A Character Theory Free Proof of
Burnside’s p
a
q
b
Theorem
by
Ben L. Adovasio
Submitted in Partial Fulfillment of the Requirements
for the Degree of
Master of Science
in the
Mathematics
Program
YOUNGSTOWN STATE UNIVERSITY
May, 2012
A Character Theory Free Proof of Burnside’s p
a
q
b
Theorem
Ben L. Adovasio
I hereby release this thesis to the public. I understand that this thesis will be made
available from the OhioLINK ETD Center and the Maag Library Circulation Desk
for public access. I also authorize the University or other individuals to make copies
of this thesis as needed for scholarly research.
Signature:
Ben L. Adovasio, Student Date
Approvals:
Dr. Neil Flowers, Thesis Advisor Date
Dr. Tom Wakefield, Committee Member Date
Dr. Eric Wingler, Committee Member Date
Peter J. Kasvinsky, Dean of School of Graduate Studies & Research Date
c©
Ben L. Adovasio
2012
ABSTRACT
In 1904, George Burnside [2] proved that any group G with |G| = p
a
q
b
where p
and q are primes and a and b are positive integers is solvable. Burnside accomplished
this through the use of character theory, i.e., the interaction between a group and a
vector space.
Since then, group theorists began to try to prove this theorem without the use of
charactertheory. Theywantedaproofthatreliedonlyongrouptheoreticalprinciples.
This was finally achieved in 1972 by Helmut Bender [1].
However, in 1970, David M. Goldschmidt [3] supplied a group theoretic proof of
Burnside’s Theorem but only when the order of the group, G, was odd. Then in 1972,
Hiroshi Matsuyama [4] supplied a group theoretic proof of Burnside’s Theorem when
the order of the group, G, was even. Ironically, Bender’s and Matsuyama’s results
occurred independently and simultaneously. Therefore, both papers were published
even though Bender’s proof was more general.
The goal of this paper is to present the background knowledge and the more
general proof of Burnside’s Theorem.
iv
I would like to thank my friends and family for their support. I would especially like
to thank my advisor, Dr. Flowers, for all his insight and for inspiring me to learn
about group theory.
v
Contents
1 Preliminaries 1
2 Solvable Groups 12
3 Nilpotent Groups 23
4 Groups Acting on Groups 33
5 Burnsides p
a
q
b
Theorem 47
vi
1 Preliminaries
In this section, we will introduce some background concepts and ideas. These
ideas will build up the tools needed for the proof of Burnside’s Theorem. We begin
by introducing the idea of a group.
Definition 1.1. A group is a nonempty set G along with a binary operation ∗ such
that
1. (closure): a∗b ∈ G for all a,b ∈ G;
2. (associativity): (a∗b)∗c = a∗(b∗c) for all a,b,c ∈ G;
3. (identity): there exists e ∈ G such that e∗a = a∗e = a for all a ∈ G;
4. (inverse): for all a ∈ G there exists b ∈ G such that a∗b = b∗a = e.
Definition 1.2. Let (G,∗) be a group. A subset H ⊆ G is called a subgroup of G
if (H,∗) is a group. We write H ≤ G.
Theorem 1.1 (Subgroup Test). Let G be a group and ∅ negationslash= H ⊆ G. Then H ≤ G if
and only if ab
−1
∈ H for all a,b ∈ H.
Definition 1.3. Let G be group, a ∈ G, and H ≤ G. Then the following are subgroups
of G:
1. Z(G)={g ∈ G | gx= xg for all x ∈ G}. We call this the center of G.
2. 〈a〉 = {a
n
| n ∈ Z}. We call this the cyclic subgroup generated by a.
3. C
G
(a)={g ∈ G | ag = ga}. We call this the centralizer of a.
4. N
G
(H)={g ∈ G | gHg
−1
∈ H}. We call this the normalizer of H.
1
Definition 1.4. Let G be a group, H ≤ G and g ∈ G. The left coset of H in G
containing g is
gH = {gh| h ∈ H}.
Theorem 1.2. Let G be a group, H ≤ G, and a,b ∈ G. Then aH = bH if and only
if b
−1
a ∈ H.
Definition 1.5. Let G
1
and G
2
be groups and φ : G
1
→ G
2
. Then φ is a homomor-
phism if
φ(ab)=φ(a)φ(b) for all a,b ∈ G.
If, in addition, φ is 1-1 and onto, we call φ an isomorphism and we write G
1
∼
= G
2
.
Theorem 1.3. Let G
1
, G
2
be groups and φ : G
1
→ G
2
be a homomorphism. Define
the kernel of φ by
kernφ= {g ∈ G
1
| φ(g)=1}.
Then kern φ ≤ G
1
.
Definition 1.6. Let G be a group and H ≤ G. Then H is a normal subgroup of
G if ghg
−1
∈ H for all g ∈ G and for all h ∈ H. We write H triangleleftequalG.
Theorem 1.4. Let G be a group and H triangleleftequalG. Define
G
H
= {gH | g ∈ G}.
Then
G
H
is a group under the operation
aHbH = abH for all aH,bH ∈
G
H
.
2
We call
G
H
the quotient group.
Lemma 1.1. Let G be a group. Then
G
{1}
∼
= G.
Proof. Define φ : G →
G
{1}
by φ(g)=g{1} for all g ∈ G.Wewanttoshowthatφ is
a homomorphism. Let a,b ∈ G.Then
φ(ab)=ab{1}
= a{1}b{1}
= φ(a)φ(b).
Thus, φ is a homomorphism. We want to show that φ is onto. Let g{1}∈
G
{1}
.Then
g ∈ G and φ(g)=g{1}.Thus,φ is onto. We want to show that φ is 1-1. Suppose
a,b ∈ G such that φ(a)=φ(b). Then a{1} = b{1} or b
−1
a ∈{1}.Sob
−1
a =1or
a = b.Thus,φ is 1-1. Therefore, φ is an isomorphism and so
G
{1}
∼
= G.
Lemma 1.2. Let G be a group and H ≤ G such that
|G|
|H|
=2then H triangleleftequalG.
Proof. Let g ∈ G and h ∈ H. Wewanttoshowthatghg
−1
∈ H.Ifg ∈ H then
ghg
−1
∈ H since H ≤ G.Ifg negationslash∈ H then gH negationslash=1H. Then since
|G|
|H|
=2weget
G =1H ∪gH.Nowghg
−1
∈ G and so ghg
−1
∈ 1H or ghg
−1
∈ gH.Ifghg
−1
∈ gH
then there exists h
1
∈ H such that ghg
−1
= gh
1
.Thenhg
−1
= h
1
or g = h
−1
1
h ∈ H,
which contradicts g negationslash∈ H. Therefore ghg
−1
∈ 1H = H and so H triangleleftequalG.
Theorem 1.5 (1
st
Isomorphism Theorem). Let G
1
, G
2
be groups and φ : G
1
→ G
2
be a homomorphism. Then
G
1
kernφ
∼
= φ(G
1
).
3
Proof. Let K = kernφ. Define θ :
G
1
K
→ φ(G
1
)byθ(aK)=φ(a) for all aK ∈
G
1
K
.
We want to show that θ is a homomorphism. Let aK,bK ∈
G
1
K
.Then
θ(aKbK)=θ(abK)
= φ(ab)
= φ(a)φ(b)
= θ(aK)θ(bK)
and so θ is a homomorphism. We want to show that θ is 1-1. Let aK,bK ∈
G
1
K
such
that θ(aK)=θ(bK). Then φ(a)=φ(b)orφ(b)
−1
φ(a)=1.Thus,φ(b
−1
)φ(a)=1or
φ(b
−1
a) = 1. Hence b
−1
a ∈ kernφ= K.Thus,aK = bK and so θ is 1-1. We want to
show that θ is onto. Let φ(x) ∈ φ(G
1
)wherex ∈ G
1
.ThenxK ∈
G
1
K
and θ(xK)=
φ(x). Hence, θ is onto and so θ is an isomorphism. Therefore,
G
1
kernφ
∼
= φ(G
1
).
Theorem 1.6 (2
nd
Isomorphism Theorem). Let G be a group, N triangleleftequalG, and H ≤ G.
Then
HN
N
∼
=
H
H ∩N
.
Proof. Define φ : H →
HN
N
by φ(h)=hN for all h ∈ H. We want to show that φ is
a homomorphism. Let a,b ∈ H.Then
φ(ab)=abN
= aNbN
= φ(a)φ(b)
4
and so φ is a homomorphism. We want to show that φ is onto. Let hnN ∈
HN
N
.
Then
φ(h)=hN
= hnN as (hn)
−1
h = n
−1
∈ N
and so φ is onto. We claim that the kernφ= H ∩N.Now,
h ∈ kernφ⇔ φ(h)=1N ⇔ hN =1N ⇔ 1
−1
h ∈ N ⇔ h ∈ N ⇔ h ∈ H ∩N.
Thus, kernφ = H ∩ N. By Theorem 1.5,
H
kernφ
∼
= φ(H). Thus,
H
H ∩N
∼
= φ(H)
and, since φ is onto, we get
H
H ∩N
∼
=
HN
N
.
Theorem 1.7 (3
rd
Isomorphism Theorem). Let G be a group, N triangleleftequalG, H triangleleftequalG such
that N ≤ H. Then
G/N
H/N
∼
=
G
H
.
Proof. Define φ :
G
N
→
G
H
by φ(gN)=gH for all gN ∈
G
N
. We want to show that φ
is well-defined. Let aN,bN ∈
G
N
such that aN = bN.Thena = a1 ∈ aN = bN and
so there exists n ∈ N such that a = bn.Then
φ(aN)=aH
= bnH
= bH
= φ(bN)
5
and so φ is well-defined. We want to show that φ is a homomorphism. Let aN,bN ∈
G
N
.Then
φ(aNbN)=φ(abN)
= abH
= aHbH
= φ(aN)φ(bN)
and so φ is a homomorphism. We want to show that φ is onto. Let gH ∈
G
H
.Then
gN ∈
G
N
and φ(gN)=gH.Thus,φ is onto. We claim that the kernφ=
H
N
. Then,
gN ∈ kernφ⇔ φ(gN)=1H ⇔ gH =1H ⇔ g ∈ H ⇔ gN ∈
H
N
.
Thus, kernφ =
H
N
. By Theorem 1.5,
G/N
kernφ
∼
= φ(G/N)or
G/N
H/N
∼
= φ(G/N) and,
since φ is onto, we get
G/N
H/N
∼
=
G
H
.
