For $H \leq G$, showing that $N_G(H)/C_G(H) \leq \text{Aut}(H)$

Question

This question probably has a very simple answer!

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

Here is the proposition referenced:

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.

Answer 1

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 $C_G(H) \subseteq N_G(H)$.