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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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    $\begingroup$ You are right, but if $g$ is in $C_G(H)$, then a fortiori $g$ is in $N_G(H)$, so $C_G(H) = C_{N_G(H)}(H)$. (You should write out the definitions of the two sets that I'm claiming are equal and convince yourself that they are the same.) $\endgroup$ – William DeMeo Mar 7 '12 at 8:01
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    $\begingroup$ $C_G(H)$ is certainly contained in $N_G(H).$ Hence $C_{N_G(H)}(H)$ is the same as $C_G(H).$ $\endgroup$ – Geoff Robinson Mar 7 '12 at 8:03
  • $\begingroup$ Thank you Geoff and William. It is now very clear that $C_X(H) = C_G(H) \cap X$, and so $C_{N_G(H)}(H) = C_G(H) \cap N_G(H) = C_G(H)$. $\endgroup$ – Antonio Vargas Mar 7 '12 at 8:18
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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 $C_G(H) \subseteq N_G(H)$.

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