Solution: Let $G$ can be generated by $b$, that is, $G = \left\langle b \right\rangle $ for some $b \in G$. Since $b \in G = \left\langle a \right\rangle $, there exists $m \in \mathbb{Z} $ such that $b = a^m$. Since, $a\in G$ so there exists $l \in \mathbb{Z} $ such that $a = b^l$. Thus, \[ b = a^m \implies b^l = a^{ml } \implies a = a^{ml}. \] Since $G$ is of infante order, and $a$ is a generator, the above is possible if and only if $ml = 1$. This means $m,l \in \{ 1, -1 \} $. Therefore, \[ b = a \text{ or } b = a^{-1} . \]

Now it only remains to show that $G = \left\langle a^{-1} \right\rangle $. We will show that $a \in \left\langle a^{-1} \right\rangle, $ which will prove the claim. Since $a = \left( a^{-1} \right)^{-1} $, it is clear that $a \in \left\langle a^{-1} \right\rangle $.