Let $a$ be a strictly positive integer.
There exists one-to-one correspondence between the composition series of $(H_i)_{1\leq i\leq n}$ of the group $\mathbb{Z}/a\mathbb{Z}$ and the sequences $(s_i)_{1\leq i\leq n}$ of strictly positive integers such that $a=s_1...s_n$ and
$$[H_{i-1}:H_{i}]=s_i,\text{ for } 1\leq i\leq n .$$
Furthermore, $(H_i)$ is Jordan-Hölder if and only if $(s_i)$ is a sequence of primes.
Let $\mathfrak{P}$ denote the set of prime numbers.
Suppose $a=\prod_{p\in\mathfrak{P}} p^{\nu_{p}(a)}$, where $(\nu_{p}(a))_{p\in\mathfrak{P}}$ is a sequence of strictly positive integers with finite support.
Given this notation, i.e. $a=\prod_{p\in\mathfrak{P}} p^{\nu_{p}(a)}$, how can I apply the above to obtain a Jordan-Hölder composition series of $\mathbb{Z}/a\mathbb{Z}$? The problem is that the result above refers to plain sequences $(s_i)$ without multiplicity, and yet here I have a factorization with multiplicity---how can I turn the sequence $(p^{\nu_{p}(a)})_{p\in\mathfrak{P}}$ into one that iterates all the factors one by one, according to their multiplicities?
For example, given $36=2^{2}3^{2}$, I want to turn the sequence $(2^{2},3^{2})$ into $(2,2,3,3)$.
I only want to be able to associate a Jordan-Hölder series to a factorization of the form $\prod_{p\in\mathfrak{P}} p^{\nu_{p}(a)}$; therefore I'll be happy with any other solution.
$C_a = \mathbb{Z}/a\mathbb{Z}$ is a cyclic group of order $a$, so every subgroup is cyclic of order that divides $a$. Suppose that $a=p_1\dots p_n$ (not necessarily distinct). Then you just pick the sequence of cyclic groups
$$ 1 \lhd C_{p_1} \lhd C_{p_1p_2} \lhd \dots \lhd C_{p_1p_2\dots p_{n-1}} \lhd C_a$$
In your example, this would correspond to the sequence $(2,2,3,3)$
$$ 1 \lhd C_2 \lhd C_{4} \lhd C_{12} \lhd C_{36}$$
which, if $C_a = \langle x \rangle$, corresponds to the groups generated by $$ \langle x^{36} \rangle \lhd \langle x^{18}\rangle \lhd \langle x^{9}\rangle \lhd \langle x^{3}\rangle \lhd \langle x^1\rangle $$