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The question is as follows:

A chief factor $\frac{K}{L}$ of a finite group $G$ is said to be central if $\frac{K}{L} \subseteq Z(\frac{G}{L})$. Show that all chief series for $G$ contain the same number of central factors.

$\textbf{Definitions:}$

We know that a chief series of a group G is an invariant (normal) series $$\{ e \} = G_r \subseteq \cdots G_1 \subseteq G_0 = G$$ of G such that $ G_i \subset G_{i-1}$ and if $G_i \subseteq N \subseteq G_{i-1}$ with $N \vartriangleleft G$, then either $G_{i-1}=N $ or $N=G_i$. The factor groups $\frac{G_{i-1}}{G_i}$ are called the chief factors.

A finite sequence $$ \{ e \} = G_n \subseteq \cdots G_1 \subseteq G_0 = G$$ of subgroups of $G$ is called subnormal series of $G$ if $G_i$ is a normal subgroup of $G_{i-1}$ for each $i$, $1 \leq i \leq n$.

A subnormal series $$ \{ e \} = G_n \subseteq \cdots G_1 \subseteq G_0 = G$$ of group $G$ is called composition series if for $1 \leq i \leq r$ , all the non trivial factor group $\frac{G_{i-1}}{G_i}$ are simple. The factor groups of this series are called composition factors.

Now let $ \{ e \} \subseteq G_n \subseteq \cdots G_1 \subseteq G_0 = G$ be a subnormal series of $G$. Then a subnormal series $ \{ e \} = H_n \subseteq \cdots H_1 \subseteq H_0 = G$ is called refinement of $G$ if every $G_i$ is one of the $H_j$’s.

$\textbf{Idea}$ $\textbf{for} $ $ \textbf{to}$ $\textbf{prove} $ $\textbf{the}$ $\textbf{statement}$:

First it is possible to prove that any two invariant (normal) series have isomorphic refinements. And then by using this we can prove that for a given Chief series, any other chief series is isomorphic to the given one.

Now my question is that will it prove the desired result in the statement of the question?

Thanks!

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  • $\begingroup$ This follows as a corrolary from some other results on series. Would that answer your question or do you need a proof from more basic principles? $\endgroup$ – Qudit Dec 13 '17 at 20:24
  • $\begingroup$ @Qudit Thanks! So you mean the answer by Colin is not correct? $\endgroup$ – Farrokh Dec 14 '17 at 12:28
  • $\begingroup$ No, I think it's fine. $\endgroup$ – Qudit Dec 14 '17 at 15:53
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Let's take two chief series $$\{ e \} = G_r \subseteq \cdots G_1 \subseteq G_0 = G$$ $$\{ e \} = H_r \subseteq \cdots H_1 \subseteq H_0 = G$$ As you've said, these series are isomorphic up to a permutation $\pi:\{0,\ldots, r\}\rightarrow \{0,\ldots, r\}$ such that $G_{i-1}/G_i$ is isomorphic to $H_{\pi(i)-1}/H_{\pi(i)}$. We'll call the isomorphisms $f_i:G_{i-1}/G_i\rightarrow H_{\pi(i)-1}/H_{\pi(i)}$.

The important point is that for any $g\in G$, $f_i\left(gkG_i\right)=gf_i\left(kG_i\right)$, which in particular means that $f_i\left(gG_i\right)=gf_i\left(G_i\right)$ for any $g\in G$.

Using your definition, chief factors $\frac{G_{i-1}}{G_i}$ are called central if $\frac{G_{i-1}}{G_i}\subset \mathcal{Z}\left(\frac{G}{G_{i}}\right)$. Elements $kG_i\in\mathcal{Z}\left(\frac{G}{G_{i}}\right)$ are defined by the property that $gkG_i=kgG_i$ for all $g\in G$.

That means that, assuming $\frac{G_{i-1}}{G_i}$ is central, $$gkH_{\pi(i)}=gkf_i\left(G_i\right)=gf_i\left(kG_i\right)=f_i\left(gkG_i\right)=f_i\left(kgG_i\right)=kf_i\left(gG_i\right)=kgf_i\left(G_i\right)=kgH_{\pi(i)}$$ The outermost left and right hand side show that $H_{\pi(i)-1}/H_{\pi(i)}\subset \mathcal{Z}\left(G/H_{\pi(i)}\right)$, i.e. $H_{\pi(i)-1}/H_{\pi(i)}$ is a central factor.

So what we have obtained there is that factors $\frac{G_{i-1}}{G_i}$ of the first chief series are central if and only if factors $f\left(\frac{G_{i-1}}{G_i}\right)=H_{\pi(i)-1}/H_{\pi(i)}$ of the second chief series are central. Since $\pi$ is one to one, $\{i \in \{1,\ldots, r\}:\frac{G_{i-1}}{G_i}\text{ is central}\}$ and $\{\pi(i) \in \{1,\ldots, r\}:\frac{H_{\pi(i)-1}}{H_\pi(i)}\text{ is central}\}$ must be the same size.

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  • $\begingroup$ Many thanks for your knowledgeable answer! $\endgroup$ – Farrokh Dec 15 '17 at 5:43
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You want to prove that any two normal series for $G$ have $G$-equivariantly isomorphic refinements, where $G$ acts on each normal factor by conjugation: that is, you can refine them to normal series

$G_0 \le G_1 \le \dots \le G_n$

$H_0 \le H_1 \le \dots \le H_n$

such that there is a permutation $\pi$ of $\{1,\dots,n\}$ and a $G$-equivariant isomorphism from $G_{i}/G_{i-1}$ to $H_{\pi(i)}/H_{\pi(i) -1}$ for each $i$. The proof is similar to the basic Schreier refinement theorem, except you keep track of the $G$-actions.

So if both of the series you start with are chief series (i.e. as normal series, they cannot be properly refined), then up to reordering they have the same factors with the same multiplicities, regarded as equivalence classes of $G$-groups (i.e. groups equipped with an action of $G$ by automorphisms). In particular, being central (i.e. having trivial $G$-action) is a property that is accounted for, so the number of central factors must be the same.

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    $\begingroup$ Many thanks! Can you please explain more? $\endgroup$ – Farrokh Dec 10 '17 at 16:36

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