I have this proposition in the book I'm studying:
Consider $n>2$ on $\mathbb{N}$ and $p$ a prime number such that $p\nmid n$ and $p$ is not equivalent to $1$ modulo $n$. If the factorization of $n$ on $\mathbb{N}^*$ is $$n=2^\alpha q_1^{\alpha_1} q_2^{\alpha_2} \cdots q_k^{\alpha_k},$$ with $\alpha\geq 0$, $k\geq 0$ and for $k>0$, the $q_i$, $1\leq q_i \leq k$, are prime numbers two to two distinct and $\alpha_i$ are not null, we get the results down to the degree $m$ of $n$-ésima cyclotomic extension over $\mathbb{F}_p$: \begin{align*} (k=0, \alpha =2) &\Rightarrow m=2 \\ (k=0, 2< \alpha) &\Rightarrow m=2^{\alpha -2} \\ (1 \leq k, 0 \leq \alpha <2) &\Rightarrow m=\text{lcm}(o(\overline{p}_{(i)}); 1 \leq i \leq k) \\ (1 \leq k, \alpha =2) &\Rightarrow m=\text{lcm}(2, o(\overline{p}_{(i)}); 1 \leq i \leq k) \\ (1 \leq k, 2< \alpha) &\Rightarrow m=\text{lcm}(2^{\alpha-2}, o(\overline{p}_{(i)}); 1 \leq i \leq k), \end{align*} with $\overline{p}_{(i)}$ the class of equivalence of $p$ modulo $q{_i^{\alpha_i}}$ and the $o(\overline{p}_{(i)})$ is the order of $\overline{p}_{(i)}$ in the multiplicatif group $G_{q{_i^{\alpha_i}}}$.
For the demonstration of this, the autor use the fact that: $$G_n\simeq G_{2^\alpha}\times G_{q_1^{\alpha_1}}\times \cdots \times G_{q_k^{\alpha_k}}$$ but I did not understand why the lcm appeared in there.
$\Bbb{F}_p(\zeta_n)$ is the splitting field of $x^n-1$ which divides $x^{p^k-1}-1$ iff $n |p^k-1$ iff $ order(p\bmod n)\ |\ k$.
For $m=ab,\gcd(a,b)=1$ then $$order(p\bmod n) = lcm(order(p\bmod a),order(p\bmod b))$$