Chapter 2-Composition Series

Question 1

What is a series of the group G?
Define the following terms for a series of G:
• length of the series
• terms of the series
• factors of the series
• a proper series
• a normal series
• (proper) refinement of a series
• a composition series
• composition factors of G

Answer 1

A series of G is a chain of subgroups:
1=G0◀G1◀…◀Gn=G
The length is then n and the terms are the Gi.
The factors are the quotient groups
Gi/Gi-1
A proper series is one where there are no repeating groups
A normal series is one where Gi◀G for each i
A refinement is a longer chain which includes the original chain. It is proper if there is a new subgroup in the chain.
A composition series is a series with no proper refinement.
The composition facts are then just the factors of the series

Question 2

Prove that any proper series of G has a refinement which is a composition series

Answer 2

Take any refinement and delete repetitions. This is a longer proper refinement and so doing this forever cannot work. So it must stop at a composition series.

Question 3

Prove that a series is a composition series IFF all it’s factors are simple

Answer 3

By definition a proper series is one whose factors are non trivial. If Gi/Gi-1 is not simple, then it has a normal subgroup of the form: H/Gi-1. But then H is normal in Gi and Gi is normal in H, so we get a longer sequence by adding H:
Gi-1◀H◀Gi

Conversely if our series has a proper refinement, in Hi’s, then let r be the max so that Hr is not equal to any Gi. Then see that Hr+1 = Gi for some i and Gi-1 = Hj for some j<r. Finally, use this to show that Gi/Gi-1 is not simple

Question 4

When are two series equivalent?

Answer 4

Let 1=G0◀G1◀…◀Gn=G
& 1=H0◀H1◀…◀HM=G

Be two series for G. They are equivalent if n=m and each factor group is ISOM to a factor group in the other series.

Question 5

What is the Jorden-Holder Theorem for two composition series?
How do you prove it?

Answer 5

J-H Thm states that any two composition series are equivalent.

Proof is by induction on |G|, as follows:
Base case is easy.
Assume |G| >1 and Thm is true for groups of smaller size.
Let S1 and S2 be two composition series for G, with penultimate terms K and L respectively. Then G/K and G/L are simple. But then we see:
K<G so KL is either K or G
But similarly, KL is either L or G.
If LK~= G then L=K and we can use induction
If LK=G then use the ISOM theorem to show:
K/K^LΞG/L and L/K^LΞG/K
Then look at the truncation series at K^L. Create S3 by adding K and G on to the end of this and S4 by adding L and G. Finally, it is obvious that S3~S4, so all need to show is S1~S3 and S2~S4. But this can be done again by induction.