# Is the kernel of this group action the centralizer?

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?

• The centraliser is always contained in the normaliser. – Daniel Fischer Oct 6 '13 at 18:14

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