2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$C_a = \mathbb{Z}/a\mathbb{Z}$ 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 $$ \langle x^{36} \rangle \lhd \langle x^{18}\rangle \lhd \langle x^{9}\rangle \lhd \langle x^{3}\rangle \lhd \langle x^1\rangle $$

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.