Results about Composition Series of Finite Groups we should all know and love

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EnglishMuon
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What?!?! Finite Groups is not enough to get u excited for this thread?!? Well I guess then I need a

COOL TITLE TO GET UR ATTENTION?!?!


Well this is mostly a form of procrastination from actual revision, but I thought it would be more productive to waffle here than join physicsmaths in watching prison break for the 10th straight time.

Anyway when dealing with finite groups, I feel we should get some feeling for how 'fiddle can be turned into definiteness' and come up with concise results for future use and peace of mind.

So the intention of this thread is my to work through some hopefully partially coherent proofs of results on this area whilst either discussing with those interested, or, writing some strange scribbles as the rest of tsr leaves this thread to go about its strange business and eventually die.

So let's start with some definitions and basic proofs, leading up to the Jordan-Holder Theorem.

(There is a reason for the title colour- you may want to read https://blog.kissmetrics.com/color-psychology/ if interested). (And the comic sans? no, that was to be memey)

Obligatory tags for max publicity: Zacken Duke Glacia Number Nine Jeremy Corbyn


prepare yourselves for some mad group action right here...
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EnglishMuon
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Prelim. Definitions

For all following posts let  G be any group.

An  \Omega -series of  G is a finite sequence of subgroups  e = G_{0} \unlhd G_{1} \unlhd ... \unlhd G_{n} =G such that  G_{i+1} \unlhd G_{i} for all i. (Note this does not mean we have to have factors aside from second to last which are normal in  G . In such a case with normal factors, we call it an  \Omega -subnormal series .)

Similarly to refinements in analysis, we say a series A is a refinement of another series B if all the factors of B are in A.

Two series A and B are  \Omega -isomorphic if there exists a bijection between the factors such that each factor in one series is mapped to an isomorphic factor in the other.
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Number Nine
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first we define what a group is
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EnglishMuon
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Zassenhaus's Lemma:

Let  A_1, A_2, B_1, B_2 be  \Omega -subgroups of  G with  A_1 \lhd A_2, B_1 \lhd B_2 . Let  D_{ij} = A_{i} \cup B_{j} .

Then  A_1 D_{21} \lhd A_1 D_{22}, B_1 D_{12} \lhd B_1 D_{22} , and

 A_1 D_{22} / A_1 D_{21} \simeq B_1 D_{22} / B_1 D_{12} .

Proof:

Conjugating  A_1 D_{21} , B_1 D_{12}  by  a_1 h, a_1 \in A_1 . h \in D_{21} and simplifying gives the result for the first part.

2nd part follows from the second isomorphism theorem.
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EnglishMuon
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The Schreier Refinement Theorem:

Any two  \Omega -series of  G have  \Omega -isomorphic refinements.

Proof:

Let  e=H_0 \lhd H_1 \lhd ... \ H_m = G, e=K_0 \lhd K_1 \lhd ... \ K_n = G be two given series.

Define  H_{ij}=H_{i} (H_{i+1} \cap K_{j}), K_{ij}=K_{j} (H_{i} \cap K_{j+1}) .

Then by Zassenhaus's lemma  H_{ij} \lhd H_{i, j+1}, K_{ij} \lhd K_{i+1, j} and  H_{i,j+1}/H_{ij} \simeq K_{i+1,j}/K_{ij} , so the series  \{ H_{ij} \}, \{ K_{ij} \} are  \Omega -isomorphic (where i, j run over all possible values).

Note they are refinements of the original series as  H_{i,0} = H_i (H_{i+1} \cap e)=H_{i} for example.

(The new series is a sort of lexicographic ordering on the i and j indices, filling the gap between any two previous terms with another series.)
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EnglishMuon
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Composition Series

We call an  \Omega -series an  \Omega Composition series if there is no proper refinement of the series. Such a series always exists for  G is finite (simply as we have the weak upper bound for number of factors as the order of  G ), but not always for the infinite case.

For example consider  G = ( \mathbb{Z}, + ) . Since  G is abelian, if the last (nontrivial) factor of one series is  n \mathbb{Z} we may add on  2n \mathbb{Z} for instance.

An  \Omega -series is an  \Omega Composition series iff all of its factors are simple

Proof:

Forwards: Let  e=H_0 \lhd H_1 \lhd ... \lhd H_n = G be a given series.

