# How to find composition series of Z_n^* of length k?

1. Sep 27, 2015

### jackmell

Hi,

I'd like to find easily-accessible composition series of an interesting lengths say 10 or so (to study some theorems about subgroup series experimentally). I'm thinking the integer mod unit groups, that is
$\mathbb{Z}_n^*=Z_0\rhd Z_1\rhd Z_2\rhd \cdots \rhd Z_9\rhd Z_{10}=\{1\}$. Would this be a relatively simple set of groups to do so with?

That's seems an interesting question: What is the smallest $n$ such that $\mathbb{Z}_n^*$ has a composition series of length $10$? Or for that matter, what is the set of minimum such sizes for a range of $n$? I have no idea and to make matters worst, I do now know how to easily find the set $\{Z_i\}$ other than number-crunching or I guess figure out maybe how to do it in GAP.

Is there a standard way to find these series? May be a GAP command to do so. Don't know yet -- not an easy program to use I think.

Jack

Edit:

Ok just thought of something: A normal subgroup $K\lhd G$ is a maximal normal subgroup iff $G/K$ is simple. Ok then, they're all normal in $Z_n^*$ so then I guess just find the largest one in $Z_n^*$, then find the largest one in that one, largest one in that one and so on until I get to 1. Not sure GAP can handle this though for large groups.

. . . I'm thinking in the millions to get it to 10.

Here's a start:

Code (Text):

gap> DisplayCompositionSeries(Group((1,2,3,4,5,6,7,8,9,10),(1,2)));

G (2 gens, size 3628800)

| Z(2)

S (8 gens, size 1814400)

| A(10)

1 (0 gens, size 1)

I interpret that to mean: $S_{10}\rhd A_{10}\rhd \{1\}$ and I think in fact we can write $S_n\rhd A_n\rhd \{1\}$. Can't figure out how to code integer mod groups though.

Last edited: Sep 27, 2015
2. Sep 27, 2015

### mathwonk

I think the same set of notes I inked for you on finding all groups of small order also classifies in detail the decompostion of all groups of units mod n, as an interesting spoecial example of the theiorem on decomposing finite abelian groups. I'll try to find a page number. see pages 48-50 of these notesn for a product decomposition, which should yield what you want: (but I guess you still need to figure out how many prime factors p-1 has for p a given prime.)

http://alpha.math.uga.edu/~roy/844-2.pdf

Last edited: Sep 27, 2015
3. Sep 27, 2015

### jackmell

Thanks for that mathwonk. However, that reference refers to the decomposition of an integer mod group into it's p-sylow subgroups. I don't see how that's related to the composition series of a group. Take for example:

$\mathbb{Z}_{10!}^*\cong \mathbb{Z}_2\times \mathbb{Z}_2\times\mathbb{Z}_2\times \mathbb{Z}_{2^2}\times\mathbb{Z}_{2^6}\times \mathbb{Z}_3\times\mathbb{Z}_{3^3}\times \mathbb{Z}_{5}$

Now suppose I wanted to compute a (solvable) composition series for that group. Are you suggesting I can use it's decomposition into p-groups to obtain the composition series?

Last edited: Sep 27, 2015
4. Sep 27, 2015

### mathwonk

yes. it looks as if there would be a series of length 16. i.e. for an abelian group the factors of the composition series are all cyclic of prime order. right?

5. Sep 27, 2015

### jackmell

Ok thanks. Afraid I'm not seeing the connection between the p-group decomposition and the group's composition series. Perhaps if we take a smaller group. For example consider $\mathbb{Z}_{100}^*$.

Ok, now the divisors of this group are $1,2,4,5,10,20,25,50,100$ and then using the Theorem I quoted above about maximal normal subgroups $K$ inducing $G/K$ to be simple, then am I correct in saying a composition series for this group would be:

$Z_{100}^* \rhd H_{50} \rhd H_{25} \rhd H_5 \rhd \{1\}$

where the $H_x$ is some maximal (normal) subgroup of the one before it. Alright, now the p-group decomposition for the group is:

$\mathbb{Z}_{100}^*\cong \mathbb{Z}_2 \times \mathbb{Z}_{2^2} \times \mathbb{Z}_5$

How are these two related?

Edit: I made a mistake, the order of $\mathbb{Z}_{100}^*$ is 40 and the divisors of 40 are: $1,2,4,5,8,10,20,40$ so that I think I could write the composition series as:

$\mathbb{Z}_{100}^*\rhd H_{20} \rhd H_{10} \rhd H_{5}\rhd \{1\}$

Also, now I see I think where the prime order of the factors are coming from: in order for the factors $H_i/H_{i+1}$ to be simple, then necessarily, we must have that $|H_i/H_{i+1}|$ is prime (I think). And in the case I wrote above, indeed the factor sizes are 2,2, 2, and 5 (I think).

Last edited: Sep 27, 2015
6. Sep 27, 2015

### mathwonk

yes you are right. a composition series for Zp x Zq x Zr, where p,q,r are all prime, for example is Zp, ZpxZq, ZpxZqxZr, of length 3, the number of prime factors of the order of the group. so for abelian groups, the existence of a composition series, with unique quiotients,. is the world's hardest proof of unique prime factorization of integers.

7. Sep 28, 2015

### jackmell

Ok thanks. Think I figured out an example of a composition series of length 10: Note that $\mathbb{Z}_n^*$ is cyclic for powers of odd primes. Then consider $\mathbb{Z}_{3^{10}}^*=\mathbb{Z}_{59049}^*$ for which $2$ generates the group. Then I believe the following is a composition series for $\mathbb{Z}_{3^{10}}^*$:

$\mathbb{Z}_{3^{10}}^*=Z_0\rhd \big<2\big>\rhd \big<2^{3}\big>\rhd \big<2^{3^2}\big>\rhd\big<2^{3^3}\big>\cdots\rhd \big<2^{3^9}\big>\rhd \big<2^{3^{10}}\big>=\{1\}$

Pretty sure that's correct. But that's my reason for looking for one: now I can experiment with it and figure out why it is and why other properties of composition series are what they are. Working with an actual test case helps me understand the theory better. :)

Edit:

Wrong again: $|\mathbb{Z}_{3^{10}}^*|=39366$. I need to work on this -- will get back later when I figure it out.

Last edited: Sep 28, 2015
8. Sep 30, 2015

### jackmell

The group $\mathbb{Z}_{1024}^*$ I think is a small integer mod group with a composition series of length 10:

$\mathbb{Z}_{1024}^*\rhd \big<a^2\big>\rhd \big<a^4\big>\rhd \big<a^8\big>\rhd\cdots\rhd \big<a^{10}\big>=\{1\}$ where $a$ is any member of this cyclic group which generates the group.