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

  • $\begingroup$ 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. $\endgroup$
    – spring
    Nov 24 '19 at 23:55
  • $\begingroup$ How do you propose to turn $(p^{\nu_{p}(a)})_{p\in\mathfrak{P}}$ into the sort of sequence you use in your answer? $\endgroup$
    – spring
    Nov 24 '19 at 23:57
  • 1
    $\begingroup$ If I get what you are asking, by writing each $p \in \mathcal{B}$ as $(p, p, \dots, p)$ ($\nu_p(a)$ times) $\endgroup$ Nov 24 '19 at 23:58
  • $\begingroup$ 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? $\endgroup$
    – spring
    Nov 25 '19 at 0:01
  • 1
    $\begingroup$ 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! $\endgroup$ Nov 25 '19 at 0:16

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