Theorem 1.8. Let G be a group and N triangleleftequalG. Define φ : G →
G
N
by φ(g)=gN for
all g ∈ G. We call φ the natural map. The following are true:
1. φ is a homomorphism
2. kernφ= N
3. If H ≤ G, then φ(H)=
HN
N
4. If H ≤ G, then φ
−1
parenleftBig
HN
N
parenrightBig
= HN
5. If L ≤
G
N
, then L =
K
N
where N ≤ K ≤ G
6
Proof. For (1), let a,b ∈ G.Then
φ(ab)=abN
= aNbN
= φ(a)φ(b).
Thus, φ is a homomorphism. For (2), let n ∈ kernφ. Then,
n ∈ kernφ⇔ φ(n)=1N ⇔ nN =1N ⇔ 1
−1
n ∈ N ⇔ n ∈ N.
Thus, kernφ = N. For (3), φ(H) ⊆
HN
N
.Leth ∈ H,n ∈ N,andφ(h) ∈ φ(H).
Then
φ(h)=hN
= h1N
∈
HN
N
.
Hence, φ(H) ⊆
HN
N
. Next,
HN
N
⊆ φ(H). Let hnN ∈
HN
N
.Then
φ(h)=hN
= hnN as h
−1
hn ∈ N.
Hence, hnN ∈ φ(H)andso
HN
N
⊆ φ(H). Therefore, φ(H)=
HN
N
. For (4),
7
HN ⊆ φ
−1
parenleftBig
HN
N
parenrightBig
.Lethn ∈ HN.Then
φ(hn)=hnN
∈
HN
N
.
Thus, hn ∈ φ
−1
parenleftBig
HN
N
parenrightBig
and HN ⊆ φ
−1
parenleftBig
HN
N
parenrightBig
. Next, we want to show that
φ
−1
parenleftBig
HN
N
parenrightBig
⊆ HN.Letg ∈ φ
−1
parenleftBig
HN
N
parenrightBig
.Thenφ(g) ∈
HN
N
or gN ∈
HN
N
.Thus,
there exists h ∈ H and n ∈ N such that gN = hnN.Theng = g1 ∈ gN = hnN and
so there exists n
1
∈ N such that g = hnn
1
∈ HN.Thus,φ
−1
parenleftBig
HN
N
parenrightBig
⊆ HN and so
φ
−1
parenleftBig
HN
N
parenrightBig
= HN. For (5), we know φ
−1
(L) ≤ G.Ifn ∈ N then φ(n)=1N ∈ L
and so n ∈ φ
−1
(L). Thus N ≤ φ
−1
(L). We claim that
φ
−1
(L)
N
= L.LetgN ∈ L.
Then φ(g) ∈ L and so g ∈ φ
−1
(L). Hence, gN ∈
φ
−1
(L)
N
and so L ≤
φ
−1
(L)
N
.Let
xN ∈
φ
−1
(L)
N
.Thenx ∈ φ
−1
(L)andsoφ(x) ∈ L.Butφ(x)=xN and so xN ∈ L.
Thus,
φ
−1
(L)
N
≤ L and so L =
φ
−1
(L)
N
.
Theorem 1.9. Let G be any group and S ⊆ G. Define
〈S〉 = {s
n
1
1
s
n
2
2
···s
n
k
k
| s
i
∈ S,n
i
∈ Z, for all 1 ≤ i ≤ k,k ∈ Z
+
}.
Then 〈S〉≤G and is called the subgroup generated by S.
Proof. Let s ∈ S.Thens = s
1
∈〈S〉 and so 〈S〉 negationslash= ∅.Let
s
n
1
1
s
n
2
2
···s
n
k
k
,r
m
1
1
r
m
2
2
···r
n
l
l
∈〈S〉
8
where s
i
∈ S and r
i
∈ S for all i and n
i
∈ Z and m
i
∈ Z for all i and k,l ∈ Z
+
.Then
(s
n
1
1
s
n
2
2
···s
n
k
k
)(r
m
1
1
r
m
2
2
···r
n
l
l
)
−1
= s
n
1
1
s
n
2
2
···s
n
k
k
r
−m
l
l
r
−m
l−1
l−1
···r
−m
1
∈〈S〉.
Hence 〈S〉≤G by the Subgroup Test.
Definition 1.7. Let G be a group, a,b ∈ G, H ≤ G, and K ≤ G. Then
1. [a,b]=aba
−1
b
−1
is called the commutator of a and b
2. [H,K]=〈{[h,k] | h ∈ H and k ∈ K}〉 is called the commutator subgroup
generated by H and K
3. G
prime
= 〈{[a,b] | a,b ∈ G}〉 is called the commutator subgroup of G
Lemma 1.3. Let G be a group, N triangleleftequalG, H ≤ G, and a,b ∈ G. Then
1. [a,b]=1if and only if ab = ba
2. G
prime
triangleleftequalG
3.
G
G
prime
is abelian
4.
G
N
is abelian if and only if G
prime
≤ N
5. If G
prime
≤ H then H triangleleftequalG
Proof. For (1),
[a,b]=1⇔ aba
−1
b
−1
=1⇔ ab = ba.
9
For (2), let x ∈ G
prime
and g ∈ G. Then, x =
k
producttext
i=1
[a
i
,b
i
]andso
gxg
−1
= g(
k
productdisplay
i=1
[a
i
,b
i
])g
−1
=
k
productdisplay
i=1
g[a
i
,b
i
]g
−1
=
k
productdisplay
i=1
[ga
i
g
−1
,gb
i
g
−1
]
∈ G
prime
and so G
prime
triangleleftequalG. For (3), let aG
prime
,bG
prime
∈
G
G
prime
. Then,
[aG
prime
,bG
prime
]=[a,b]G
prime
=1G
prime
as 1
−1
[a,b]=[a,b] ∈ G
prime
.
Therefore,
G
G
prime
is abelian. For (4),
G
N
is abelian ⇔ [aN,bN]=1N for alla,b ∈ G
⇔ [a,b]N =1N ⇔ [a,b] ∈ N ⇔ G
prime
≤ N sinceN ≤ G.
For (5), let g ∈ G and h ∈ H.Then[h
−1
,g] ∈ G
prime
≤ H and so [h
−1
,g] ∈ H.Let
[h
−1
,g]=h
1
where h
1
∈ H.Thenh
−1
g(h
−1
)
−1
g
−1
= h
1
.Thus,h
−1
ghg
−1
= h
1
implying ghg
−1
= hh
1
∈ H. Therefore, H triangleleftequalG.
10
Definition 1.8. Let G be a group and p a prime. Then G is called a p-group if
|G| = p
r
for some r ∈ Z
+
∪{0}.
Lemma 1.4. Let G be a group and H triangleleftequalG. Then Z(H)triangleleftequalG.
Theorem 1.10 (Cauchy’s Theorem for Abelian Groups). Let G be abelian and p be
a prime such that p ||G|. Then G has an element of order p.
Definition 1.9. The group consisting of the set S
n
of all permutations on A =
{1,2,...,n}, under the operation of permutation multiplication is called the sym-
metric group of degree n.
Definition 1.10. Let G be a group and S negationslash= {} be a set. Then G acts on S if there
exists a homomorphism φ : G → Sym(S).
Definition 1.11. Let G be a group, S be a set, and a ∈ S. The orbit of S con-
taining a is
Ga = {ga| g ∈ G}.
Definition 1.12. A group G acts transitively on a set S, if there is only one orbit;
i.e., S = Ga for all a ∈ S; i.e., for all c,d ∈ S there exists g ∈ G such that cg = d.
Definition 1.13. Let G be a group, p be a prime, and n ∈ Z
+
∪{0} such that p
n
||G|
but p
n+1
negationslash ||G|. Then
1. |G|
p
= p
n
is called the p
th
part of G.
2. A subgroup H ≤ G is called a sylow p-subgroup if |H| = |G|
p
.
3. Syl
p
(G) is the set of all sylow p-subgroups of G.
11
Theorem 1.11 (Sylow’s Theorem). Let G be a group, p be any prime, H ≤ G be a
p-group, and n
p
= |Syl
p
(G)|. Then
1. Syl
p
(G) negationslash= {}
2. There exists P ∈ Syl
p
(G) such that H ≤ P. Moreover, G acts transitively on
Syl
p
(G) by conjugation
3. n
p
||G| and n
p
≡ 1(modp).
2 Solvable Groups
We next need to introduce what it means for a group to be solvable and will
discover some important properties about solvability.
Definition 2.1. A group G is solvable if there exists a subnormal series
G = G
0
trianglerightequalG
1
trianglerightequalG
2
trianglerightequal···trianglerightequalG
n
=1
such that
G
i
G
i+1
is abelian for all 0 ≤ i ≤ n−1.
Example 2.1. S
3
is a solvable group.
Proof. Consider the subnormal series
S
3
trianglerightequalA
3
trianglerightequal1.
Now,
vextendsingle
vextendsingle
vextendsingle
S
3
A
3
vextendsingle
vextendsingle
vextendsingle =
|S
3
|
|A
3
|
=
6
3
=2andso
S
3
A
3
∼
= Z
2
is abelian. Next,
vextendsingle
vextendsingle
vextendsingle
vextendsingle
A
3
{1}
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
|A
3
|
|{1}|
=3and
so
A
3
{1}
∼
= Z
3
is abelian. Therefore S
3
is solvable.
12
Lemma 2.1. Let G be an abelian group. Then G is solvable.
Proof. Consider the subnormal series
G = G
0
trianglerightequal1.
Then by Lemma 1.1 we know
G
{1}
∼
= G. Since G is abelian,
G
{1}
is abelian and so G
is solvable.
Example 2.2. The abelian groups Z
n
and Z
a
×Z
b
×···×Z
c
are solvable groups by
Lemma 2.1.
Lemma 2.2. Let G be solvable and H ≤ G. Then H is solvable.
Proof. Since G is solvable we know there exists a subnormal series
G = G
0
trianglerightequalG
1
trianglerightequalG
2
trianglerightequal···trianglerightequalG
n
=1
such that
G
i
G
i+1
is abelian. Consider the series
H = H
0
≥ H ∩G
1
≥ H ∩G
2
≥···≥H ∩G
n
=1.
We want to show that H ∩G
i+1
triangleleftequalH ∩G
i
.Letx ∈ H ∩G
i+1
and g ∈ H ∩G
i
.Then
gxg
−1
∈ G
i+1
since x ∈ G
i+1
and G
i+1
triangleleftequalG
i
.Also,gxg
−1
∈ H since g ∈ H and x ∈ H.
Thus, gxg
−1
∈ H ∩G
i+1
. Hence H ∩G
i+1
triangleleftequalH ∩G
i
for all 0 ≤ i ≤ n−1. Therefore,
13
H = H
0
trianglerightequalH ∩G
1
trianglerightequalH ∩G
2
trianglerightequal···trianglerightequalH ∩G
n
= 1 is a subnormal series. Now
H ∩G
i
H ∩G
i+1
=
H ∩G
i
H ∩G
i
∩G
i+1
∼
=
(H ∩G
i
)G
i+1
G
i+1
by 2
nd
Isomorphism Theorem
≤
G
i
G
i+1
.
