# Composition length of cyclic groups.

• Kreizhn
In summary, the composition length of a finite cyclic group G is equal to the number of prime factors of |G|, and this result can be extended to all finite solvable groups if we consider that all composition factors are cyclic of prime order. This can be shown by refining a cyclic or abelian series, keeping in mind that simple abelian groups are cyclic p-groups.
Kreizhn

## Homework Statement

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.

## 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.

Hi Kreizhn!

Kreizhn said:

## Homework Statement

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.

## 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.

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

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.

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?

micromass said:
Hi Kreizhn!
So, isn't $\ell(G)$ simply the number of prime factors of |G| (counting multiplicity)? Isn't this what you need?
Absolutely, but is that just the answer? I mean, it seems more an implicit characterization than an explicit one.

micromass said:
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?

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."

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

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

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?

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.

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."

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!

micromass said:
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.
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.

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!
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.

Kreizhn said:
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.

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

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.

Yes, that's very good!

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

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

Kreizhn said:
But why must the refinement of an abelian series be abelian?

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

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 occurred 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:
Kreizhn said:
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?
You are correct. That is not sufficient reason.

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

Kreizhn said:
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 occurred at the beggining? Namely
$$G = G_0 \supseteq H \supseteq K \supseteq G_1 \supseteq \cdots$$?

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.

Hurkyl said:
You are correct. That is not sufficient reason.

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

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.

Kreizhn said:
Sorry, why is G abelian? We only assumed it was finite and solvable.
Whoops, I missed the change of premise.

micromass said:
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.

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?

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.

Kreizhn said:
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?

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

Hurkyl said:
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.

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?

Kreizhn said:
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?
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.

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

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.

## 1. What is the composition length of a cyclic group?

The composition length of a cyclic group is equal to its order. This means that the composition length is equal to the number of elements in the group.

## 2. How do you determine the composition length of a cyclic group?

To determine the composition length of a cyclic group, you can simply count the number of elements in the group. Alternatively, you can use the formula n/φ(n), where n is the order of the group and φ(n) is the Euler totient function.

## 3. Is the composition length of a cyclic group always finite?

Yes, the composition length of a cyclic group is always finite. This is because a cyclic group is generated by a single element and thus has a finite number of elements.

## 4. Can the composition length of a cyclic group be greater than its order?

No, the composition length of a cyclic group cannot be greater than its order. This is because the composition length is equal to the number of elements in the group, and a cyclic group has a finite number of elements.

## 5. How does the composition length of a cyclic group affect its structure?

The composition length of a cyclic group is a measure of its complexity. A longer composition length indicates a more complex structure, while a shorter composition length indicates a simpler structure. Additionally, the composition length can also affect the group's subgroups and quotient groups.

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