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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ $C_G(H)$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.