Is $N_G(S)=N_G(\langle S\rangle)$?

Question

Related question: What would be a counterexample of $N_G(T)\not\subset N_G(S)$?

Let $G$ be a group.

Let $S$ be a subset of $G$.

Then, is $N_G(S)=N_G(\langle S\rangle)$?

I have proved that $N_G(S)\subset N_G(\langle S\rangle)$, but I'm not sure about the other direction. Is there a counterexample for this?

Thank you in advance:)

Answer 1

Consider $D_3=\langle a,b\mid a^2=b^3=1,ab=b^2a\rangle$, and $S=\{1,b\}$. Then $a\in N(\langle S\rangle)\setminus N(S)$.