# Refining a normal series into a composition series

• fishturtle1
In summary: But this means that the normal subgroup is a proper normal subgroup, and so it's clear that ##G## has a composition series.In summary, if a group has a composition series, then it cannot have an infinite normal series.
fishturtle1
Homework Statement
Suppose ##G## has a composition series. Then any normal series of ##G## can be refined into a composition series.
Relevant Equations
A normal series of ##G## is
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
where ##G_i \trianglelefteq G_{i+1}##. The factors of this normal series are the quotient groups ##G_i / G_{i+1}##. The length of a normal series is the number of distinct factors. A composition series is a normal series whose factors are all simple. Equivalently, a composition series is a normal series of maximal length.
Attempt: Consider an arbitrary normal series ##G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_n = 1##. We will refine this series into a composition series. We start by adding maximal normal subgroups in between ##G_0## and ##G_1##. If ##G_0/G_1## is simple, then we don't have to do anything. Choose maximal normal subgroup ##G_{0,0}## where ##G_0 \ge G_{0,0} \ge G_1##. Then ##G_0/G_{0,0}## is simple. If ##G_{0.0}/G_1## is also simple, we can stop. Otherwise we keep choosing maximal normal subgroups. Eventually?, we get
$$G = G_0 \ge G_{0,0} \ge G_{0,1} \ge \dots \ge G_{0, k_0} \ge G_1 \ge G_2 \ge \dots \ge G_n = 1$$
where ##G_0/G_{0,0}, G_{0,0}/G_{0,1}, \dots, G_{0, k_0-1}/G_{0, k_0}, G_{0, k_0}/G_1## are simple. But how do we know we eventually find a group ##G_{0, k_0}##? Is it because we're guaranteed a composition series exists?

Can we have a normal series whose length is greater than the length of a composition series?

No, to your last question. All composition series have the same length (Jordan-Hölder). Given that, you only have to verify two things:
1. Any non simple factor allows a refinement.
2. The repeated process of refining stops, i.e. there is no way, that refining a normal series is endless without ever reaching a composition series. Since ##G## is not assumed finite, it could be that ##G## has a composition series, plus a second normal series which remains only normal infinitely.
If it stops, then you can conclude with Jordan-Hölder that both series must be equivalent.

fishturtle1

Let ##G = G_0 \ge G_1 \ge \dots \ge G_n = 1## be a normal series.

1) If ##G_0/G_1## is a non simple factor, then there exists a proper normal subgroup ##G_{0,0}/G_1 \triangleleft G_0/G_1##. By correspondence theorem, ##G_0 \triangleright G_{0,0} \triangleright G_1##, i.e., we've found a refinement for ##G_0 \ge G_1##.

2) We're given that ##G## has a composition series of length, say, ##m##. We want to show the process in 1) eventually stops. Assume by contradiction it never stops. Then we could have a refinement of the above series

$$G = H_0 \ge H_1 \ge \dots \ge H_k = 1$$ with at least ##m+1## simple factors. I'm not sure where to go from here...

How could you find the ##H## refinement? What if they are completely different, i.e. no common factors? Of course it cannot happen that we have a finite composition series and simultaneously an infinite normal series, but why? What if all factors are different?

fishturtle1
I don't think I can find the ##H## refinement.

Let ##G = G_0 \ge G_1 \ge \dots \ge G_n = 1## be a finite composition series and assume by contradiction there exists an infinite normal series ##G = H_0 \ge H_1 \ge \dots \ge 1##. I thought maybe to insert ##G_1, G_2, \dots , G_{n-1}## into the ##H## series but that doesn't work...

Why do we need to consider if all the factors are different?

Well, we know the result: any two composition series are equivalent and if a group has one, there cannot be an infinite normal series. The difficulty is to say why? Let's see.

