Proof. We will explicitly determine if the group \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} can be generated by one of its own elements. Recall that we have the group (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, +) and that
\begin{align*} \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} = \{(0,0),(1,0),(0,1),(1,1)\}. \end{align*}
\begin{align*} (1,0) + (1,0) = (0,0), \quad (0,0) + (1,0) = (1,0). \end{align*}
\begin{align*} \langle (0,0) \rangle &= \{(0,0)\} \\ \langle (0,1) \rangle &= \{(0,0),(0,1)\} \\ \langle (1,1) \rangle &= \{(0,0),(1,1)\} \end{align*}