Theorem $7$ (Decomposition into prime factors). Let $\mathfrak{P}$ denote the set of prime numbers and $a$ be a strictly positive integer. There exists one and only one family $(\nu_{p}(a))_{p\in\mathfrak{P}}$ of integers $>0$ such that the set of $p\in\mathfrak{P}$ with $\nu_{p}(a)\ne 0$ is finite and $$a=\prod_{p\in\mathfrak{p}} p^{\nu_{p}(a)}.$$
Bourbaki Algebra p.51
Bourbaki says:
...for every family $(\nu_{p}(a))_{p\in\mathfrak{P}}$ satisfying the conditions of Theorem $7$ $\nu_{p}(a)$ is, for all $p\in\mathfrak{P}$, equal to the number of factors of a Jordan-Holder series of $\mathbb{Z}/a\mathbb{Z}$ isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The uniqueness of the family $(\nu_{p}(a))_{p\in\mathfrak{P}}$ therefore follows from the Jordan-Holder theorem.
Can someone please help me understand why this is so? Why does uniqueness follow from Jordan-Holder theorem?
Edit:
A Jordan-Holder series of a group with operators $G$ is a strictly decreasing composition series $\Sigma$ such that there exists no strictly decreasing composition series distinct from $\Sigma$ and finer than $\Sigma$.
Theorem $6$ (Jordan-Holder). Two Jordan-Holder series of a group with operators are equivalent.
Let $G$ be a group, and let $$G = G_1 \triangleright G_2 \triangleright G_3 \triangleright \cdots \triangleright G_{n} \triangleright G_{n+1} = 1$$ be a composition series of $G$ (Jordan-Hölder series).
Here $Q_i := G_i/G_{i+1}$ is simple for all $i = 1,\ldots,n$, and $|G| = |Q_1||Q_2| \cdots |Q_{n}|$.
In the case where $G = \mathbb{Z}/a\mathbb{Z}$, you always have $Q_i \cong \mathbb{Z}/p_i\mathbb{Z}$ for some prime $p_i$, and so $a = p_1p_2 \cdots p_n$.
Conversely when $a = p_1p_2 \cdots p_{n}$ for primes $p_i$, you can construct a composition series of $G = \mathbb{Z}/a\mathbb{Z}$ with $Q_i \cong \mathbb{Z}/p_i\mathbb{Z}$.
Hence if $a = p_1 p_2 \cdots p_n = p_1' p_2' \cdots p_m'$ for some primes $p_i, p_j'$, by Jordan-Hölder $m = n$ and $\{p_1, \ldots, p_n\} = \{p_1', \ldots, p_m'\}$.