Topics In Algebra 2nd edition, Section 2.5 exercise 13, "If $a \in G$, define $N(a)=\{x \in G \mid xa=ax \} \dots N(a)$ is usually called the normalizer or centralizer of $a$ in $G$."
Usually the normalizer of $a$ $$N_G(a)=\{x \in G \mid xax^{-1} \in H \}$$ EDIT: Should be: Normalizer of $H$ is: $N_G(H)=\{x \in G \mid xHx^{-1} = H \}$.
And centralizer of $a$ $$C_G(a)=\{x \in G \mid xa=ax\}$$
For a general subset $S\subseteq G$, you define (see e.g. Wikipedia) $$C_G(S) := \{g\in G\ |\ gs=sg\text{ for all } s\in S\},\qquad N_G(S) := \{g\in G\ |\ gS=Sg\}.$$ Taking $S=\{a\}$ to be a singleton, you see that both coincide and are given by $\{g\in G\ |\ ga=ag\}$.
Note, however, that already $N_G(\langle a\rangle)\neq C_G(\langle a\rangle)=C_G(\{a\})$ in general; e.g., in the dihedral group $D_n$ you have $srs^{-1} = r^{-1}$, where $s$ is a reflection and $r$ is a rotation, so $s\in N_{D_n}(\langle r\rangle)$ but $s\notin C_{D_n}(\{r\})$.