Let $G$ be a group and $H$ be a subgroup of $G$. When is $\rm{Aut}(H)$ a subgroup of $\rm{Aut}(G)$?
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11$\begingroup$ Generally speaking a question of this type has a shot at a reasonable answer if there is a canonical map $\text{Aut}(H) \to \text{Aut}(G)$ and you want to ask when this map is injective. But in this case there is no such map. Instead there is a canonical map the other way from the subgroup of automorphisms of $G$ preserving $H$ to the group of automorphisms of $H$. So it's not clear to me what you want: do you just want to know when there exists, abstractly, an injection of $\text{Aut}(H)$ into $\text{Aut}(G)$? Do you want this injection to land only in the subgroup of $\text{Aut}(G)$... $\endgroup$ – Qiaochu Yuan Jan 20 '12 at 6:30
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5$\begingroup$ ...which preserves $H$ and to respect the action on $H$? (That would require that you could somehow extend automorphisms of $H$ to automorphisms of $G$ in some reasonable way.) There are lots of things you could want and I doubt there is an easy answer to any of the possible forms of this question. $\endgroup$ – Qiaochu Yuan Jan 20 '12 at 6:31
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3$\begingroup$ @Ali Gholamian: Please consider registering in the site; that will make it easier to keep track of your questions. $\endgroup$ – Arturo Magidin Jan 20 '12 at 16:17
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$\begingroup$ @Ali Gholamian: As above, there is a lot of thing this qestion brings here. You certainly know that $\mathbb{Z}_{6}$ and $\mathbb{Z}_{3}$ know just one group as their Automorphism Group. $\endgroup$ – Mikasa Jan 21 '12 at 15:14
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1$\begingroup$ A partial but also quite trivial answer to this question is: If $\text{Out}(H)=0$, then $\text{Aut}(H)=\text{Inn}(H) \leq \text{Inn}(G) \leq \text{Aut}(G)$. $\endgroup$ – Nex Oct 17 '15 at 21:47
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The restriction of an automorphism of G to H is an automorphism of H requires that H is a characteristic sub-groupe in G, in this case the restriction morphism of Aut (G) to Aut (H) has a sense and is an epimorphism. In this cas it is hoped that Aut (G) is a direct product of Aut (H) ie the exacte sequence
I said that a sufficient condition for the $Aut(H)$ to be injected into $Aut(G)$ is that $H$ a characteristic subgroup in $G$ and the exact short sequence: $ ker(rest) \hookrightarrow Aut(G)\rightarrow Aut(H)$(this last morphism is the epimorphism restriction) split. Then a partial answer that I deduced from the last comment given to this issue as similar question in terms of exact short sequence of groups is this: under the condition H to be a direct factor of G and simultaneously a characteristic subgroup in G, then $Aut(H)$ is injected into $Aut(G)$. This injection may in no case be canonical (as Aut (.) Is not a functor), but the restriction is canonical and has a sense here because H supposed characteristic.
Prove: Suppose $G=H\times K$ and $H$ characteristic in $G$, so the restriction morphism $Aut(G)\rightarrow Aut(H)$ is an epimorphism $f\mapsto f_H $, it can viewed in this situation as the composite morphism $p_H\circ f\circ i_H : H\hookrightarrow H\times K \rightarrow H\times K\times \rightarrow H $ where $i_H$ the canonical monomorphism,$f\in Aut(H\times K)$ and $p_H$ the canonical projection. Since $H$ characteristic in $H\times K$ then $K$ is also characteristic in $H\times K$ because for any $f\in Aut(H\times K)$ the composite morphism $p_K\circ f$ left via the naturel projection $p_K:H\times K\rightarrow K$. and so we obtained the direct product $Aut(H\times K)\simeq Aut(H)\times Aut(K)$ with the isomorphism restriction $f\mapsto (f_H,f_K)$ and of course the already exact short sequence split.
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$\begingroup$ there are error in the kernel i am sorry, The kernel is not o but the sub-group of aut(G) that have restrection to H the identite of H. so the conclusion is not true but speking Aut(H) sub-groupe of Aut(G) requires H characteristic and the obouve exacte sequence split, thanks $\endgroup$ – m.idaya May 23 '16 at 20:45
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$\begingroup$ It is not true that the restriction of an automorphism of $G$ to $H$ being an automorphism of $H$ requires $H$ to be characteristic. $\endgroup$ – Morgan Rodgers May 23 '16 at 21:08
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$\begingroup$ I mean all automorphisms of G for have the morphism from Aut (G) in the Aut(H) by restriction. I have problem to translat my idea in english. Sorry and thanks $\endgroup$ – m.idaya May 23 '16 at 21:16
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$\begingroup$ As a non-trivial example, take G a finite nilpotent group and H a p-Sylow subgroup of G then $Aut(H)$ is a direct product factor of $Aut(G)$ $\endgroup$ – m.idaya May 26 '16 at 6:37
