Let $\mu_n:=\{\xi \in \mathbb{C}^* \ / \ \xi^n=1\}$ and $\mu_n^*$ be the set of elements of order exactly $n$.
We can also write : $\mu_n:=\{{\exp(\frac{2ik\pi}{n}) , \ k\in[\mid0,n-1 \mid]}\}$ and $\mu_n^*:=\{\exp(\frac{2ik\pi}{n}) , \ k\in[\mid0,n-1 \mid] \ \text{and} \ \gcd{(k,n)}=1\}$.
Let consider in $\mathbb{C}[X]$ the following polynomial : $X^n-1$.
Then we have that : $\displaystyle X^n-1= \prod_{k=0}^{n-1}(X-\exp\left(\frac{2ik\pi}{n}\right))= \prod_{\xi \in \mu_n}(X-\xi)$.
Moreover, note that we can consider the next partition : $\displaystyle \mu_n = \bigsqcup_{d\mid n}\mu_d^*$. Indeed, a reasoning on orders of elements gives that the two sets are the same and the definition of the order ensures that every $\mu_d^*$ for $d\mid n$ are pairwise disjoints.
Hence $\displaystyle X^n -1 =\prod_{\xi \in \mu_n}(X-\xi) = \prod_{\xi \in \underset{d\mid n}{\bigsqcup}\mu_d^*}(X-\xi)\overset{?}{=}\prod_{d\mid n}\prod_{\xi \in \mu_d^*}(X-\xi)$.
How can I justify the last equality ? It is really similar to the phenomenon with sums and the packets summation property. It seems to be just the associativity of the multiplication. However I meet difficulties to see how I can exploit the big index of the product.
Is this linked with group actions ?
Thanks in advance !
In general, for finite sets $A$ and $B$ and $a_i\in\mathbb C$ for each $i\in A\sqcup B$, you have $\prod_{i\in A\sqcup B}a_i=\prod_{i\in A}a_i\cdot\prod_{i\in B}a_i$. (which you can check straight-forwardly, by inducting on the size of $A$ and using $\prod_{i\in A}a_i=a_j\cdot\prod_{i\in A\setminus\{j\}}a_i$.)
Now we further observe that $\prod_{i\in A\sqcup B\sqcup C}a_i=\prod_{i\in A}a_i\cdot\prod_{i\in B}a_i\cdot\prod_{i\in C}a_i$, and in fact this generalizes to any number of disjoint finite sets by inducting on the number of sets ($A_1,\dots,A_N$):
$$\prod_{i\in\sqcup_{k=1}^NA_k}a_i=\prod_{k=1}^N\prod_{i\in A_k}a_i.$$