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Let $H \leq G$, $N := N_G(H) = \{g \in G\mid gHg^{-1} = H\}$ and $C:=C_G(H)= \{g \in G \mid \forall h \in H: ghg^{-1} = h\}$.

I have shown that $C \lhd N$. How can I show that $N/C$ is isomorphic with a subgroup of $Aut(H)$?

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You should think of a homomorphism $N\to\operatorname{Aut}(H)$; the definition of the normalizer is a clue. Then (if you've chosen the right thing), the kernel will be $C$, and you can use the first isomorphism theorem.

(As an aside, this would also show that $C$ is normal in a very neat way, as the kernels of homomorphisms are normal.)

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Great thanks ! I thinks this will work: Let $\phi: N \rightarrow Aut(H): n \mapsto i_n$ where $i_n:H \rightarrow H: h \mapsto nhn^{-1}$. –  André Dec 4 '12 at 18:25
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Say $\phi:N \to \operatorname{Aut}(H)$ is given by $\phi(g) = \sigma_g$, where $\sigma_g$ is our inner automorphism $\sigma_g(x) = gxg^{-1}$. The kernel of $\phi$ is obviously $C$. By the first isomorphism theorem, we know that $N/C \cong \operatorname{Im}(\phi) \le \operatorname{Aut}(H)$.

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Indeed. That is what I have found, too :) –  André Dec 4 '12 at 18:27
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