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Is the statement that $ \operatorname{Aut}(\operatorname{Hol}(Z_n)) \cong \operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.

This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group

OP of that question provided us with the complete list of groups $G$, such that $|G| \leq 506$ and $ \operatorname{Aut}(G) \cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?

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  • $\begingroup$ Very interesting question! Running this code on MAGMA shows no "small" counterexamples for i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for; $\endgroup$ – AnalysisStudent0414 Jul 26 '18 at 13:55
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    $\begingroup$ This is true. The key is that $n$ is odd,so the centeris trivial. If $\phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}\phi(k)$ does to $Z_n$. $\endgroup$ – Steve D Jul 26 '18 at 14:02
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Here's a proof, which is really an expansion of my comment above.

Let $A=\langle a\rangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=A\rtimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.

Now let $\phi$ be an automorphism of $G$. Then $\phi(A)=A$, and so there exists $k\in K$ such that $\phi(a)=a^k$. Now $H=\phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $g\in K$, there's a unique element of the form $a^?g\in H$. In particular, consider $\iota\in K$, the inversion map. If $a^r\iota\in H$, then setting $m=-r(n+1)/2$, it is easy to check that $\iota\in a^mHa^{-m}$. By looking at $[\iota,ga^s]$, we see that $C_G(\iota)=K$, and so $a^mHa^{-m}=K$. Thus $\phi$ acts on $G$ exactly like conjugation by $ka^m$, so $\phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.

Edit: I might have glossed over one too many details in the end above. Let $\psi\in Aut(G)$ be conjugation by $ka^m$, and let $\alpha=\phi\psi^{-1}$. Then we've shown $\alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $g\in K$, we have

\begin{align} a^g &= \alpha(a^g)\\ &= a^{\alpha(g)} \end{align} and thus $g$ and $\alpha(g)$ are two automorphisms of $A$ with the same action, meaning $\alpha(g)=g$. Thus $\alpha$ fixes $K$ pointwise, and since $G=AK$, $\alpha$ is the identity map.

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