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I have read the first answer for this question Centralizer of a finite normal subgroup has finite index and I did not understood, why if $|Aut(H)| < \infty$, then $kernel = C_G(H)$ must have finite index?

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    $\begingroup$ Because of the reason stated there. The index of the kernel is the cardinality of the image. $\endgroup$ – Tobias Kildetoft Apr 2 at 19:07
  • $\begingroup$ Given the homomorphism $T:G\to Aut(H)$ ( In this case was the one given by conjugation T(g)h=ghg^{-1}). By the Fundamental theorem of homomorphisms tells you that $G/kernel=G/C_G(H)$ is isomorphic to the image (in this case a subgroup of $Aut(H)$, hence finite). So, $G/C_G(H)$ is finite, which means exactly that $C_G(H)$ has finite index. $\endgroup$ – Julian Mejia Apr 2 at 19:26

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