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

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#1
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.

(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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#2
Prelim. Definitions

For all following posts let be any group.

An -series of is a finite sequence of subgroups such that for all i. (Note this does not mean we have to have factors aside from second to last which are normal in . In such a case with normal factors, we call it an -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 -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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4 years ago
#3
first we define what a group is
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#4
Zassenhaus's Lemma:

Let be -subgroups of with . Let .

Then , and

.

Proof:

Conjugating by and simplifying gives the result for the first part.

2nd part follows from the second isomorphism theorem.
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#5
The Schreier Refinement Theorem:

Any two -series of have -isomorphic refinements.

Proof:

Let be two given series.

Define .

Then by Zassenhaus's lemma and , so the series are -isomorphic (where i, j run over all possible values).

Note they are refinements of the original series as 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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#6
Composition Series

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

For example consider . Since is abelian, if the last (nontrivial) factor of one series is we may add on for instance.

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

Proof:

Forwards: Let be a given series.

Suppose is not simple for some k. Then by 1-1 correspondence of normal groups in properly containing and normal subgroups of , we can find another subgroup between 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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4 years ago
#7
alright rob
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#8
Jordan-Holder Theorem

Let S be a composition series and T be any other -series. Then there exists a refinement of T, T'that is a composition series and S, T' are 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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#9
Proof of Fundamental Theorem of Arithmetic using Jordan-Holder Theorem

Let be given. Let's assume is a product of primes since this is easy to prove by induction and WLOG let for primes.

Then are composition series of the group

So by Jordan-Holder theorem the factors and similarly are -isomorphic, hence the are the in some order.
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#10
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

.

where is the embedding map of into and is the quotient map of from , so in some loose sense we have ' '.

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

This can be extended in the obvious way for n terms, such that any 3 consecutive terms behave as in the example above.
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#11
Splitting Lemma

We say that the short exact sequence splits if there is a homomorphism such that is the identity map on .

The Splitting Lemma asserts TFAE:

Such an exists.
Similarly a exists with identity on

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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4 years ago
#12
(Original post by EnglishMuon)

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

Such an exists.
Similarly a exists with identity on
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

splits but is not the direct product of and
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#13
(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

splits but is not the direct product of and
Well yeah it depends on how you interpret this ' ' 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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#14
If and are finitely generated modules, then so is

We may look at this as the short exact sequence , with the standard inclusion/quotient maps.

Then the map , one from each class, is a homomorphism satisfying the Splitting Lemma, hence the result follows.
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#15
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 to be the inductive sequence

, where with the usual commutator bracket.

is nilpotent if for some finite .
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#16
Properties of Lower Central Series

If , then

Proof

Holds for by definition. Assume .

Then and it follows by induction

If is a surjective homomorphism, then

Proof

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

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

Proof

Follows immediately from the second claim.

Proof

is characteristic . So .
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#17
Finite p-Groups are Nilpotent

Proof

Induct on order of group: Let and suppose true for all smaller p-groups.

has a non trivial centre hence is a smaller p-group and hence nilpotent.

By our previous results, the quotient map takes . So if for some , we must have .

But is abelian so hence and is nilpotent with lower central series with atmost distinct steps.

Note that groups of order p are abelian hence nilpotent, so we are done.
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#18
Derived Series

The derived series of is defined to be the series and .

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 for all .

A perfect group is a group with .
So for a finite group (or any group for which the derived series terminates), the last group will be perfect.
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#19
there's no better way
than to prove Tychonov's theorem
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#20
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.
0
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