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In Dummit and Foote, they state

"... let the group $N_G(A)$ (normalizer) act on the set $A$ by conjugation. It is easy to check that the kernel of this action is the centralizer $C_G(A)$."

From what I understand, the kernel of this action is the set $$ \{ g \in N_G(A) : g a g^{-1} = a \text{ for all } a \in A \}. $$ However, the centralizer of $A$ is $$ \{ g \in G : g a g^{-1} = a \text{ for all } a \in A \}. $$ If this is true, I see that the kernel is a subset of the centralizer, but how are they equal?

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    $\begingroup$ The centraliser is always contained in the normaliser. $\endgroup$ – Daniel Fischer Oct 6 '13 at 18:14
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We always have $C_G(A) \subset N_G(A)$, since

$$N_G(A) = \{g\in G : gAg^{-1} = A\} = \left\{ g \in G : \bigl(\forall a \in A\bigr)\bigl(gag^{-1}\in A\bigr)\right\},$$

and of course a group element that fixes $A$ elementwise under conjugation fixes $A$ as a set.

So the kernel of the group action is

$$N_G(A) \cap C_G(A) = C_G(A).$$

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