# Jordan-Hölder composition series of the additive group $\mathbb{Z}/a\mathbb{Z}$ of integers modulo a

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 $$\langle x^{36} \rangle \lhd \langle x^{18}\rangle \lhd \langle x^{9}\rangle \lhd \langle x^{3}\rangle \lhd \langle x^1\rangle$$

• I know that; I want to know how to do this for a factorization that includes multiplicity. The point precisely is to turn a sequence of distinct factors with multiplicity into a sequence of not necessarily distinct factors without multiplicity. Nov 24 '19 at 23:55
• How do you propose to turn $(p^{\nu_{p}(a)})_{p\in\mathfrak{P}}$ into the sort of sequence you use in your answer? Nov 24 '19 at 23:57
• If I get what you are asking, by writing each $p \in \mathcal{B}$ as $(p, p, \dots, p)$ ($\nu_p(a)$ times) Nov 24 '19 at 23:58
• Yes, but I would like a formal solution to this.---For example, if I gave you the factorization $\prod p^{\nu_{p}(a)}$, what would be the associated Jordan Hölder series? Nov 25 '19 at 0:01
• There is a unique prime factorization of $p^{\nu_p(a)}$ as a product of $p$ $\nu_p(a)$ times. This one is even immune to permutations! Nov 25 '19 at 0:16