Let $H \leq G$. Define a map $f: N(H) \rightarrow Aut(H)$ given by $f(g) = \phi_g$, where $\phi_g$ is the inner automorphism of H induced by $g$: $\phi_g(h) = ghg^{-1} \forall h \in H$. That this map $f$ is a homomorphism is clear, but I have trouble trying to see why $Kerf = C(H)$. Can someone explain this to me? $N(H)$ is the normalizer of $H$ in $G$ and $C(H)$ the centralizer.
Question about the N/C theorem
5
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abstract-algebra
group-theory
1 Answers
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So, we have $g \in \ker f$ iff $\phi_g = \mathrm{id}_H$, that is iff $\phi_g(h) = h$ for all $h \in H$, so $ghg^{-1} = h$ for all $h$, which means $gh = hg$ for all $h \in H$. This holds exactly iff $g \in C(H)$ by definition of the centralizer.
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0Clear now! thanks a lot!! – 2012-10-20