Solution: Given that $N$ is a normal subgroup of order $2$. Let $N = \{ e,h \} $. For any $g\in G$, we have \[ ghg^{-1} = h \implies gh = hg. \] Thus $h$ commutes with every element of $G$ and hence, $N = Z(G)$, where $Z(G)$ denotes the centre of $G$.

Now let us prove a general statement. If $G/Z(G)$ is cyclic, then $G$ is abelian. Since $G/Z(G)$ is cyclic, there exists $g\in G$ such that $\left\langle g Z(G) \right\rangle = G / Z(G)$. For any $x \in G$, there exists $k\in \mathbb{Z}$ such that $x Z(G) = g^k Z(G)$. Thus, \[ x Z(G) = g^k Z(G) \implies g^{-k} x \in Z(G) \implies x = g^k z \text{ for some $z \in Z(G)$}. \] In other words, given any element $x \in G$ there exists $z \in Z(G)$ and $k\in \mathbb{Z} $ such that $x = g^k z$. For any $x_1, x_2 \in G$, let $x_1 = g^{k_1}z_1$ and $x_2 = g^{k_2}z_2$ for some $z_1,z_2 \in Z(G)$ and $k_1, k_2 \in \mathbb{Z} $. Thus, \begin{align*} x_1 x_2 & = \left( g^{k_1}z_1 \right) \left( g^{k_2} z_2 \right) = \left( g^{k_2} z_2 \right) \left( g^{k_1} z_1 \right) = x_2 x_1. \end{align*} Thus, $G$ is abelian.