Since
G
i
G
i+1
is abelian we get
H ∩G
i
H ∩G
i+1
is abelian for all 0 ≤ i ≤ n − 1. Thus H is
solvable.
Lemma 2.3. Let G be solvable and N triangleleftequalG. Then
G
N
is solvable.
Proof. Since G is solvable we know there exists a subnormal series
G = G
0
trianglerightequalG
1
trianglerightequalG
2
trianglerightequal···trianglerightequalG
n
=1
such that
G
i
G
i+1
is abelian for all 0 ≤ i ≤ n−1. Taking the image of this series under
the natural map we get
G
N
=
G
0
N
≥
G
1
N
N
≥
G
2
N
N
≥···≥
G
n
N
N
=1N.
14
We claim that
G
i+1
N
N
triangleleftequal
G
i
N
N
.Letg
i+1
n
1
N ∈
G
i+1
N
N
and g
i
n
2
N ∈
G
i
N
N
.Then
(g
i
n
2
N)(g
i+1
n
1
N)(g
i
n
2
N)
−1
=(g
i
n
2
N)(g
i+1
n
1
N)(n
−1
2
g
−1
i
N)
= g
i
n
2
g
i+1
n
1
n
−1
2
g
−1
i
N
= g
i
n
2
g
−1
i
g
i
g
i+1
g
−1
i
g
i
n
1
n
−1
2
g
−1
i
N
= g
i
n
2
g
−1
i
g
i
g
i+1
g
−1
i
N since g
i
n
1
n
−1
2
g
−1
i
∈ N
∈
G
i+1
N
N
since g
i
n
2
g
−1
i
∈ G
i+1
and g
i
g
i+1
g
−1
i
∈ N
Thus,
G
N
=
G
0
N
trianglerightequal
G
1
N
N
trianglerightequal
G
2
N
N
trianglerightequal···trianglerightequal
G
n
N
N
=1N
is a subnormal series. Then
G
i
N/N
G
i+1
N/N
∼
=
G
i
N
G
i+1
N
by 3
rd
Isomorphism Theorem
=
G
i
G
i+1
N
G
i+1
N
∼
=
G
i
G
I
∩G
i+1
N
by 2
nd
Isomorphism Theorem
∼
=
G
i
/G
i+1
G
I
∩G
i+!
N/G
i+1
by 3
rd
Isomorphism Theorem
Since quotients of abelian groups are abelian, we get
G
i
N/N
G
i+1
N/N
is abelian for all
0 ≤ i ≤ n−1. Therefore,
G
N
is solvable.
Theorem 2.1. Let G be a p-group. Then G is solvable.
Proof. Use induction on |G|.If|G| =1thenG = {1} is abelian and therefore
solvable. Assume the theorem holds for all p-groups of order less than |G|. Without
15
loss of generality, G negationslash= 1. Since G is a p-group we know Z(G) negationslash=1.Then
vextendsingle
vextendsingle
vextendsingle
vextendsingle
G
Z(G)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
|G|
|Z(G)|
< |G| and
G
Z(G)
is a p-group. Thus
G
Z(G)
is solvable by induction and so
there exists a subnormal series
G
Z(G)
=
G
0
Z(G)
trianglerightequal
G
1
Z(G)
trianglerightequal···trianglerightequal
G
n
Z(G)
= Z(G)
such that
G
i
/Z(G)
G
i+1
/Z(G)
is abelian for all 0 ≤ i ≤ n−1. Taking the pre-image of this
series under the natural map we get
G = G
0
trianglerightequalG
1
trianglerightequal···trianglerightequalZ(G)trianglerightequal1.
Then
G
i
G
i+1
∼
=
G
i
/Z(G)
G
i+1
/Z(G)
is abelian by the 3
rd
Isomorphism Theorem and
Z(G)
{1}
∼
=
Z(G) which is abelian. Therefore, G is solvable and every p-group is solvable by
induction.
Definition 2.2. Let G be a group. Define the derived series of G by
G
(0)
= G,G
(1)
=(G
(0)
)
prime
= G
prime
,G
(2)
=(G
(1)
)
prime
= G
primeprime
and inductively define
G
(n)
=(G
(n−1)
)
prime
.
By Lemma 1.3 we have a subnormal series
G = G
(0)
trianglerightequalG
(1)
trianglerightequalG
(2)
trianglerightequalG
(3)
trianglerightequal···
16
Theorem 2.2. Let G be a group. Then G is solvable if and only if there exists n ∈ Z
+
such that G
(n)
=1.
Proof. (⇐) Suppose there exists n ∈ Z
+
such that G
(n)
= 1. Consider the derived
series
G = G
(0)
trianglerightequalG
(1)
trianglerightequalG
(2)
trianglerightequal···trianglerightequalG
(n)
=1
Then
G
(i)
G
(i+1)
=
G
(i)
(G
(i)
)
prime
is abelian by Lemma 1.3 for all 0 ≤ i ≤ n−1. Therefore, G is
solvable. (⇒) Suppose G is solvable. Then there exists a subnormal series
G = G
0
trianglerightequalG
1
trianglerightequalG
2
trianglerightequal···trianglerightequalG
n
=1
such that
G
i
G
i+1
is abelian. We claim that G
(i)
≤ G
i
. Use induction on i.Ifi =0then
G
(0)
= G ≤ G = G
0
. Suppose G
(i)
≤ G
i
.WewanttoshowG
(i+1)
≤ G
i+1
.Now
G
(i+1)
=(G
(i)
)
prime
≤ (G
i
)
prime
by induction hypothesis
≤ G
i+1
since
G
i
G
i+1
is abelian and by Lemma 1.3
Thus, the claim holds. Hence G
(n)
≤ G
n
=1andsoG
(n)
=1.
Theorem 2.3. Let G be a group and H triangleleftequalG such that H and
G
H
are solvable. Then
G is solvable.
Proof. SinceH and
G
H
aresolvablethenthereexistm,n ∈ Z
+
suchthatH
(m)
=1and
parenleftBig
G
H
parenrightBig
(n)
=1H.Then
G
(n)
H
H
=1H by the claim in Lemma 2.3. Let ahH ∈
G
(n)
H
H
wherea ∈ G
(n)
andh ∈ H.ThenahH =1H andso,byTheorem1.2,1
−1
ah = ah ∈ H
17
and so there exists h
1
∈ H such that ah = h
1
or a = h
1
h
−1
∈ H.Thus,G
(n)
≤ H.
Now by the claim in Lemma 2.3 we get G
(n+m)
=(G
(n)
)
(m)
≤ H
(m)
=1. Thus,
G
(n+m)
=1andsoG is solvable.
Definition 2.3. Let G be a group and φ : G → G be a map. Then φ is an automor-
phism if φ is 1-1, onto, and a homomorphism. Let
Aut(G)={φ | φ is an automorphism}.
Definition 2.4. Let G be a group and H ≤ G. Then H is a characteristic sub-
group of G if φ(H) ≤ H for all φ ∈ Aut(G). We write H char G.
Lemma 2.4. Let G be a group, H ≤ G, and K ≤ G such that H char K and K char
G. Then H char G.
Proof. Let φ ∈ Aut(G). Since K char G we know φ(K) ≤ K.Ifx,y ∈ K such that
φ(x)=φ(y) then since φ is 1-1 we get x = y.Thus,|φ(K)| = |K| and so φ(K)=K.
But then φ|
K
∈ Aut(K)sinceHcharKwegetφ|
K
(H) ≤ H and so φ(H) ≤ H.Thus,
H char G.
Lemma 2.5. Let G be a group, H ≤ G, and K ≤ G such that H char K and KtriangleleftequalG.
Then H triangleleftequalG.
Proof. Let g ∈ G and h ∈ H. We want to show that ghg
−1
∈ H. Define φ
g
: K → K
by φ
g
(k)=gkg
−1
for all k ∈ K. First, we need to show that φ
g
is a homomorphism.
18
Let x,y ∈ K. Then,
φ
g
(xy)=gxyg
−1
= gxg
−1
gyg
−1
= φ
g
(x)φ
g
(y).
Next, we need to show that φ
g
is 1-1. If φ
g
(x)=φ
g
(y)thengxg
−1
= gyg
−1
implying
x = y. Finally, we need to show that φ
g
is onto. Let x ∈ K. Since K triangleleftequalG we know
g
−1
xg =(g
−1
)x(g
−1
)
−1
∈ K and φ
g
(g
−1
xg)=g(g
−1
xg)g
−1
= x.Thus,φ
g
∈ Aut(K).
Since H char K we get φ
g
(h) ∈ H or gxg
−1
∈ H. Therefore, H triangleleftequalG.
Lemma 2.6. Z(G) char G.
Proof. Let φ ∈ Aut(G), z ∈ Z(G), and g ∈ G. We want to show that φ(z) ∈ Z(G).
Since φ ∈ Aut(G) there exists g
1
∈ G such that φ(g
1
)=g.Now,
φ(z)g = φ(z)φ(g
1
)
= φ(zg
1
) since φ ∈ Aut(G)
= φ(g
1
z) since z ∈ Z(G)
= φ(g
1
)φ(z)
= gφ(z).
Thus, φ(z) ∈ Z(G)andsoZ(G)charG.
19
Definition 2.5. A group G is characteristically simple if {1} and G are its only
characteristic subgroups.
Definition 2.6. Let G be a group and {H
i
}
n
i=1
be a collection of subgroups of G. We
say G = H
1
×H
2
×···×H
n
if
1. G =
n
producttext
i=1
H
i
2. H
i
∩
producttext
jnegationslash=i
H
i
=1for all 1 ≤ i ≤ n
3. H
i
triangleleftequalG for all 1 ≤ i ≤ n
Theorem 2.4. Let G be a characteristically simple group. Then G = G
1
×G
2
×···×
G
n
where G
i
s are simple isomorphic groups.
