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If $H$ is a subgroup of $G$, then if $N(H) = \{g \in G~|~gHg^{-1}=H\} ; ~~C(H) = \{g \in G ~|~ ghg^{-1}=h ~\forall~h \in H\}$, then I have proved that :

$\Psi: N(H) \rightarrow \operatorname{Aut}(H)$ given by $\Psi(n) = nhn^{-1} ~\forall~h \in H,~ n \in N(H)$ is a homomorphism and its kernel is $C(H)$

Hence, by the first isomorphism theorem,

$N(H)/ C(H) ~\approx~ \Psi[N(H)]$

I am confused how is $\Psi[N(H)]$ a subgroup of $\operatorname{Aut}(H)$ as per the statement of the $N/C$ theorem.

I know that $Inn(H)$ is a subgroup of $\operatorname{Aut}(H)$

Is $\Psi[N(H)]$ a subgroup of $\operatorname{Inn}(H)$ by any chance?

Help will be appreciated. Thank you.

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    $\begingroup$ Conjugations acting on $H$ by elements of $G$ need not act like conjugations by elements of $H$, so they may not be inner in ${\rm Aut}(H)$. $\endgroup$
    – anon
    May 29, 2014 at 20:49
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    $\begingroup$ $\Psi(n)( h_1 h_2 ) = n (h_1 h_2) n ^{-1} = n( h_1) n^{-1} n (h_2) n^{-1} = \Psi(n)(h_1) \cdot \Psi(n)(h_2)$, so $\Psi(n)$ is always a homomorphism from $H$ to somewhere. If $n \in N(H)$, then the homomorphism takes $H$ to $H$. As @mezhang mentioned, it is easy to find an inverse, so $\Psi(n)$ is even an automorphism. $\endgroup$ May 29, 2014 at 20:57

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Basically, $\Psi$ is telling you to think of elements of $N(H)$ as things that you can conjugate stuff in $H$ by to get back stuff in $H$. Conjugating stuff in $H$ by things outside $N(H)$ will generally not result in something in $H$, so only conjugation by elements of $N(H)$ can give you automorphisms of $H$.

Since you're defining these automorphisms by conjugation, of course anything in $C(H)$ will act trivially. Hence, $N(H)/C(H)$ gives you a subgroup of $\text{Aut}(H)$, in the sense that you have a homomorphism $N(H)\rightarrow\text{Aut}(H)$ which has kernel $C(H)$, so you have an injection $N(H)/C(H)\hookrightarrow\text{Aut}(H)$. (ie, any element $n\in N(H)$ gives you an automorphism of $H$, but $nc$ for any $c\in C(H)$ will give you the same automorphism. Ie, the automorphism determined by $n$ does not depend on the part that comes from $C(H)$).

$\Psi(N(H))$ is not a subgroup of $\text{Inn}(H)$, because $\text{Inn}(H)$ only allows conjugation by elements of $H$. Unless $N(H) = H\cdot C(H)$, $\Psi(N(H))$ will in general be bigger than $\text{Inn}(H)$.

For example, if you embed a group $G$ of size $n$ via the left regular representation inside $S_n$, then $\text{Aut}(G) = N_{S_n}(G)/C_{S_n}(G)$.

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  • $\begingroup$ The last claim you make: is it obvious? Do you have a reference? $\endgroup$
    – Eric Auld
    Sep 7, 2016 at 18:56
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    $\begingroup$ @EricAuld Well you definitely get an injection from $N_{S_n}(G)/C_{S_n}(G)\rightarrow Aut(G)$, so it suffices to show that it's surjective. Given any $\sigma\in Aut(G)$, via the embedding of $G$ in $S_n$, we can think of $\sigma$ as an element of $S_n$. Now I believe if you work this out you'll find that conjugation in $S_n$ via $\sigma$ induces precisely the automorphism $\sigma$ on the image of $G$ embedded inside $S_n$ $\endgroup$
    – oxeimon
    Sep 7, 2016 at 19:26
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    $\begingroup$ @oxeimon, I think that your regarding $\sigma$ as an element of $S_n$ comes not from embedding $G$ in $S_n$, but from identifying $G$ with the set $n = \{0, \dotsc, n - 1\}$ that's permuted by $S_n$. That is, automorphisms of $G$ are in particular permutations of $G$ (that happen to have a lot of extra structure). $\endgroup$
    – LSpice
    Mar 9, 2020 at 23:07
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    $\begingroup$ @LSpice You're right! Though it seems I can no longer edit the comment. Hopefully if my mistake confuses anyone they'll see your comment and understand what I meant. $\endgroup$
    – oxeimon
    Mar 9, 2020 at 23:46

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