Suppose  H_{k+1} / H_{k} is not simple for some k. Then by 1-1 correspondence of normal groups in  H_{k+1} properly containing  H_{k} and normal subgroups of  H_{k+1} , we can find another subgroup between  H_k, H_{k+1} and add to our series. So the series is not a composition series.

Backwards: By the same correspondence, if all the factors are simple, we can find any more terms to add.
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username1258398
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alright rob
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EnglishMuon
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Jordan-Holder Theorem

Let S be a composition series and T be any other  \Omega -series. Then there exists a refinement of T, T'that is a composition series and S, T' are  \Omega isomorphic.

Proof

By the Schreier Refinement Theorem, we may find a common refinement T', S' which are isomorphic. But S is a composition series so S=S' and so T' is isomorphic to S.
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EnglishMuon
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Proof of Fundamental Theorem of Arithmetic using Jordan-Holder Theorem

Let  n \in \mathbb{N} be given. Let's assume  n is a product of primes since this is easy to prove by induction and WLOG let  n = p_1 p_2 ... p_n = q_1 q_2 ... q_n for  p_i, q_i primes.

Then  1 \lhd \mathbb{Z} / p_1 \mathbb{Z} \lhd \mathbb{Z} / p_1 p_2 \mathbb{Z} \lhd ... \lhd \mathbb{Z} / n \mathbb{Z}, 1 \lhd \mathbb{Z} / q_1 \mathbb{Z} \lhd \mathbb{Z} / q_1 q_2 \mathbb{Z} \lhd ... \lhd \mathbb{Z} / n \mathbb{Z} are composition series of the group  G= ( \mathbb{Z} / n \mathbb{Z}, \times).

So by Jordan-Holder theorem the factors ( \mathbb{Z} / p_1 p_2 ....p_{r} \mathbb{Z}) / ( \mathbb{Z} / p_1 p_2 ....p_{r-1} \mathbb{Z}) \simeq \mathbb{Z} / p_{r} \mathbb{Z} and similarly  \mathbb{Z} / q_{r} \mathbb{Z} are \Omega -isomorphic, hence the  p_i are the  q_i in some order.
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EnglishMuon
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Short Exact Sequences

A Short Exact Sequence is essentially a neat way of showing the relationships between morphisms of elements in a given category that contain the notion of kernels.

More explicitly, the simplest (nontrivial) exact short exact sequence is of the form

 0 \rightarrow A \rightarrow^{f} B \rightarrow^{g} C \rightarrow 0 .

where  f is the embedding map of  A into  B and  g is the quotient map of  A from  B , so in some loose sense we have '  C \simeq B/A '.

This occurs for lots of familiar objects we are used to-  A,B.C could all the groups, rings, modules, ... .

This can be extended in the obvious way for n terms,  A_1 , ... , A_n such that any 3 consecutive terms behave as in the example above.
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EnglishMuon
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Splitting Lemma

We say that the short exact sequence 0 \rightarrow A \rightarrow^{f} B \rightarrow^{g} C \rightarrow 0 splits if there is a homomorphism  h: C \rightarrow B such that  gh is the identity map on  C .

The Splitting Lemma asserts TFAE:

Such an  h: C \rightarrow B exists.
Similarly a  k: B \rightarrow A exists with  kf identity on  B
 B \simeq A \bigoplus C

For familiar objects such as modules this seems very intuitive, so I just attach an abstract nonsense proof below.
https://math.stackexchange.com/quest.../753182#753182
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RichE
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(Original post by EnglishMuon)

This occurs for lots of familiar objects we are used to-  A,B.C could all the groups, rings, modules, ... .
(Original post by EnglishMuon)
Splitting Lemma
The Splitting Lemma asserts TFAE:

Such an  h: C \rightarrow B exists.
Similarly a  k: B \rightarrow A exists with  kf identity on  B
 B \simeq A \bigoplus C
I'm pretty sure the above is wrong for groups. Rather the splitting of the exact sequence is equivalent to B being a semi-direct product of A and C.

For example

 e \to A_3 \to S_3 \to S_3/A_3 \to e

splits but  S_3 is not the direct product of  A_3 and  S_3/A_3
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EnglishMuon
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(Original post by RichE)
I'm pretty sure the above is wrong for groups. Rather the splitting of the exact sequence is equivalent to B being a semi-direct product of A and C.

