Here is the theorem:
Here is the proof:
I underlined in red the parts I did not understand. For the first one, why is it $0\leq i-j<1$? For the second one, why is $0\leq r\leq n-1$ significant? Sorry, my professor wrote this proof a year ago and I am trying to fully grasp what they were trying to say, since I am still a beginner...!
On
For the first one, since there is no natural number with $a^n=e$, the fact that $a^{i-j}=e\implies i-j$ isn't a natural number. On the other hand $i-j$ is a whole number. The only whole number that isn't natural is $0$...
For the second part, $0\le r\le n-1$, by the division algorithm. Also, there is a typo. It should say: Let $a^k\color{red}{\in} \langle a\rangle$...
For the first part: you know that there is no positive integer $n\in \mathbb{N}$ such that $a^n=e$. Actually, this also implies that there is no non-zero integer $n\in \mathbb{Z}$ such that $a^n=e$ (do you see why?). But you find out that if $a^i=a^j$ for some integers $i,j$, then $a^{i-j}=e$. According to the above, the only possibility of this occuring is when $i-j=0$, that is $i=j$.
For the second part: take any integer $k$ and look at the element $a^k$. Actually, the value of this element only depends on the congruence class of $k$ modulo $n$ (the order of $a$). In other words, let me write down the Euclidean division of $k$ by $n$: $k=nq+r$, with $q,r$ two integers such that $0\leq r < n$. Then, we have $$a^k=a^{nq+r}=(a^{n})^qa^r=e^qa^r=a^r$$ But $r\in \{0,1,\ldots ,n-1\}$. So $a^k\in \{a^0=e,a^1=a,a^2,\ldots ,a^{n-1}\}$. This shows that any element of the form $a^k$ (eg for $k$ equal $7$ or $k$ equal $-1000000000$ or any other integer you like) can be computed as $a^r$, for some small $r$ just between $0$ and $n-1$. It simplifies things a lot! The sequence of powers of $a$ only takes, periodically, a finite number of values (exactly $n$ distinct values).