# Homework Help: Composition length of cyclic groups.

1. Jun 27, 2011

### Kreizhn

1. The problem statement, all variables and given/known data
Let G be a finite cyclic group and $\ell(G)$ be the composition length of G (that is, the length of a maximal composition series for G). Compute $\ell(G)$ in terms of |G|. Extend this to all finite solvable groups.

3. The attempt at a solution

Decompose |G| into its prime factors, say
$$|G| = p_1^{r_1} \cdots p_n^{r_n}.$$
Since G is cyclic, it has a subgroup of order corresponding to every divisor of |G|, and this subgroup is normal. Further, since quotients of cyclic groups are cyclic, in order to ensure that the quotients are simple we need to give them prime order. Consequently, the maximal composition series is given by
$$\ell(G) = \sum_{i=1}^n r_i$$
since for each step, we choose a prime $p_{i_j}$ and create the normal subgroup of order $p_1^{r_1} \cdots p_{i_j}^{r_{i_j}-1} \cdots p_n^{r_n}$ which then has prime index making its quotient simple.

Now problem one: It's unclear to me how to write $\ell(G)$ in terms of |G|. I feel like the totient should be involved somewhere.

Problem two: I'm not too sure how to extend this to finite solvable groups. I can show things like the group is solvable if and only if all composition factors (the quotients of the comp series) are cyclic. It seems to me that I could probably use this, though it's not clear to me how.

2. Jun 27, 2011

### micromass

Hi Kreizhn!

So, isn't $\ell(G)$ simply the number of prime factors of |G| (counting multiplicity)? Isn't this what you need?

That's not all of it. A group is solvable iff all composition factors are cyclic of prime order. Can you do something with this?

3. Jun 27, 2011

### Kreizhn

Absolutely, but is that just the answer? I mean, it seems more an implicit characterization than an explicit one.

I have a theorem that states, for a finite group

All composition factors are cyclic $\Leftrightarrow$ G admits a cyclic series terminating in identity $\Leftrightarrow$ G admits and abelian series terminating in identity $\Leftrightarrow$ G is solvable

Primality wasn't explicitly included. Admittedly, there is a part of the proof that I never fully understood. There is a statement that admitting an abelian series implies composition factors are cyclic by simply refining the abelian series to a composition series and keeping in mind that simple abelian groups are cyclic p-groups. Maybe this is the statement of primality, although hidden?

Anyway, assuming that we can use the fact that the composition factors are cyclic of prime order, then it seems like we would do something similar. The only problem I foresee is that we cannot guarantee the existence of a subgroup of prime index. Thus the answer would be somewhat hand-waving, in that I would say "If we can make a subgroup of prime index, great. Otherwise, do something else."

4. Jun 27, 2011

### micromass

Well, I don't really know what they're looking for, but I think you can't really do much simpler.

The composition factors are always simple by definition, right? So if the composition factors are cyclic, then they are cyclic and simple. And the only cyclic and simple groups have prime order.

It wouldn't be handwaving, since we know that a composition series exist. So we can make a series of subgroups such that everything you want it satisfies. It can be made completely rigorous!

5. Jun 27, 2011

### Kreizhn

Ah yes, this makes sense, the primality is implied. Though I still don't see the implication that "admits abelian implies cyclic composition factors" since wouldn't it be possible for the refinement from abelian to composition to contain composition factors that are simple but not cyclic? I guess it's a little off topic though.

Sorry, I didn't mean that the proof would be hand-wavy, just that the statement of $\ell(G)$ in terms of |G| would be somewhat shady. But I think I see what you're saying, and I would like to redact that statement.

Okay, let's give it a try:

Let G be a finite solvable group. By theorem, it's composition factors are all cyclic (and hence of prime order by simplicity), and so we can write
$$G = G_0 \supseteq G_1 \supseteq \cdots \supseteq G_n = {e}, \qquad |G_{i+1}/G_i | = p$$
Now we claim that every prime, including its multiplicity, must appear. This is probably better done with induction, but since the series is finite, I'm just going to argue by recursion. We note that $|G_0/G_1| = p_1$ so that $|G| = |G_0| = p_1|G_1|$. Similarly, |G_1| = p_2 |G_2|, and this terminates at $|G_{n-1}| = p_n |G_n| = p _n$ since $G_n = {e}$. Hence we have
$$|G| = \prod_{i=1}^n p_i$$
by the fundamental theorem of arithmetic, the prime decomposition of |G| is unique and hence we pick up all of the prime factors of G including their multiplicity.

