Refining a normal series into a composition series

  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
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  • #3
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Thanks for the reply!

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...
 
  • #4
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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?
 
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  • #5
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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?
 
  • #6
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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##.
 
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  • #7
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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...

Thank you for your time and patience on this thread.
 
  • #8
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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.
 
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