Proof of Pocklingtons theorem: why do we have $\text{ord}_n(a) \mid (q-1)$ but $\text{ord}_n(a) \nmid (n-1)/p_i$?

Question

Proof of Pocklingtons theorem: why do we have $\text{ord}_n(a) \mid (q-1)$ but $\text{ord}_n(a) \nmid (n-1)/p_i$ ? And why does $p_i^{\alpha_i} \mid \text{ord}_n(a)$ ?

I see that $\text{ord}_n(a)$ must divide $n-1$, since the order of an integer $\pmod n$ always divide any power $k$ of $a$ such that $a^k \equiv 1 \pmod n$.

But why are $\text{ord}_n(a) \mid (q-1)$ and $\text{ord}_n(a) \nmid (n-1)/p_i$ true ? And why does $p_i^{\alpha_i} \mid \text{ord}_n(a)$ ?

Answer 1

I believe there is a typo; the proof makes complete sense if $ord_n(a)$ is replaced by $ord_q(a)$.

Indeed:

1. $ord_q(a) \mid q-1$ due to Fermat's little theorem
2. $ord_q(a) \mid n-1$ by (1)
3. $ord_q(a) \nmid \frac{n-1}{p_i}$ by (2) (since $\gcd(a^{\frac{n-1}{p_i}}-1, n)=1$, we have $a^{\frac{n-1}{p_i}} \not \equiv 1 \pmod{q}$)
4. Now from $ord_q(a) \mid n-1=p_i^{\alpha_i}\frac{n-1}{p_i^{\alpha_i}}$ and $ord_q(a) \nmid \frac{n-1}{p_i}=p_i^{\alpha_i-1}\frac{n-1}{p_i^{\alpha_i}}$, we must have $p_i^{\alpha_i} \mid ord_q(a)$. (If $p_i^k \|ord_q(a)$ for $k<\alpha_i$ then $ord_q(a) \mid p_i^{\alpha_i}\frac{n-1}{p_i^{\alpha_i}} \Rightarrow ord_q(a) \mid p_i^{\alpha_i-1}\frac{n-1}{p_i^{\alpha_i}}$, a contradiction)