Solution:

Suppose that n is a division of m then there exist an integer q, such that $q=\frac{m}{n}$ or, $n q=m$.

Now, $a^m=\left(a^n\right)^q=e^q=e$

If $O(a)=n$ and n is a divisor of m then $a^m=e$.

Now, $\quad a^m=e$

$ \Rightarrow \quad O(a)=m $

Hence, $m=n q$ i.e. n is a divisor of q.

m is an integer and n is a positive integer. Therefore, by divisor algorithm there exist integer q and r. such that $m=n q+r$ where $\leqslant r \leqslant n$.

$ a^m=a^{n q+r} $

$ \begin{aligned}\\ &\Rightarrow\left(a^n\right)^a a^\gamma \\\\ &\Rightarrow e^q \cdot a^\gamma=e a^\gamma=a^\gamma\\ \end{aligned}\\ $

Therefore, $a^m=e$ i.e. $a^r=e$.

Since, $0 \leq r=n$ therefore, $a^r=e$ suggest that r must be equal to zero, otherwise $o(a) \neq h$. Thus, if $O(a)=n$ then there exist not positive integer r < n satisfying $a^\gamma=e$.