For example

 e \to A_3 \to S_3 \to S_3/A_3 \to e

splits but  S_3 is not the direct product of  A_3 and  S_3/A_3
Well yeah it depends on how you interpret this '  \bigoplus ' in each of the cases. The splitting lemma holds for anything abelian so other cases you have to specify separately I suppose.
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EnglishMuon
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If  N and  M/N are finitely generated modules, then so is  M

We may look at this as the short exact sequence  0 \rightarrow N \rightarrow^{f} M \rightarrow^{g} M/N \rightarrow 0 , with  f, g the standard inclusion/quotient maps.

Then the map  h( m + N) = m \in M , one  m from each class, is a homomorphism satisfying the Splitting Lemma, hence the result follows.
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EnglishMuon
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Nilpotent Groups and Lower Central Series

The kind of thing thats happens when we construct Lower Central Series, is effectively mashing successive elements together multiple times, and sees what filters out; if after a finite number of steps everything is removed, we call the group Nilpotent.

Precisely, we define the Lower Central Series of a group  G to be the inductive sequence

 \gamma_1 (G)=G, \gamma_{i+1} = [ \gamma_i , G ] , where  [ A, B ] = \{ [ a, b ] : a \in A, b \in B \} with  [ . , . ] the usual commutator bracket.

 G is nilpotent if  \gamma_j = \{ e \} for some finite  j .
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EnglishMuon
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Properties of Lower Central Series

If  H \leq G , then  \gamma_i (H) \leq \gamma_i (G) \ \forall i

Proof

Holds for  i =1 by definition. Assume  \gamma_k (H) \leq \gamma_k (G) .

Then  [ \gamma_k (H), H ] \leq [ \gamma_k (H), G ] \leq [ \gamma_k (G), G ] and it follows by induction

If  \phi : G \Rightarrow K is a surjective homomorphism, then  \phi ( \gamma_i (G))= \gamma_i (K)

Proof

Again, follows from induction by noting that  \phi distributes over the commutator of any two elements.

 \gamma_i (G) is a Characteristic subgroup of  G (i.e. it is invariant under any automorphism of  G )

Proof

Follows immediately from the second claim.

 \gamma_{i+1} (G) \leq \gamma_i (G)

Proof

\gamma_i (G) is characteristic  \Rightarrow \gamma_i (G) \lhd G . So  xyx^{-1}y^{-1} = xx' \in \gamma_i (G) \forall x \in \gamma_i (G) y \in G .
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EnglishMuon
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Finite p-Groups are Nilpotent

Proof

Induct on order of group: Let  |G| = p^n and suppose true for all smaller p-groups.

 G has a non trivial centre  Z(G) hence  G/ Z(G) is a smaller p-group and hence nilpotent.

By our previous results, the quotient map  \phi takes  \gamma_i (G) \rightarrow \gamma_i (G/Z(G)) . So if  \gamma_k (G/Z(G)) = \{ e \} for some  k , we must have  \gamma_k (G) \leq ker \phi =Z(G) .

But  Z(G) is abelian so  [ Z(G), G ] = \{e \} hence  \gamma_{k+1} (G) \leq [ Z(G), G ] \Rightarrow  \gamma_{k+1} (G) = \{ e \} and  G is nilpotent with lower central series with atmost  k+1 distinct steps.

Note that groups of order p are abelian hence nilpotent, so we are done.
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EnglishMuon
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Derived Series

The derived series of  G is defined to be the series  G^{0} = G and  G^(i+1)= [ G^(i), G^(i) ] .

Note that the difference between the derived series and the lower central series is that for derived series, we are only mashing together elements inside the previous step with themselves, ignoring the elements removed in previous stages, whereas we still include the whole group in the mashing with lower central.
Hence we have  G^{(i)} \leq \gamma_i (G) for all  i .

A perfect group is a group  G with  G^{(1)} = G .
So for a finite group (or any group for which the derived series terminates), the last group will be perfect.
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EnglishMuon
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there's no better way
to start your day
than to prove Tychonov's theorem
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EnglishMuon
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turns out conway writes topology poems as well:

If E’s closed and bounded, says Heine–Borel,
And also Euclidean, then we can tell
That, if it we smother
With a large open cover,
There’s a finite refinement as well.
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