Chapter 3 - Nilpotent & Soluble Groups

Question 1

For a factor of a series of G define what it means to be a central factor.
Prove that H/K is central IFF
[H,G]<K

Answer 1

Given a series of G containing successive terms K, H, then the factor H/K is called a central factor of G if K◀G and H/K < Z(G/K)

Proof:
H/K is a central factor
IFF KgKh=KhKg for all g in G, h in H
IFF K[g,h]=K for all ...
IFF [H,G]<K then K◀G. But then [h,g] in K implies h^-1•(g^-1•h•g) in K, in particular for all h in K and g in G. But then (g^-1•h•g) in K for same elements and so K◀G

Question 2

Define a central series
Define a nilpotent group
Define an abelian series
Define a soluble group
Which is weaker out of the following types of series:
Abelian, central, normal

Answer 2

A central series is one in which all factors are central factors
A nilpotent group is just a group with a central series
An abelian series is a series of G whose factors are all abelian groups.
A group is soluble if it has an abelian series.
Central => Normal
Central => abelian
Abelian series need not be normal.

Question 3

Define, for a prime p, a p-group

Prove that all p-groups have non trivial centres

Answer 3

A p-group is a group whose order is a power of p.
Every conjugacy class of G divides the order, and the identity element has conjugacy class size of 1. So if we sum up all the sizes of the conjugacy classes we get the size of G, which is a power of p. but then there must be at least (p-1) elements whose conjugacy class has size 1. And so each if these is in the centre, and so |Z(G)|>=p so is clearly non trivial.

Question 4

Prove that any p-group is nil-potent

Answer 4

We use induction on |G|.
Base case is easy. Assume G is a non trivial p-group and that all p-groups smaller than G are nil-potent.
Then let Z=Z(G). Z is non trivial and so G/Z is a nil-potent p-group, hence it has a central series:
Z/Z◀G0/Z◀…◀Gn/Z=G/Z
for some groups Gi/Z◀G/Z. But then each Gi◀G, and
(Gi/Z)/(Gi-1/Z)=< Z(G/Gi-1)
But then the series:
1◀Z◀G1◀…◀Gn=G

Question 5

When is a p-group an elementary abelian group?

Answer 5

If G Ξ Cp x Cp x … x Cp

Question 6

Prove the following technical lemma:
LEMMA: Suppose B=<=G. Then:
1) as SETS   A^BC=B(A^C)
2) if B◀A then B^C◀A^C and
 A^C/B^C Ξ B(A^C)/B
3) if B◀A and C◀G then BC◀AC and AC/BC Ξ Α/Β(A^C)

Answer 6

1) clearly B(A^C) < A^BC as sets. Now if a in A^BC then a=bc for some b in B, c in C. So c=b^-1•a is in both A & C, so a=bc is in B(A^C) as we need
2) if B normal in A then use an appropriate ISOM theorem on A, A^C and B, noting that B^(A^C)=B^C◀A^C. This immediately gives the result
3) if B◀A and C◀G then AC, BC<BC. So BC◀AC

Then take AC, A and BC in an ISOM theorem to get a close result and apply 1).

Question 7

Prove that if G is soluble then so is any subgroup or quotient

Answer 7

Let G have an abelian series:

G0◀…◀Gn=G

Let H<Gi/Gi-1, but any subgroup of an abelian group is abelian. So H is soluble.

Let Ki=Gi•K/K. Then we use our ISOM theorems and technical lemma to show the factors are a quotient of Gi/Gi-1 by some appropriate group, and so are abelian.

Question 8

Prove that if G is nilpotent then so is any subgroup or quotient

Answer 8

Let G have an abelian series:

G0◀…◀Gn=G

Let H<Hi-1.
But this is true for the series for G, which is central so we can show this series for H is also central.
Let Ki=K•Gi/K. This gives a series for G/K. Then show:
[Gi•K/K, G/K]=[Gi,G]•K/K
This is basically enough. Why?

Question 9

Prove that the direct product of two nilpotent/soluble groups is nilpotent/soluble.

Answer 9

H,K nilpotent/soluble. So let Hi, Ki give two central//abelian series for H,K respectively. By adding repetitions we can assume they have the same length. Then we see Hi x Ki is a series for H x K. But then:
(Hi x Ki)/(Hi-1 x Ki-1) Ξ Hi/Hi-1 x Ki/Ki-1. So soluble case is now obvious. But then this is clearly in the required centre now so nilpotent case is also done.

Question 10

Prove that for K◀G:

K & G/K soluble then G is also soluble

Why does this mean each group has a unique largest normal soluble group?

Answer 10

Pick an abelian series for K, and one for G/K. By ISOM theorem we can lift the series for G/K to an abelian series between K and G and then add in the one for K.
But the if H, K are both normal in G and soluble, we see that HK/ΚΞK/H^K and so HK is soluble from above. But then H•K has same order as H and K, and so HK=H=K
This is called the soluble radical group.