$H\lhd G$ implies $C_G(h)/C_H(h)$ is cyclic

Question

The actual question is as follows:

Let $G$ be a finite group of odd order, $H$ a subgroup of $G$ of index 3. Let $h\in H$. Prove that $C_H(h) \lhd C_G(h)$ and $C_G(h)/C_H(h)$ is cyclic (where $C_G(h)$ is the centralizer of $h$ in $G$).

Since $H$ is of smallest possible index, we have that $H\lhd G$, and I have shown that $C_H(h) \lhd C_G(h)$ by working through the definitions, but I am unsure how to show the quotient group is cyclic.

Answer 1

$$ \frac{C_{G}(h)}{C_{H}(h)} = \frac{C_{G}(h)}{H \cap C_{G}(h)} \cong \frac{H C_{G}(h)}{H}, $$ and the latter is a subgroup of $G/H$.