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Let $G$ be a group and $H$ a characteristic subgroup of $G$ (that is, invariant under all automorphisms of $G$). Let $\phi \in \mathcal{Aut}(H)$, we'll call $\widetilde{\phi}$ an automorphism of $G$ such that

$$\widetilde{\phi}(h) = \phi(h) \quad \forall h \in H$$

Or in other words, such that $\widetilde{\phi}_H = \phi$, where $\widetilde{\phi}_H$ denotes its restriction to $H$.

Can we construct a $\widetilde{\phi}$ for every $\phi \in \mathcal{Aut}(H)$?

Or in other words, are there automorphisms of $H$ which are not restrictions to $H$ of an automorphism of $G$?

It's easy to show that if the answer is no, then there is a bijection $\mathfrak{A}: \mathcal{Aut}(H) \xrightarrow{\sim} \mathcal{Aut}(G)$ such that $\mathfrak{A}(\phi) = \widetilde{\phi}$. Could this fact be used to disprove the claim above by contradiction?

seldon
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  • The part at the end in italics doesn't look right. Take $H$ to be the trivial subgroup of a group $G$ with nontrivial automorphism group. Then the answer to the question preceding the italics is "no". But the claim in italics does not hold. – halrankard Jul 17 '20 at 13:25

2 Answers2

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The answer is no. One counterexample is the group $G = {\rm AGL}(1,8)$, which has order $56$. It has a normal (and therefor characteristic) Sylow $2$-subgroup $P$, which is elementary abelian of order $8$.

Now ${\rm Aut}(G)$ is the group ${\rm A \Gamma L}(1,8)$, which is a solvable group of order $168$ and has $G$ as a normal subgroup of index $3$. However, ${\rm Aut}(P)$ is the group ${\rm PSL}(3,2)$, which coincidentally also has order $168$, but it is a nonabelian simple group. In fact only $21$ of the $168$ automorphisms of $P$ extend to automorphisms of $G$.

Derek Holt
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  • Thank you for answering, I will check out the groups you mentioned but this seems the answer I was looking for :) – seldon Dec 23 '15 at 19:20
  • If $\mathcal{Aut}(G)$ has 168 elements, how can $\mathcal{Aut}(P)$ have only 21 elements which extends to automorphism of $G$? I mean, shouldn't al least all automorphisms of $G$ be automorphisms of $P$ too? – seldon Dec 24 '15 at 11:10
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    $P$ is abelian, so the $8$ inner automorphisms of $G$ induced by conjugation by elements of $P$ induce the trivial automorphism of $P$. So the group consisting of the automorphisms of $P$ that are induced by automorphisms of $G$ has order $|{\rm Aut}(G)|/8 = 168/8=21$. – Derek Holt Dec 24 '15 at 11:31
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If $H$ is characteristic in $G$ then every automorphism of $G$ is an automorphism of $H$. Thus, the question reduces to answering whether $A(H) \subset A(G)$ is true. Consider $Gh_1$ and $Gh_2$ where $h_1$ and $h_2$ are in $H$. Then $g_1h_1g_2h_2=g_1g_2'h_1h_2$. This shows that the right-regular representation of $H$ is a subgroup of the right-regular representation of $G$. Thus the holomorph of $H$ is a subgroup of the holomorph of $G$. Since $A(H)$ and $A(G)$ fix the identity in the respective holomorphs, $A(H) \subset A(G)$.

EDIT: The claim that the holomorph of $H$ is a subgroup of the holomorph of $G$ is indeed incorrect, for the normalizer of the right regular representation of $H$ in $S_G$(The symmetric group on elements of $G$) need not necessarily be contained in the normalizer of the right regular representation of $G$ in $S_G$.

vnd
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  • Thank you for answering, but I have not studied representation theory yet, thus I struggle to follow your argument. If you can write it down in a simpler form, it would be great, otherwise I will come back to this answer when I'm familiar with those concepts. – seldon Dec 23 '15 at 19:05
  • With $\subset$ you mean strictly subset right? Or you meant $\subseteq$? – seldon Dec 23 '15 at 19:06
  • from what i know $\subset$ means possible equality – vnd Dec 23 '15 at 19:11
  • for the downvote, I would be interested to "learn" which particular statement earned me that – vnd Dec 23 '15 at 19:13
  • I don't know the answer to what earned you the downvote, but you seem to be suggesting that the answer to the question asked is yes, whereas in fact it is no. I think your claim that the holomorph of $H$ is a subgroup of the holomorph of $G$ might be at fault! – Derek Holt Dec 23 '15 at 19:16
  • I'm not the downvoter, btw to mean "subset or equal" I believe $\subseteq$ is more widespread so I'd suggest to edit the answer – seldon Dec 23 '15 at 19:18
  • Let us accept the down vote, it is legitimate more often than not. I was only stating what I thought was right. I will look into why my claim as stated by @Derek Holt might be wrong – vnd Dec 23 '15 at 19:22