Proof. Let 1 negationslash= G
1
triangleleftequalG such that |G| is minimal and H =
n
producttext
i=1
G
i
such that
1. G
i
∼
= G
1
for all 1 ≤ i ≤ n
2. G
i
triangleleftequalG for all 1 ≤ i ≤ n
3. G
i
∩
producttext
jnegationslash=i
G
j
=1forall1≤ i ≤ n
4. n is maximal
Clearly, H triangleleftequalG since G
i
triangleleftequalG for all 1 ≤ i ≤ n.IfH is not a characteristic subgroup
of G then there exists φ ∈ Aut(G)and1≤ i ≤ n such that φ(G
i
) negationslash≤ H. Since
G
i
triangleleftequal G and φ ∈ Aut(G)weknowφ(G
i
) triangleleftequal G.Alsoφ(G
i
)
∼
= G
i
∼
= G
1
and so
φ(G
i
)
∼
= G
1
.NowH ∩ φ(G
i
) triangleleftequal G and H ∩ φ(G
i
) <φ(G
i
). Thus, |H ∩ φ(G
i
)| <
|φ(G
i
)| = |G
i
| = |G
1
|. Hence, H ∩ φ(G
i
) = 1 by the minimality of |G
1
|.Butthen
φ(G
i
) ∩
n
producttext
i=1
G
i
= φ(G
i
) ∩ H = 1. Therefore, the subgroups {G
1
,G
2
,···,G
n
,φ(G
i
)}
20
satisfy (1), (2), and (3), a contradiction, since n is maximal. Therefore, H char G.
Since G is characteristically simple we get G = H =
n
producttext
i=1
G
i
= G
1
× G
2
×···×G
n
where G
i
’s are isomorphic groups. Suppose N triangleleftequalG
i
for some 1 ≤ i ≤ n.Ifj negationslash= i and
x ∈ G
i
and y ∈ G
j
then, xyx
−1
y
−1
∈ G
j
∩ G
i
≤ G
j
∩
n
producttext
inegationslash=j
G
i
= 1. Hence xy = yx.
Now let g
1
g
2
···g
n
∈ G where g
i
∈ G
i
for all 1 ≤ i ≤ n and n ∈ N. Then,
g
1
g
2
···g
n
n(g
1
g
2
···g
n
)
−1
= g
1
g
2
···g
n
ng
−1
n
g
−1
n−1
···g
−1
1
= g
i
ng
−1
i
∈ N since N triangleleftequalG
i
.
Thus, N triangleleftequalG.But,|N| < |G
i
| = |G
1
|. Hence, N =1orN = G
i
by the minimality of
|G
1
|. Therefore, each G
i
is simple for all 1 ≤ i ≤ n.
Definition 2.7. Let G be a group and N ≤ G. Then N is a minimal normal
subgroup of G if
1. N triangleleftequalG
2. If there exists a L ≤ N such that LtriangleleftequalG then L =1or L = N.
Definition 2.8. A group G is called an elementary abelian p-group if G
∼
=
Z
p
×Z
p
×···×Z
p
where p is a prime.
Theorem 2.5. Let G be a group and N be a minimal normal subgroup of G. Then
N is an elementary abelian p-group for some prime p or N = N
0
× N
1
×···×N
n
where N
i
s are nonabelian simple isomorphic groups.
21
Proof. If K char N, then by Lemma 2.5, since NtriangleleftequalG we get KtriangleleftequalG.Butthen,K =1
or K = N since N is a minimal normal subgroup. Hence, N is characteristically
simple. Then, by Theorem 2.4, N = N
1
× N
2
×···×N
n
where N
i
’s are simple
isomorphic groups.
Case 1 N
i
is nonabelian for all 0 ≤ i ≤ n.ThenN = N
1
×N
2
×···×N
n
and N
i
sare
nonabelian simple isomorphic groups.
Case 2 N
i
s are abelian for all 0 ≤ i ≤ n.ThenN
i
is simple and abelian for all
0 ≤ i ≤ n. Then the only subgroups of N
i
are {1} and N
i
for all 0 ≤ i ≤ n.
If N
i
is not a p-group then there exists a prime q such that q ||N
i
| and q negationslash= p.
By Sylow’s Theorem there exists Q ∈ Syl
q
(N
i
). Then Q ≤ N
i
and Q negationslash=1and
Q negationslash= N
i
.Thus,N
i
is a p-group for some prime p.Let|N
i
| = p
n
.Ifn>1then
by Cauchy’s Theorem for Abelian Groups, there exists 1 negationslash= x ∈ N
i
such that
x
p
= 1. Then, 〈x〉≤N
i
and |〈x〉| = p<|N
i
|. Therefore, 〈x〉 negationslash=1and〈x〉 negationslash= N
i
.
Hence, n =1and|N
i
| = p. Now we know that N
i
is cyclic and so N
i
∼
= Z
p
.
Thus, N
∼
= Z
p
×Z
p
×···×Z
p
is an elementary abelian p-group.
Therefore, N is an elementary abelian p-group for some prime p or N = N
0
×N
1
×
···×N
n
where N
i
’s are nonabelian simple isomorphic groups.
Theorem 2.6. Let G be solvable and N be a minimal normal subgroup of G. Then
N is an elementary abelian p-group for some prime p.
Proof. By Theorem 2.5, N is an elementary abelian p-group for some prime p or
N = N
1
× N
2
×···×N
n
such that N
i
s are simple nonabelian isomorphic groups.
Hence N
1
is simple. Then the only subnormal series N
1
has is N
1
trianglerightequal1 by simplicity.
22
But,
N
1
{1}
∼
= N is nonabelian. Therefore, N
1
is not solvable. But, N
1
≤ G and G is
solvable, a contradiction. Thus, N is an elementary abelian p-group for some p.
3 Nilpotent Groups
We now introduce the idea of nilpotent groups. This allows us to explore
important properties of nilpotent groups and will let us build the structures of these
groups.
Definition 3.1. Let G is a group. Define the upper central series of G by
Z
0
(G)=1,Z
1
(G)=Z(G),
Z
2
(G)
Z
1
(G)
= Z
parenleftbigg
G
Z
1
(G)
parenrightbigg
,
Z
3
(G)
Z
2
(G)
= Z
parenleftbigg
G
Z
2
(G)
parenrightbigg
,···
and inductively define
Z
n
(G)
Z
n−1
(G)
= Z
parenleftbigg
G
Z
n−1
(G)
parenrightbigg
for all n ∈ Z
+
.
Lemma 3.1. Let G be a group. Then Z
i
(G)triangleleftequalG for all i and Z
i
(G) ≤ Z
i+1
(G) for
all i.
Proof. Use induction on i.Ifi =0,thenZ
0
(G)={1}triangleleftequalG. Assume Z
n
(G)triangleleftequalG.
Then
Z
n+1
(G)
Z
n
(G)
= Z
parenleftbigg
G
Z
n
(G)
parenrightbigg
triangleleftequal
G
Z
n
(G)
and so taking pre-images we get Z
n+1
(G)triangleleftequalG.
Hence, Z
i
(G) ≤ Z
i+1
(G) for all i.
Definition 3.2. A group G is nilpotent if there exists n ∈ Z
+
∪{0} such that
G = Z
n
(G).
23
Definition 3.3. Let G be a group. Define the lower central series of G by
K
0
(G)=G,K
1
(G)=[K
0
(G),G]=[G,G]=G
prime
,K
2
(G)=[K
1
(G),G],···
and inductively define
K
n
(G)=[K
n−1
,G].
Lemma 3.2. Let G be group. Then K
i
(G)triangleleftequalG for all i and K
i+1
(G) ≤ K
i
(G) for
all i.
Proof. Use induction on i.Ifi =0thenK
0
(G)=GtriangleleftequalG. Suppose K
i
(G)triangleleftequalG. Then,
since GtriangleleftequalG we get K
i+1
(G)=[K
i
(G),G]triangleleftequalG as conjugation is a homomorphism.
Next, we know that K
i
(G)triangleleftequalG.Thus,K
i+1
(G)=[K
i
(G),G] ≤ K
i
(G) for all i.
Theorem 3.1. Let G be a group. Then G is nilpotent if and only if there exists
n ∈ Z
+
∪{0} such that K
n
(G)=1.
Proof. (⇒)LetG be nilpotent. Then there exists n ∈ Z
+
∪{0} such that Z
n
(G)=G.
We claim that K
i
(G) ≤ Z
n−i
(G) for all i. Use induction on i.Ifi =0thenK
0
(G)=
G ≤ G = Z
n
(G)=Z
n−0
(G). Suppose K
i
(G) ≤ Z
n−i
(G). Then,
K
i+1
(G)=[K
i
(G),G]
≤ [Z
n−i
(G),G] since K
i
(G) ≤ Z
n−i
(G)
≤ Z
n−i−1
(G) since
Z
n−i
(G)
Z
n−i−1
(G)
= Z
parenleftbigg
G
Z
n−i−1
(G)
parenrightbigg
= Z
n−(i+1)
(G).
Thus, the claim hold by induction. But then, K
n
(G)=Z
n−n
(G)=Z
0
(G)=1and
24
so K
n
(G)=1.(⇐) Suppose there exists a n ∈ Z
+
∪{0} such that K
n
(G)=1.We
claim that K
n−i
(G) ≤ Z
i
(G) for all i. Use induction on i.Ifi =0thenZ
0
(G)=1≥
1=K
n
(G)=K
n−0
(G). Suppose K
n−i
(G) ≤ Z
i
(G). Now, [K
n−i−1
,G]=K
n−i
(G) ≤
Z
i
(G). Hence,
K
n−i−1
(G)Z
i
(G)
Z
i
(G)
≤ Z(
G
Z
i
(G)
)=
Z
i+1
Z
i
(G)
. Taking pre-images we get
K
n−i−1
(G) ≤ K
n−i−1
(G)Z
i
(G) ≤ Z
i+1
(G)orK
n−(i+1)
(G) ≤ Z
i+1
(G). Thus, the claim
holds. But then, Z
n
(G) ≥ K
n−n
(G)=K
0
(G)=G and so Z
n
(G)=G and so G is
nilpotent.
Lemma 3.3. Let G be a group, NtriangleleftequalG, and H ≤ G such that N ≤ H.If
H
N
≤ Z
parenleftBig
G
N
parenrightBig
if and only if [G,H] ≤ N.
Proof.
H
N
≤ Z
parenleftBig
G
N
parenrightBig
⇔ [hN,gN]=N for all h ∈ H and for all g ∈ G.Then
hNgN(hN)
−1
(gN)
−1
= N ⇔ hNgNh
−1
g
−1
= N ⇔ hgh
−1
g
−1
N = N ⇔ [h,g]=N
⇔ [h,g] ∈ N ⇔ [G,H] ≤ N.
Theorem 3.2. Let G be nilpotent. Then Z(G) negationslash=1.
Proof. Suppose Z(G) = 1. Since G is nilpotent there exists n ∈ Z
+
∪{0} such that
Z
n
(G)=G.NoticeZ
1
(G)=Z(G) = 1. Suppose Z
i
(G) = 1. Then
Z
i+1
(G)
Z
i
(G)
= Z
parenleftbigg
G
Z
i
parenrightbigg
= Z
parenleftbigg
G
{1}
parenrightbigg
∼
= Z(G)=1.
