Refining a normal series into a composition 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?

 

[fresh_42]

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]

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...

 

[fresh_42]

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?

 

[fresh_42]

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 $G_i$.

 

[fishturtle1]

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.

 

[fresh_42]

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.