6. Jun 27, 2011

### micromass

Well, the only abelian simple groups are the cyclic groups of prime order!

Yes, that's very good!

7. Jun 27, 2011

### Kreizhn

But why must the refinement of an abelian series be abelian?

8. Jun 27, 2011

### micromass

Because quotients and subgroups of abelian groups are abelian. Or do I misunderstand your question?

9. Jun 27, 2011

### Kreizhn

Well, I'm just thinking like this. Say our group admits an abelian series terminating in identity, say
$$G =G_0 \supseteq G_1 \supseteq \cdots \supseteq G_n = \{ e\}$$.
Now each quotient $G_{i}/G_{i+1}$ is abelian by assumption.

Refine this to a composition series, and say at some point we have subgroups H,K pop up as follows:
$$G = G_0 \supseteq \cdots \supseteq G_{r-1} \supseteq H \supseteq K \supseteq G_{r} \supseteq \cdots \supseteq G_n = \{e\}$$

Now why must $G_{r-1}/H$ be abelian? We know that $G_{r-1}/G_r$ is abelian but this doesn't mean that either $G_{r-1}$ or $G_r$ is abelian right?

Edit: Is $G_{r}/H$ a subgroup of $G_{r-1}/G_r$? That would work for this case, but then what if the H,K occured at the beggining? Namely
$$G = G_0 \supseteq H \supseteq K \supseteq G_1 \supseteq \cdots$$

Edit 2: Yes, that should be $H/G_{r-1}$. So many indices to keep track of, it hurts my head.

Last edited: Jun 27, 2011
10. Jun 27, 2011

### Hurkyl

Staff Emeritus
You are correct. That is not sufficient reason.

The most obvious reason is because G is abelian....

11. Jun 27, 2011

### micromass

Since H is a subgroup of $G_{r-1}$, we know that $H/G_r$ is a subgroup of $G_{r-1}/G_r$ and thus abelian.

12. Jun 27, 2011

### Kreizhn

Sorry, why is G abelian? We only assumed it was finite and solvable.

Edit: Sorry. In this case we actually only assumed it admitted an abelian series terminating in identity.

13. Jun 27, 2011

### Hurkyl

Staff Emeritus
Whoops, I missed the change of premise.

14. Jun 27, 2011

### Kreizhn

Ah yes, that goes back to my edit from that previous post. But does that still work if the refinement occurs at the beginning of the tower?

15. Jun 27, 2011

### Hurkyl

Staff Emeritus
Things are coming back to me. Aren't the first few composition factors of
$$G_0 \supseteq G_1 \supseteq G_2 \supseteq \cdots$$​
the same as the composition factors of
$$G_0 / G_3 \supseteq G_1 / G_3 \supseteq G_2 / G_3 \supseteq 1$$​
? So we can simplify the problem from a whole big composition series to just the part that matters.

16. Jun 27, 2011

### micromass

I don't see why not, just take r=1...

17. Jun 27, 2011

### Kreizhn

I'm not sure that the normality transcends its immediate neighbours though. Namely, if $H \subseteq K \subseteq G$ with H normal in K and K normal in G, it is not necessary that H be normal in G, so that composition needn't make sense. Unless of course I missed something in the definition of composition series?

18. Jun 27, 2011

### Hurkyl

Staff Emeritus
Ah, you're right. I'm far too spoiled by commutative algebra. (I was remembering echoes of composition series for modules, not for groups) Fortunately, mm answered the question.

19. Jun 27, 2011

### Kreizhn

Ah, I missed this post amongst the flurry of postings and from getting bumped to page 2. Okay, I guess everything works out. Thanks for your patience.