Then,
vextendsingle
vextendsingle
vextendsingle
vextendsingle
Z
i+1
(G)
Z
i
(G)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=1or
|Z
i+1
(G)|
|Z
i
(G)|
=1andso|Z
i+1
(G)| = |Z
i
(G)|.But,Z
i
(G) ≤
Z
i+1
(G)andsoZ
i+1
(G)=Z
i
(G) = 1. Thus, by induction Z
i
(G) = 1 for all i.But
then we get G = Z
n
(G) = 1, a contradiction. Therefore, Z(G) negationslash=1.
25
Theorem 3.3. Let G be nilpotent and 1 negationslash= H triangleleftequalG. Then H ∩Z(G) negationslash=1.
Proof. Since G is nilpotent there exists n ∈ Z
+
such that Z
n
(G)=G. Define
H
0
= H,H
1
=[H
0
,G]=[H,G]
and inductively define
H
n
=[H
n−1
,G].
Since H triangleleftequalG we get H = H
0
≥ H
1
≥ H
2
≥···. We claim that H
i
≤ Z
n−i
(G) for all
i.Ifi =0thenH
0
= H ≤ G = Z
n
(G)=Z
n−0
(G). Assume H
i
≤ Z
n−i
(G). Then,
H
i
Z
n−i−1
(G)
Z
n−i−1
(G)
≤
Z
n−i
(G)
Z
n−i−1
(G)
= Z
parenleftbigg
G
Z
n−i−1
(G)
parenrightbigg
.
ByLemma3.3,[H
i
Z
n−i−1
(G),G] ≤ [Z
n−i−1
(G)]. Hence,[H
i
,G] ≤ [H
i
Z
n−i−1
(G),G] ≤
Z
n−i−1
(G). Thus, H
i+1
=[H
i
,G] ≤ Z
n−i−1
(G)=Z
n−(i+1)
(G). Therefore, the claim
holdsbyinduction. ButthenH
n
=[H
n−1
,G] ≤ Z
n−n
(G)=Z
0
(G)=1andsoH
n
=1.
Let 0 ≤ k ≤ n be minimal such that H
k
=1.ThenH
k−1
negationslash=1and1=H
k
=[H
k−1
,G]
and so 1 negationslash= H
k−1
≤ H ∩Z(G).
Theorem 3.4. Let G be nilpotent and H ≤ G. Then H<N
G
(H).
Proof. SinceGisnilpotentthereexistsn ∈ Z
+
suchthatZ
n
(G)=G.Letibeminimal
such that Z
i
(G) negationslash≤ H.ThenZ
i−1
(G) ≤ H.Also,[H,Z
i
(G)] ≤ [G,Z
i
(G)] ≤ Z
i−1
(G)
since
Z
i
(G)
Z
i−1
(G)
= Z
parenleftbigg
G
Z
i−1
(G)
parenrightbigg
.Thus,[H,Z
i
(G)] ≤ H. Hence, Z
i
(G) ≤ N
G
(H)\H
and so H<N
G
(H).
26
Definition 3.4. Let G be a group and M ≤ G. Then M is a maximal subgroup if
1. M negationslash= G
2. whenever there exists H ≤ G such that M ≤ H ≤ G then H = M or H = G
Theorem 3.5. Let G be nilpotent and M be a maximal subgroup of G. Then MtriangleleftequalG.
Proof. Since M is a maximal subgroup of G we know M<G. By Theorem 3.4,
M<N
G
(M) ≤ G. Hence G = N
G
(M) by the maximality of M. Therefore, M triangleleftequalG.
Lemma 3.4. Let G be a group, P ∈ Syl
p
(G), and N triangleleftequalG. Then P ∩N ∈ Syl
p
(N).
Lemma 3.5 (Frattini Argument). Let G be a group, N triangleleftequalG, P ∈ Syl
p
(G). Then,
G = N
G
(P ∩N)N.
Proof. Clearly, N
G
(P ∩N)N ⊆ G since G is a group. Let g ∈ G. Since N triangleleftequalG and
P ∈ Syl
p
(G)byLemma3.4P ∩N ∈ Syl
p
(N). Since N triangleleftequalG and P ∩N ≤ N we get
g
−1
P ∩ Ng ≤ g
−1
Ng = N.Now,|g
−1
(P ∩ N)g| = |P ∩ N| and so g
−1
(P ∩ N)g ∈
Syl
p
(N). BySylow’sTheoremthereexistsn ∈ N suchthatng
−1
(P∩N)gn
−1
= P∩N.
Hence ng
−1
∈ N
G
(P∩N) and so there exists x ∈ N
G
(P∩N) such that ng
−1
= x.But
then g = x
−1
n ∈ N
G
(P ∩N)N.Thus,G ⊆ N
G
(P ∩N)N. Hence G = N
G
(P ∩N)N.
Lemma 3.6. Let G be nilpotent and P ∈ Syl
p
(G) then P triangleleftequalG.
Proof. Suppose P is not normal in G.ThenN
G
(P) <G. Hence, there exists a
maximal subgroup M of G such that N
G
(P) ≤ M. Since G is nilpotent, by Theorem
3.5, M triangleleftequalG.Now,P ≤ N
G
(P) ≤ M and so P ≤ M. By Lemma 3.5,
G = N
G
(P ∩M)M = N
G
(P)M = M.
27
Thus, G = M, a contradiction, since M is maximal. Hence, P triangleleftequalG.
Lemma 3.7. Let G be a group, H triangleleftequalG, K triangleleftequalG such that H and K are nilpotent.
Then HKtriangleleftequalG and HK is nilpotent.
Proof. Use induction on |G|. Since H triangleleftequalG and K triangleleftequalG clearly HKtriangleleftequalG.IfHK <G
then H triangleleftequal HK and K triangleleftequal HK.Also,H and K are nilpotent. Thus, by induction
HK is nilpotent. We may assume G = HK. Since H is nilpotent by Theorem 3.2,
Z(H) negationslash=1.LetN =[Z(H),K].
Case 1 If N =1.Then[Z(H),K]=1.Butalso[H,Z(H)] = 1. Hence, [G,Z(H)] = 1
since G = HK.Thus1negationslash= Z(H) ≤ Z(G). Thus,
G
Z(G)
is a group and
vextendsingle
vextendsingle
vextendsingle
vextendsingle
G
Z(G)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
|G|
|Z(G)|
< |G| since Z(G) negationslash= 1. Since, HtriangleleftequalG and KtriangleleftequalG we get
HZ(G)
Z(G)
triangleleftequal
G
Z(G)
and
KZ(G)
Z(G)
triangleleftequal
G
Z(G)
. Then,
HZ(G)
Z(G)
∼
=
H
H ∩Z(G)
by 2
nd
Isomorphism Theorem
KZ(G)
Z(G)
∼
=
K
K ∩Z(G)
by 2
nd
Isomorphism Theorem
and
HZ(G)
Z(G)
is nilpotent by induction hypothesis.
28
Case 2 If N negationslash= 1. Since K triangleleftequalG we know N ≤ K. Also, since H triangleleftequalG by Lemma 1.4
Z(H)triangleleftequalG. Therefore, since K triangleleftequalG we get N =[Z(H),G]triangleleftequalG. Hence N triangleleftequalK.
Since N negationslash=1andK is nilpotent we get 1 negationslash= N ∩Z(K). Now Z(H)triangleleftequalG implies
N ≤ Z(H). Thus, we get 1 negationslash= N ∩Z(K) ≤ Z(H)∩Z(K). Since G = HK we
have Z(H) ∩ Z(K) ≤ Z(G). Thus, Z(G) negationslash= 1 and we get HK is nilpotent by
Case 1.
Lemma 3.8. Let G be a group and NtriangleleftequalG such that N ≤ Z
i
(G) for all i ∈ Z
+
. Then
Z
i
parenleftBig
G
N
parenrightBig
=
Z
i
(G)
N
for all i.
Theorem 3.6. Let G be nilpotent and N triangleleftequalG. Then
G
N
is nilpotent.
Proof. Since G is nilpotent there exists n ∈ Z
+
∪{0} such that Z
n
(G)=G. We claim
that
Z
i
(G)N
N
≤ Z
i
parenleftBig
G
N
parenrightBig
for all i. Use induction on i.Ifi =0then
Z
0
(G)N
N
=
1N
N
=
N
N
=1N
= Z
0
parenleftbigg
G
N
parenrightbigg
.
Assumetheclaimholds. SinceZ
parenleftbigg
G
Z
i
(G)
parenrightbigg
=
Z
i+1
(G)
Z
i
(G)
, byLemma3.3, [G,Z
i+1
(G)] ≤
Z
i
(G). Then
[G,Z
i+1
(G)]N
N
≤
Z
i
(G)N
N
. Since N triangleleftequal G we get [G,Z
i+1
(G)]N =
[G,Z
i+1
(G)N]. But,
bracketleftbigg
G,Z
i+1
(G)N
N
bracketrightbigg
=
bracketleftbigg
G
N
,
Z
i+1
(G)N
N
bracketrightbigg
.
29
Therefore,
bracketleftbigg
G
N
,
Z
i+1
(G)N
N
bracketrightbigg
≤
Z
i
(G)N
N
≤ Z
i
parenleftBig
G
N
parenrightBig
. Then by Lemma 3.8,
Z
i+1
(G/N)N
N
Z
i
(G/N)
≤ Z
parenleftbigg
G/N
Z
i
(G/N)
parenrightbigg
or
Z
i+1
(G/N)N
N
Z
i
(G/N)
≤
Z
i+1
(G/N)
Z
i
(G/N)
and taking pre-images we get
Z
i+1
(G/N)N
N
≤ Z
i+1
(G/N). Thus the claims holds
by induction. Then Z
n
parenleftBig
G
N
parenrightBig
≤
Z
n
(G)N
N
=
GN
N
=
G
N
. Therefore, Z
n
parenleftBig
G
N
parenrightBig
=
G
N
.
Hence,
G
N
is nilpotent.
Theorem 3.7. Let G be nilpotent. Then G =
producttext
P where the product runs over all
P ∈ Syl
p
(G) and p ||G|.
Proof. By Lemma 3.6 P triangleleftequal G for all P ∈ Syl
p
(G). Therefore,
producttext
P ≤ G where
P ∈ Syl
p
(G)andp ||G|. SinceP∩
producttext
Pnegationslash=Q
Q =1forallP whenq negationslash= pwhereQ ∈ Syl
q
(G)
we get |
producttext
P| =
producttext
|P| = |G|.Thus,G =
producttext
P.
Definition 3.5. Let G be a group. Define the fitting group F(G)=
producttext
NtriangleleftequalG
N and N
nilpotent. Then F(G) is the unique maximal normal nilpotent subgroup of G.
