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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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    @QiaochuYuan One consider a similar question in term of shoer exact sequence (in various category):let $0\to A \to B \to C \to 0$ is a short exact sequence. when Aut(A) is embedable in Aut B?2016-03-09

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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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    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)$2016-05-26