Chapter 4 - Optimal Series

Question 1

Probe the following PROPOSITION:

G is soluble IFF all it’s composition factors are of prime order.

Answer 1

Proof:
If G is soluble then all it’s comp factors are soluble, as they are subgroups of quotients. But they are also simple, as they are composition factors. But then if H is any soluble simple group, we have H’ is normal in H and strictly less than H, so has to be 1 as H is simple. So H is abelian. And so it must have prime order by our classification if simple abelian groups.
Conversely, if all comp factors have prime order they are all cyclic and so simple and abelian. So G is soluble.

Question 2

Define the derived series o a group. How is it useful for characterising soluble groups?

Answer 2

Set G’0= G, and then
Gi’=[Gi-1’, Gi-1’]. Then, Gi-1’charGi’ for all i, so by induction Gi’charG for all i. So they definitely form a normal series.
Then G is soluble IFF Gi’=1 for some i.
Proof:
Obviously G is soluble if some Gi’=1
Conversely, suppose G is soluble. Then we get a series Gj of length n which is abelian. By induction, we show Gi’0 then Gi+1’=(Gi’)’<Gn-i-1 since Gn-i/Gn-i-1 is abelian.
So in particular, The derived series is always the shortest abelian series.

Question 3

What are the minimal normal subgroups of a soluble groups?

Answer 3

Any minimal normal subgroup of a soluble group is an elementary abelian p-group. Why?

Well let G be soluble, N it’s minimal normal subgroup. Then N is soluble and so it’s derived subgroup must be trivial, since this will be a normal subgroup of G (note N’charN, N◀G so N’◀G). So N is abelian. So N is a product of cyclic groups of prime power order. Then for any prime p dividing the order of N, set M to be all the elements in N of order p. Then this is a characteristic subgroup of N, and so is normal in G. But it is non trivial. So M=N.

Question 4

Define the lower central series and the upper central series of G.

Answer 4

The lower central series is:

Γ1(G)=G,
Γi+1(G)=[Γi(G), G]

Then G=Γ1(G)▶Γ2(G)▶…

The upper central series is:

Z0(G)=1
Zi(G) is the subgroup of G such that Zi(G)/Zi-1(G) = Z( G/Zi-1(G) )

Then Zi-1(G)charZi(G) by induction and question on example sheet. So this gives a series for G if some j has Zj(G)=G

Question 5

Proof the following proposition:

G is nilpotent
IFF Γn(G)=1 for some n
IFF Zn(G)=G for some n

Moreover, if G is any nilpotent group with a central series

1=G1◀…◀Gn=G then have:

Γn-i+1(G)<Zi(G)

And Γr-i+1(G)=1 IFF Zr(G)=G

Answer 5

If the lower central series terminates at 1 and/or the upper central series terminates at G then G is certainly soluble, as the second is obviously a central series as is the first after application of a lemma.
Now assume G is nilpotent, Gi a central series. So
Gi/Gi-1 < Z( G/Gi-1) and Gi◀G.
By induction on i we show:
Γi+1(G)=<Zi(G)
The obvious proof now does everything.

Question 6

What is the nilpotency class of G?

Answer 6

Smallest integer n such that

Zn(G) =G

Question 7

Outline the proof of the following classification of nilpotent groups:

G is nilpotent IFF G is the direct product of p-groups for various primes p.

Answer 7

First you prove that if G is a nilpotent group and H a proper subgroup, then H is properly contained in its normaliser in G. Do this using the upper central series to find an element in the normaliser of H NOT in H.
Then use the frattini argument to show that if P is any Sylow p-group, then taking the normaliser of P in G twice is the same as doing it once. So: let N=NG(P), then NG(N)=N. But by the first part, this means G=NG(P). So that, if G is nilpotent, then all Sylow p-groups are normal in G.
This is enough! As all Sylow p-groups then have direct product equal to G.