Definition 3.6. Let G be a group and p be a prime. Define O
p
(G) by O
p
(G)=
producttext
PtriangleleftequalG
P
and P is a p-group. Then O
p
(G) is the unique maximal normal p-subgroup of G.
Definition 3.7. Let G be a group and p be a prime. Define O
p
prime(G)=
producttext
QtriangleleftequalG
Q and Q
is a p
prime
-subgroup. Then O
p
prime(G) is the unique maximal normal p
prime
-subgroup of G.
30
Theorem 3.8. Let G be a group. Then F(G)=
producttext
O
p
(G) where p ||G|.
Proof. Asp ||G|andO
p
(G)isap-group,weknowO
p
(G)isnilpotent. SinceO
p
(G)triangleleftequalG
we get O
p
(G) ≤ F(G). By Sylow’s Theorem there exists P ∈ Syl
p
(F(G)) such that
O
p
(G) ≤ P. By Lemma 3.6 we know P triangleleftequal G. Hence since P is a p-group we get
P ≤ O
p
(G). Thus, O
p
(G)=P ∈ Syl
p
(F(G)). Since F(G) is nilpotent, by Theorem
3.7, F(G)=
producttext
P =
producttext
O
p
(G).
Lemma 3.9. Let G be a group and H triangleleftequalG. Then C
G
(H)triangleleftequalG.
Lemma 3.10. Let G be a group, H ≤ G, K ≤ G, and L ≤ G such that [H,K]=1.
Then [H,KL]=[H,L].
Lemma 3.11. Let G be a group, H ≤ G, K ≤ G, and L ≤ G such that K ≤ H.
Then H ∩KL= K(H ∩L).
Theorem 3.9. Let G be solvable. Then C
G
(F(G)) ≤ F(G).
Proof. Let F = F(G)andC = C
G
(F). Suppose C negationslash≤ F. By Lemma 3.9, since F triangleleftequalG
we know CtriangleleftequalG. Then,
CF
F
triangleleftequal
G
F
. Also, since C negationslash≤ F we know
CF
F
negationslash=1F. Then there
exists 1 negationslash=
N
F
≤
CF
F
such that
N
F
is a minimal normal subgroup of
G
F
. Since G is
solvable we know
G
F
is solvable. Hence by Theorem 2.6,
N
F
is an elementary abelian
p-group. Hence,
parenleftBig
N
F
parenrightBig
prime
=
N
prime
F
=1andsoN
prime
≤ F. Since
N
F
≤
CF
F
we get N ≤ CF.
But then, N = N ∩CF = F(N ∩C). We claim that K
i
(N) ≤ K
i−1
(F) for all i ≥ 1.
Use induction on i.Ifi =1then,K
1
(N)=[N,N]=N
prime
≤ F = K
0
(F). Assume
31
K
i
(N) ≤ K
i−1
(F) for all i ≥ 1. Then
K
i+1
(N)=[K
i
(N),N]
≤ [K
i−1
(F),N]
=[K
i−1
(F),F(N ∩C)]
=[K
i−1
(F),F] by Lemma 3.10
= K
i
(F)
Thus, the claim holds. Since F = F(G) in nilpotent there exists n ∈ Z
+
∪{0} such
that K
n
(F) = 1. Then K
n+1
(N) ≤ K
n
(F)=1andsoK
n+1
(N)=1. ThusN is
nilpotent. Since
N
F
triangleleftequal
G
F
we have N triangleleftequal G.Thus,N ≤ F.Butthen,
N
F
=1,a
contradiction. Therefore, C
G
(F(G)) ≤ F(G).
Lemma 3.12. Let G be a group and P ∈ Syl
p
(F(G)). Then P triangleleftequalG.
Proof. Let g ∈ G. Since P ≤ F(G)wegetgPg
−1
≤ gF(G)g
−1
. Since F(G)triangleleftequalG,
gF(G)g
−1
≤ F(G)andsogPg
−1
≤ F(G). Now F(G) is nilpotent implies P triangleleftequalF(G).
Thus by Sylow’s Theorem,
n
p
=
|F(G)|
|N
F(G)
(P)|
=
|F(G)|
|F(G)|
=1.
Also |gPg
−1
| = |P| and so gPg
−1
∈ Syl
p
(F(G)). Since n
p
=1wegetP = gPg
−1
and so P triangleleftequalG.
32
Theorem 3.10. Let G be a group, P ≤ G be a p-group, and N triangleleftequalG be a p
prime
-group.
Then
N
G
(P)
P
= N
G/N
parenleftbigg
PN
N
parenrightbigg
Proof. Let xN ∈
N
G
(P)
N
where x ∈ N
G
(P). Then
xN
parenleftbigg
PN
N
parenrightbigg
x
−1
N =
x(PN)x
−1
N
=
xPx
−1
xNx
−1
N
=
PN
N
since x ∈ N
G
(P)andN triangleleftequalG
Hence, xN ∈ N
G/N
parenleftBig
PN
N
parenrightBig
andso
N
G
(P)N
N
≤ N
G/N
parenleftBig
PN
N
parenrightBig
.LetxN ∈ N
G/N
parenleftBig
PN
N
parenrightBig
.
Then xN
parenleftBig
PN
N
parenrightBig
x
−1
N =
PN
N
and so as before we get
xPx
−1
N
=
PN
N
Taking pre-
imageswegetxPx
−1
= PN. Since N is a p
prime
-group we get P,xPx
−1
∈ Syl
p
(PN). By
Sylow’s Theorem there exists n ∈ N such that nxPx
−1
n
−1
= P or nxP(nx)
−1
= P.
Thus, nx ∈ N
G
(P). But then xN = nxN ∈
N
G
(P)N
N
.Thus,N
G/N
parenleftBig
PN
N
parenrightBig
≤
N
G
(P)N
N
. Therefore,
N
G
(P)N
N
= N
G/N
parenleftBig
PN
N
parenrightBig
.
4 Groups Acting on Groups
We know look at how groups act on groups and will prove important Theorems
about co-prime actions.
Definition 4.1. Let G and H be groups. Then G acts on H if there exists a homo-
morphism φ such that φ : G → Aut(H).
33
Theorem 4.1. Let G and H be p-groups such that G acts on H. Then there exists
1 negationslash= h ∈ H such that G = G
h
.
Proof. Since G acts on H we know G acts on S = H \{1}. Since H is a p-group and
p negationslash |1weknowp negationslash ||S|. Since G is a p-group by the Fixed Point Theorem there exists
s ∈ S such that G = G
s
.Butthen,1negationslash= s ∈ H.
Theorem 4.2. Let G be a group, A ≤ G, B ≤ G, and C ≤ G such that [A,B,C]=1
and [B,C,A]=1. Then [C,A,B]=1.
Proof. Let a ∈ A, b ∈ B,andc ∈ C.Notice
b[a
−1
,b,c
−1
]b
−1
c[b
−1
,c,a
−1
]c
−1
a[c
−1
,a,b
−1
]a
−1
=1
Now, [a
−1
,b,c
−1
]=1andsob[a
−1
,b,c
−1
]b
−1
= 1 since [A,B,C] = 1. Similarly,
c[b
−1
,c,a
−1
]c
−1
= 1since[B,C,A]=1.Thus,a[c
−1
,a,b
−1
]a
−1
=1andso[c
−1
,a,b
−1
]=
1. Therefore, [C,A,B]=1.
Theorem 4.3. Let A ≤ Aut(P) be a p
prime
-group and P be a p-group such that there
exists a subnormal series
P trianglerightequalP
1
trianglerightequalP
2
trianglerightequal···trianglerightequalP
n
=1
such that P
i
is A-invariant and A acts trivially on
P
i
P
i+1
for all 1 ≤ i ≤ n−1. Then
A acts trivially on P.
34
Proof. Use induction on |P|. Since |P
1
|≤|P| we get A acts trivially on P
1
.IfA does
not act trivially on P there exists φ ∈ A and x ∈ P such that φ(x) negationslash= x. Since A acts
trivially on
P
P
1
we get φ(xP
1
)=xP
1
or φ(x)P
1
= xP
1
. Hence, there exists a y ∈ P
1
such that φ(x)=xy. Then,
φ(φ(x)) = φ(xy)
= φ(x)φ(y)
= xyy
= xy
2
.
Since A acts trivially on P
1
we get x = φ
|φ|
(x)=xy
|φ|
.Butthen,y
|φ|
=1andso
|y|||φ|. Since P is a p-group and A is a p
prime
-group we get gcd(|y|,|φ|)=1. Thus,
|y| =1andsoy =1.Butthenφ(x)=x1=x, a contradiction. Therefore, A acts
trivially on P.
Theorem 4.4. Suppose A×B acts on P such that A is a p
prime
-group and B and P are
p-groups. If A acts trivially on C
P
(B) then A acts trivially on P.
Proof. Let C
P
(B) ≤ Q ≤ P where Q is a maximal A × B-invariant subgroup of P
such that A acts trivially on Q.IfQ<P. Now by Theorem 3.4, Q<N
P
(Q)=R
and QtriangleleftequalR. Since P and Q are A×B-invariant we know R is A×B-invariant. Thus,
A × B acts on
R
Q
and so B acts on
R
Q
.Let1Q negationslash=
S
Q
≤
R
Q
be a minimal A × B-
invariant subgroup of
R
Q
.Now,B acts on
S
Q
and they are both p-groups. Therefore,
by Theorem 4.1, 1 negationslash= C
S/Q
(B) ≤
S
Q
. Since
S
Q
is A × B-invariant we get C
S/Q
(B)
is A × B-invariant. By the minimality of
S
Q
= C
S/Q
(B). But then, [S,B] ≤ Q.
35
Now, [S,B,A] ≤ [Q,A] = 1 since A acts trivially on Q. Hence, [S,B,A]=1.
Also, [B,A] = 1 implies [B,A,S] = 1. By Theorem 4.2, [A,S,B]=1. Thus,
[A,S] ≤ C
P
(B) ≤ Q. Now we have a subnormal series StriangleleftequalQtriangleleftequal1andA acts trivially
on
S
Q
and on
Q
{1}
. Since A is a p
prime
-group and S is a p-group by Theorem 4.3, A acts
trivially on S.Now,Q<Ssince
S
Q
negationslash=1Q and S is A×B-invariant, this contradicts
the maximality of Q. Hence, P = Q and A acts trivially on P.
Definition 4.2. Let G be a group, A ≤ Aut(G), g ∈ G, and φ ∈ A. Then
1. [g,φ]=g
−1
φ(g) is the commutator of g and φ
2. [G,A]=〈{[g,φ] | g ∈ Gforallφ∈ A}〉
3. C
G
(A)={g ∈ G | φ(g)=gforallφ∈ A}.
