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This question probably has a very simple answer!

I'm trying to understand the proof of the following result from Dummit and Foote, 3ed:

cor4.4.15

Here is the proposition referenced:

prop4.4.13

I don't understand the part where Proposition 13 is applied "with $N_G(H)$ playing the role of $G$". Wouldn't this only give me that $N_G(H)/C_{N_G(H)}(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$? How does $C_G(H)$ appear?

Thanks for any help.

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    Thank you Geoff and William. It is now very clear that $C_X(H) = C_G(H) \cap X$, and so $C_{N_G(H)}(H) = C_G(H) \cap N_G(H) = C_G(H)$.2012-03-07

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Just so the question isn't "unanswered"...

Thanks to the comments left by William DeMeo and Geoff Robinson.

We have $C_X(H) = C_G(H) \cap X$ for any $X$, so $C_{N_G(H)}(H) = C_G(H) \cap N_G(H) = C_G(H)$ since $C_G(H) \subseteq N_G(H)$.