Let $\delta^{\prime}=\delta_{m}(a), \delta^{\prime \prime}=\delta_{m}(b), \delta=\delta_{m}(a b), \eta=\left[\delta_{m}(a), \delta_{m}(b)\right]$.
Sufficiency We have

So $\delta^{\prime} \mid \delta \delta^{\prime \prime}$. From this and $\left(\delta^{\prime}, \delta^{\prime \prime}\right)=1$, it follows that $\delta^{\prime} \mid \delta$. Similarly, we have
$$1 \equiv(a b)^{8} \equiv(a b)^{8 \delta^{\prime}} \equiv b^{8 \delta^{\prime}}(\bmod m)$$

So $\delta^{\prime \prime} \mid \delta \delta^{\prime}$. From this and $\left(\delta^{\prime}, \delta^{\prime \prime}\right)=1$, it follows that $\delta^{\prime \prime} \mid \delta$, and thus from $\delta^{\prime}\left|\delta, \delta^{\prime \prime}\right| \delta$ and $\left(\delta^{\prime}, \delta^{\prime \prime}\right)=1$, it follows that $\delta^{\prime} \delta^{\prime \prime} \mid \delta$. Additionally, it is clear that
$$(a b)^{\gamma^{\gamma}} \equiv 1(\bmod m),$$

So, $\delta \mid \delta^{\prime} \delta^{\prime \prime}$. Therefore, $\delta=\delta^{\prime} \delta^{\prime \prime}$.
Necessity We have
$$(a b)^{\eta} \equiv 1(\bmod m)$$

So $\delta \mid \eta$. On the other hand, it is clear that $\eta \mid \delta^{\prime} \delta^{\prime \prime}$. From this and $\delta=\delta^{\prime} \delta^{\prime \prime}$, it follows that $\eta=\delta^{\prime} \delta^{\prime \prime}$, i.e., $\left(\delta^{\prime}, \delta^{\prime \prime}\right)=1$. Proof completed.
$$\begin{array}{l} 1 \equiv(a b)^{8} \equiv(a b)^{\sigma^{6}} \\ \equiv a^{8 \delta^{\circ}}(\bmod m), \end{array}$$