Theorem 4.5. Let A ≤ Aut(P) be a p
prime
-group and P be an abelian p-group. Then
P = C
P
(A)×[P,A].
Proof. Let |A| = n and writing P additively define θ =
1
n
summationtext
φ∈A
φ.Thenθ : P → P is
a homomorphism since P is abelian. We want to show the following,
1. θφ
1
= θ for all φ
1
∈ A
2. θ
2
= θ
3. θ(P)=C
P
(A)
4. [P,A]={−x+θ(x) | x ∈ P}
5. P = θ(P)×B where B =[P,A]
36
For (1), let x ∈ P.Then
θφ
1
(X)=
1
n
summationdisplay
φ∈A
φφ
1
(x)
=
1
n
summationdisplay
φ∈A
φ(x) since P is abelian
= θ(x).
For (2), let x ∈ P. Then,
θ
2
(x)=θ(
1
n
summationdisplay
φ∈A
φ(x))
=
1
n
summationdisplay
φ∈A
θφ(x)
=
1
n
summationdisplay
φ∈A
θ(x)
=
1
n
nθ(x)
= θ(x).
For (3), let θ(x) ∈ θ(P)andφ ∈ A.Thenφθ(x)=θ(x). Hence, θ(x) ∈ C
P
(A)andso
θ(P) ≤ C
P
(A). Let x ∈ C
P
(A). Then,
θ(x)=
1
n
summationdisplay
φ∈A
φ(x)
=
1
n
summationdisplay
φ∈A
x
=
1
n
nx
= x.
37
Hence, x ∈ θ(P)andsoC
P
(A) ≤ θ(P). Thus, φ(P)=C
P
(A). For (4), let x ∈ P.
Then,
−x+θ(x)=−x+
1
n
summationdisplay
φ∈A
φ(x)
=
1
n
summationdisplay
φ∈A
−x+θ(x) since P is abelian
∈ [P,A] since −x+θ(x) ∈ [P,A] for all x ∈ P and φ ∈ A.
Hence, {−x+θ(x) | x ∈ P}⊆[P,A]. Let x ∈ P and φ ∈ A. Then,
[x,φ]=−x+φ(x)
= −x+φ(x)+0
= −x+φ(x)+θ(x+−φ(x)) by (1)
∈{x+θ(x) | x ∈ P}.
Hence, [P,A] ⊆{−x+θ(x) | x ∈ P} since {−x+θ(x) | x ∈ P} is closed. Therefore,
[P,A]={−x+θ(x) | x ∈ P}. For(5), letx ∈ P. Then, x = θ(x)+x+−θ(x) ∈ θ(P)B.
Thus, P = θ(P)B. Suppose, there exists u ∈ φ(P)∩B. Then, there exists x,y ∈ P
38
such that u = θ(x)andu = −y +θ(y). Then,
u = θ(x)
= θ
2
(x)by(2)
= θ(θ(x))
= θ(u)
= θ(−y +θ(y))
= −θ(y)+θ
2
(y)
= −θ(y)+θ(y)
=0.
Hence, θ(B)∩B =0andsoP = θ(P)B = C
P
(A)[P,A]=C
P
(A)×[P,A].
Lemma 4.1. Let G be a group and A ≤ Aut(G). Then [G,A] triangleleftequal G and [G,A] is
A-invariant.
Lemma 4.2. Let G be a group, A ≤ Aut(G), and N triangleleftequalG be A-invariant. Then A
acts on
G
N
by φ(gN)=φ(g)N for all gN ∈
G
N
and for all φ ∈ A.
Theorem 4.6. Let A ≤ Aut(P) be a p
prime
-group and P be a p-group. Then
P = C
P
(A)[P,A].
Proof. Let H =[P,A].
39
Case 1 H ≤ Z(P). Let φ ∈ A. Define α
φ
: P → [P,φ]byα
φ
=[x,φ] for all x ∈ P.If
x,y ∈ P then
α
φ
(xy)=[xy,φ]
=(xy)
−1
φ(xy)
= y
−1
x
−1
φ(x)φ(y)
= x
−1
φ(x)y
−1
φ(y) since x
−1
∈ φ(x) ≤ Z(P)
=[x,φ][y,φ]
= α
φ
(x)α
φ
(y).
Hence, α
φ
is a homomorphism. Now, Kernα
φ
= C
P
(φ)andα
φ
is onto. By
Theorem 1.5,
P
Kernα
φ
∼
= α
φ
(P). Thus,
P
C
P
(φ)
∼
= [P,φ] since α
φ
is onto. Since
[P,φ] ≤ H ≤ Z(P)weget
P
C
P
(φ)
is abelian. Hence, by Lemma 1.3, P
prime
≤
C
P
(φ). Therefore, P
prime
≤ C
P
(A). Now A acts on
P
P
prime
which is an abelian p-group.
Then,
P
P
prime
= C
P/P
prime(A)
bracketleftBig
P
P
prime
,A
bracketrightBig
= C
P/P
prime
[P,A]P
prime
P
prime
.Let
C
P
prime
= C
P/P
prime(A). Then,
P
P
prime
=
C
P
prime
[P,A]P
prime
P
prime
and taking pre-images we get P = C[P,A]P
prime
or P = C[P,A].
Now, [P
prime
,C]=P
prime
since
C
P
prime
= C
P/P
prime(A). Hence, we have a subnormal series
C trianglerightequalP
prime
trianglerightequal1
and A acts trivially on
C
P
prime
and
P
prime
{1}
. Since A is a p
prime
-group and C is a p-group
by Theorem 4.3, A acts trivially on C.Thus,P = C[P,A] ≤ C
P
(A)[P,A] ≤ P
and so P = C
P
(A)[P,A].
Case 2 H negationslash≤ Z(P). Use induction on |P|. Since P is nilpotent and H =[P,A]triangleleftequalP
40
by Theorem 3.3, 1 negationslash= H ∩ Z(P). Now, K = H ∩ Z(P) triangleleftequal P and so
P
K
is a
p-group. Also, K is A-invariant and so A acts on
P
K
.Also,
vextendsingle
vextendsingle
vextendsingle
P
K
vextendsingle
vextendsingle
vextendsingle < |P| and so
by induction
P
K
= C
P/K
(A)
bracketleftBig
P
K
,A
bracketrightBig
= C
P/K
(A)
[P,A]K
K
.Let
C
K
= C
P/K
(A).
Then,
P
K
=
C
K
[P,A]K
K
and taking pre-images we get P = C[P,A]K = C[P,A].
If P = C then
P
K
=
C
K
= C
P/K
(A). Hence,
bracketleftBig
P
K
,A
bracketrightBig
= K and so [P,A] ≤ K.
But then H =[P,A] ≤ K ≤ Z(P), a contradiction. Therefore, P negationslash= C.Thus,
C<Pand so |C| < |P|. Hence, by induction C = C
C
(A)[C,A]. But then,
P = C
C
(A)[C,A][P,A] ≤ C
P
(A)[P,A] ≤ P and so P = C
P
(A).
Theorem 4.7. Let A ≤ Aut(P) be a p
prime
-group, P be a p-group, and N triangleleftequalP be A-
invariant. Then C
P/N
(A)=
C
P
(A)N
N
.
Proof. Let cN ∈
C
P
(A)N
N
and φ ∈ A. Then, φ(cN)=φ(c)N = cN since c ∈
C
P
(A). Thus,
C
P
(A)N
N
≤ C
P/N
(A). Let
C
N
= C
P/N
(A). Then, N ≤ C ≤ P and
C is A-invariant. Also [C,A] ⊆ N. Hence, by Theorem 4.6, C = C
P
(A)[C,A] ≤
C
P
(A)N. Therefore, by taking pre-images we get C
P/N
(A)=
C
N
≤
C
P
(A)N
N
. Hence,
C
P/N
(A)=
C
P
(A)N
N
.
Definition 4.3. Let G be a group then O
p
prime(G)=
producttext
NtriangleleftequalG
N where N is a p
prime
-group and
is the largest normal p
prime
-subgroup of G.
Theorem 4.8. Let G be solvable and P ≤ G be a p-subgroup. Then, O
p
prime(N
G
(P)) ≤
O
p
prime(G).
Proof. Let A = O
p
prime(N
G
(P)) and B = O
p
prime(G).
41
Case 1 O
p
prime(G) = 1. We want to show that A = 1. Suppose, A negationslash= 1. Then, AtriangleleftequalN
G
(P)
andPtriangleleftequalN
G
(P). Hence,APtriangleleftequalN
G
(P). SinceAisap
prime
-groupandP isap-groupwe
get |A∩P| = 1. Since AtriangleleftequalN
G
(P)andPtriangleleftequalN
G
(P)weget[A,P] = 1. Since BtriangleleftequalG
we get A×P acts on B by conjugation. Now, C
B
(P) ≤ N
G
(P)andAtriangleleftequalN
G
(P).
Since BtriangleleftequalG we get [A,C
B
(P)] ≤ A∩B =1.Thus,A acts trivially on C
B
(P).
By Theorem 4.4, A acts trivially on B.Butthen,A ≤ C
G
(B)andsoC
G
(B)
is not a p-group. Since B triangleleftequal G we know C
G
(B) triangleleftequal G. Then,
C
G
(B)B
B
triangleleftequal
G
B
.
If
C
G
(B)B
B
=1wegetC
G
(B) ≤ B. But then, since B is a p-group we get
C
G
(B)isap-group, a contradiction. Thus, 1 negationslash=
C
G
(B)
B
triangleleftequal
G
B
. Hence, there
exists 1 negationslash=
N
B
≤
C
G
(B)
B
such that
N
B
is a minimal subgroup of
G
B
. Since G is
solvable by Theorem 2.3,
G
B
is solvable. By Theorem 2.6,
N
B
is an elementary
q-group. Suppose p = q. Since
N
B
triangleleftequal
G
B
we get N triangleleftequalG.Also,|N| =
|N|
|B|
and
|B| =
vextendsingle
vextendsingle
vextendsingle
N
B
vextendsingle
vextendsingle
vextendsingle|B| is a power of P and so N is a p-group. Hence, N ≤ B = O
p
(G)
and we get
N
B
= 1, a contradiction. Therefore, p negationslash= q.LetQ ∈ Syl
q
(N).
