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 $\endgroup$ – Václav Mordvinov Apr 25 '18 at 9:53
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    $\begingroup$ 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 ? $\endgroup$ – Maxime Ramzi Apr 25 '18 at 10:17
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    $\begingroup$ Well your identity yields that if $a$ belongs to it then $gag^{-1}$ does too $\endgroup$ – Maxime Ramzi Apr 25 '18 at 10:29
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    $\begingroup$ 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 $g\in G$. $\endgroup$ – Servaes Apr 25 '18 at 10:53
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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.

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    $\begingroup$ Yes, and part e follows if you accept my amendment to the question. $\endgroup$ – ancientmathematician Apr 25 '18 at 11:34
  • $\begingroup$ 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? $\endgroup$ – Václav Mordvinov Apr 25 '18 at 12:23
  • $\begingroup$ It follows from (d) : any automorphism of $A(G)$ fixes the set $I(G)$. $\endgroup$ – ancientmathematician Apr 25 '18 at 12:35
  • $\begingroup$ 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 $\mathrm{Aut}(\{\mathrm{id}_G\})$ $\endgroup$ – Václav Mordvinov Apr 25 '18 at 13:45
  • $\begingroup$ 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. $\endgroup$ – Servaes Apr 25 '18 at 14:01

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