We have ##G=G_0=H_0##. Assume ##G_1\neq H_1##. Then ##H_1/H_1\cap G_1 \cong H_1G_1/G_1 \trianglelefteq G/G_1##, because ##gh_1g_1g^{-1}=h_1'gg_1g^{-1}=h_1'g_1'## and ##H_1G_1## is normal. This means either ##H_1G_1=G_1## or ##H_1\cap G_1=H_1## or ##H_1\subsetneq G_1## in which case we can put ##G_1## between ##G=H_0## and ##H_1##, or ##H_1G_1=G.## I haven't an elegant argument why this can't happen, maybe you find one. I proceed with ##G_2## or ##H_2## and rule out more cases, but I think, there must be a better way to see it.

Edit: Maybe it is easier to show that any normal subgroup has to be one of the ##G_i##.

Last edited:
fishturtle1
fresh_42 said:
Well, we know the result: any two composition series are equivalent and if a group has one, there cannot be an infinite normal series. The difficulty is to say why? Let's see.

We have ##G=G_0=H_0##. Assume ##G_1\neq H_1##. Then ##H_1/H_1\cap G_1 \cong H_1G_1/G_1 \trianglelefteq G/G_1##, because ##gh_1g_1g^{-1}=h_1'gg_1g^{-1}=h_1'g_1'## and ##H_1G_1## is normal. This means either ##H_1G_1=G_1## or ##H_1\cap G_1=H_1## or ##H_1\subsetneq G_1## in which case we can put ##G_1## between ##G=H_0## and ##H_1##, or ##H_1G_1=G.## I haven't an elegant argument why this can't happen, maybe you find one. I proceed with ##G_2## or ##H_2## and rule out more cases, but I think, there must be a better way to see it.

Edit: Maybe it is easier to show that any normal subgroup has to be one of the ##G_i##.

That makes sense; Maybe something like this would work?

Let ##P(k)## say that if a group ##G## of order ##k## has a composition series, then it cannot have an infinite normal series.

base case: If ##k = 1##, then ##G## has a composition series ##G \ge 1##. It's clear that any normal series of ##G## has exactly one factor: ##G/1##.

inductive step: We induct on ##\vert G \vert##. Suppose ##G## has a composition series. Then any normal subgroup of ##G## also has a composition series. Consider any normal series of ##G##
$$G = G_0 \ge G_1 \ge \dots \ge 1$$
Then ##G_1## is a normal subgroup of ##G## with ##\vert G_1 \vert < \vert G \vert##. Hence, ##G_1## has a composition series. By ind. hypotheses, ##G_1## has no infinite normal series. Hence,
$$G = G_0 \ge G_1 \ge \dots \ge 1$$ must be finite in length. []

But here, I assumed if ##G## has a composition series than any normal subgroup of ##G## also has a composition series; this statement is an ex. in the textbook that I haven't proven yet so maybe this is circular...

i'll keep thinking on it...

Finite groups aren't the problem. Any refinement process will automatically stop (repetitions excluded). So finite orders are off the table.

I'm pretty sure that ##G=H_1G_1## with two proper normal subgroups and ##G/G_1=G_1H_1/G_1=H_1/G_1\cap H_1## cannot occur. I just haven't found the clue.

fishturtle1

## 1. What is a normal series?

A normal series is a sequence of subgroups in a group, where each subgroup is a normal subgroup of the previous subgroup. This means that the quotient group formed by each consecutive pair of subgroups is also a normal subgroup.

## 2. What is a composition series?

A composition series is a normal series where each quotient group is simple, meaning it has no nontrivial normal subgroups. In other words, a composition series breaks down a group into its simplest form.

## 3. Why is refining a normal series into a composition series important?

Refining a normal series into a composition series allows us to understand the structure of a group in a more simplified way. It also helps us to classify groups and determine their properties.

## 4. How do you refine a normal series into a composition series?

To refine a normal series into a composition series, we need to keep replacing the subgroups in the series with their corresponding composition factors until we reach a composition series. This process is known as composition series refinement.

## 5. Can a group have more than one composition series?

Yes, a group can have multiple composition series. However, all composition series for a given group will have the same length, meaning they will have the same number of subgroups in the series. Additionally, any two composition series for a group will have the same composition factors, although they may be in a different order.

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