Then,
QB
B
∈ Syl
q
parenleftBig
N
B
parenrightBig
. Since
N
B
is a p-group we get
N
B
=
QB
B
. Taking
pre-images we get N = QB. Since
N
B
≤
C
G
(B)
B
we get N ≤ C
G
(B). Hence,
Q ≤ N ≤ C
G
(B)B and so
QC
G
(B)
C
G
(B)
≤
C
G
(B)B
C
G
(B)
.But,
C
G
(B)B
C
G
(B)
∼
=
B
B ∩C
G
(B)
is a p-group. Thus,
QC
G
(B)
C
G
(B)
is a p-group. But,
QC
G
(B)
C
G
(B)
∼
=
Q
Q∩C
G
(B)
is
a q-group. Thus,
QC
G
(B)
C
G
(B)
=1andsoQ ≤ C
G
(B). Since N = QB we get
QtriangleleftequalN. Therefore, Q is the only Sylow q-subgroup of N by Sylow’s Theorem.
Now, since N triangleleftequalG we get QtriangleleftequalG. Since p negationslash= q we know Q is a p
prime
-group and so
Q ≤ O
p
prime(G)=1andsoN = QB = B and we get
N
B
=
B
B
= 1, a contradiction.
Thus, A =1
Case 2 O
p
prime(G) negationslash= 1. Then,
G
O
p
prime(G)
is solvable and
PO
p
prime(G)
O
p
prime
≤
G
O
p
prime
(G) is a p-group.
42
Finally, O
p
prime
parenleftbigg
G
O
p
prime(G)
parenrightbigg
= 1 by Case 1 we get O
p
prime
parenleftbigg
N
G/O
p
prime(G)
parenleftbigg
PO
p
prime(G)
O
p
prime(G)
parenrightbiggparenrightbigg
=1.
Then, O
p
prime
parenleftbigg
N
G
(P)O
p
prime(G)
O
p
prime
parenrightbigg
= 1 by Lemma 3.10. But,
O
p
prime(N
G
(P))O
p
prime(G)
O
p
prime(G)
≤
O
p
prime
parenleftbigg
N
G
(P)O
p
prime(G)
O
p
prime
parenrightbigg
andso
O
p
prime(N
G
(P))O
p
prime(G)
O
p
prime(G)
= 1whichimpliesO
p
prime(N
G
(P)) ≤
O
p
prime(G).
Definition 4.4. Let G be a group. Define the Franttini Subgroup by Φ(G)=
intersectiontext
M
where M is a maximal subgroup of G.
Theorem 4.9. Let P be a p-group. Then,
P
Φ(P)
is an elementary abelian p-group.
In particular, if Φ(P)=1then P is an elementary p-group.
Proof. Let M be a maximal subgroup of P and let x ∈ P. Since P is nilpotent we
get M triangleleftequalP by Theorem 3.5. Since M is maximal we know {1} and
P
M
are the only
subgroups of
P
M
.Thus,
P
M
∼
= Z
p
is abelian. Thus, P
prime
≤ M.Also,(xM)
p
= x
p
M =
1M since
P
M
∼
= Z
p
.Thus,x
p
∈ M but then, P
prime
≤ Φ(P)andx
p
∈ Φ(P) for all x ∈ P
which implies all the elements have order p or 1. By the Fundamental Theorem of
Finite Abelian Groups we get
P
Φ(P)
∼
= Z
p
×Z
p
×···×Z
p
is an elementary abelian
p-group. In particular, if Φ(P)=1thenP
∼
=
P
{1}
∼
=
P
Φ(P)
is an elementary p-group.
Definition 4.5. A group A acts regularly on a group G if C
G
(α)=1for all 1 negationslash=
α ∈ A.
Theorem 4.10. Suppose an elementary p-group A acts regularly on a q-group V.
Then A
∼
= Z
p
.
43
Proof. Use contradiction. Suppose A
∼
= Z
p
×Z
p
. Then all elements of A have order
P. Hence, H =
p+1
uniontext
i=1
A
i
such that |A
i
| = p for all i and A
i
∩A
j
=1foralli negationslash= j.Let
1 negationslash= v ∈ V and 1 negationslash= a
0
∈ A.Thena
0
(
producttext
a∈A
av)=
producttext
a∈A
a
0
av =
producttext
a∈A
av since as a runs
over A so does a
0
a and V is abelian. Since A acts regular on V we get
producttext
av =1.
Similarly,
producttext
a
i
∈A
i
a
i
v = 1. Hence,
1=
p+1
productdisplay
i=1
productdisplay
a
i
∈A
i
a
i
v
= v
p
productdisplay
a∈A
av since V is abelian
= v
p
1
= v
p
.
Hence, v
p
=1andso|v| = p since v negationslash=1.Butthen,p = |v|||V| which implies p | q
a
,
a contradiction. Thus, A
∼
= Z
p
.
Theorem 4.11. Let G = BV be a group such that B triangleleftequalG is a p-group and V is an
elementary abelian q-group. Then
B = 〈C
B
(U) | U ≤ V,
|V|
|U|
= q〉
Proof. Use induction on |G|.LetA = 〈C
B
(U) | U ≤ V,
|V|
|U|
= q〉.IfA<Bthen
since B is nilpotent we get A<N
B
(A). Since V ≤ N
G
(B)andV is abelian we know
V ≤ N
G
(A). Hence, V ≤ N
B
(N
B
(A)) and so VN
B
(A) ≤ G.IfVN
B
(A) <Gthen
by induction we get N
B
(A)=〈C
N
B
(A)
(U) | U ≤ V,
|V|
|U|
= q〉≤A, a contradiction.
44
Hence, G = VN
B
(A)andsoA triangleleftequal G.Thus,
G
A
=
B
A
VA
A
is a group. If A negationslash=1,
then
vextendsingle
vextendsingle
vextendsingle
G
A
vextendsingle
vextendsingle
vextendsingle < |G| and
B
A
triangleleftequal
G
A
is a p-group and
VA
A
is an elementary q-group. Hence,
by induction
B
A
= 〈C
B/A
(
U
A
) |
U
A
≤
VA
A
,
|VA|
|U|
= q〉. Since A<Bwe know
B
A
negationslash=1A. Hence, there exists
U
A
≤
VA
A
such that
|VA|
|U|
= q and C
B/A
(
U
A
) negationslash=1A.
Let U
0
∈ Syl
q
(U). Then
U
0
A
A
∈ Syl
q
(
U
A
). Since
U
A
is a q-group we get
U
A
=
U
0
A
A
Hence, C
B/A
(
U
0
A
A
) negationslash=1A. Since U
0
and
U
0
A
A
act on
B
A
inthesamewayweget
C
B
(U
0
) negationslash= 1 by Theorem 4.7 the q-group U
0
acts on the p-group
B
A
and we get
1 negationslash= C
B/A
(U
0
)=
C
B
(U
0
)A
A
. Hence, C
B
(U
0
) negationslash≤ A.Now,
q =
|VA|
|U|
=
|VA|
|U
0
A|
=
|VU
0
A|
|U
0
A|
=
|V||U
0
A|
|V ∩U
0
A|
|U
0
A|
=
|V|
|V ∩U
0
A|
.
Hence,
|V|
|V ∩U
0
A|
= q.Now,V ∩ U
0
A ≤ U
0
A and V ∩ U
0
A is a q-group. Since
U
0
∈ Syl
q
(U
0
A) by Sylow’s Theorem there exists a ∈ A such that V ∩U
0
A ≤ aU
0
a
−1
.
Then V ∩ U
0
A ≤ V ∩ aU
0
a
−1
.ButV ∩ aU
0
a
−1
≤ V ∩ U
0
A and so V ∩ U
0
A =
V∩aU
0
a
−1
. Hence,
|V|
|V ∩aU
0
a
−1
|
= q andsoC
B
(V∩aU
0
a
−1
) ≤ A.Now,C
B
(U
0
) negationslash≤ A,
a contradiction. Hence, A = 1. As, Φ(B) <Band B triangleleftequalG, we get Φ(B)triangleleftequalG.Then
Φ(B)V<BV= G. Hence, |Φ(B)| < |G| and so, by induction, Φ(B)=〈C
Φ(B)
(U) |
U ≤ V,
|V|
|U|
= q〉≤A = 1. Thus, Φ(B) = 1 which implies by Theorem 4.9 B is
an elementary abelian p-group. Let 1 negationslash= b ∈ B and 〈b
G
〉 = 〈gbg
−1
| g ∈ G〉.Then
45
〈b
G
〉triangleleftequalG and so V acts on 〈b
G
〉 by conjugation. Now, since G = BV and B is abelian
we get 〈b
G
〉 = 〈b
V
〉. Moreover, since V is abelian,
V
C
V
(b)
acts on 〈b
V
〉 regularly
by conjugation. Then, by Theorem 4.10,
V
C
V
(b)
∼
= Z
q
and so |
V
C
V
(b)
| = q.Now,
1 negationslash= b ∈ C
B
(C
V
(b)) ≤ A = 1, a contradiction. Thus, B = A = 〈C
B
(U) | U ≤ V,
|V|
|U|
〉.
Theorem 4.12. Let G = AB where A is a p-group and B is a q-group. Further
suppose there exists 1 negationslash= A
0
triangleleftequalA and 1 negationslash= B
0
triangleleftequalB such that 〈A
B
0
0
〉 is a p-group. Then
G is not simple.
Proof. Let 〈A
B
0
0
〉≤P
0
≤ G such that P
0
is maximal with respect to P
0
being a
p-group, P
0
generated by conjugates of A
0
,andB
0
≤ N
G
(P
0
). By Sylow’s Theorem
there exists P ∈ Syl
p
(G) such that P
0
≤ P.WewanttoshowthatP
0
triangleleftequalP. Suppose
not. Then, N
P
(P
0
) <P. Since P is nilpotent we get N
P
(P
0
) <N
P
(N
P
(P
0
)). Let
x ∈ N
P
(N
P
(P
0
)) \ N
P
(P
0
). Then xP
0
x
−1
negationslash= P
0
and so xP
0
x
−1
negationslash≤ P
0
. Hence, there
exists g ∈ G such that xgA
0
(xg)
−1
negationslash≤ gA
0
g
−1
≤ P
0
.LetH = 〈P
0
(xgA
0
(xg)
−1
)
B
0
〉.
Then P
0
≤ H.Also,H is generated by conjugates of A
0
and since B
0
≤ N
G
(P
0
)
we know B
0
≤ N
B
(H). Now, gA
0
g
−1
≤ P
0
≤ N
P
(P
0
)andsoxgA
0
(xg)
−1
≤
N
P
(P
0
) since x ∈ N
P
(N
P
(P
0
)). Thus, (xgA
0
(xg)
−1
)
B
0
≤ N
G
(P
0
). Therefore, H =
〈P
0
,(xgA
0
(xg)
−1
)
B
0
〉 = P
0
〈(xgA
0
(xg)
−1
)
B
0
〉. Now, since A
0
triangleleftequal A and B
0
triangleleftequal B and
G = AB we get 〈(xgA
0
(xg)
−1
)
B
0
〉≤〈A
B
0
0
〉
b
since g = ba and b ∈ B and a ∈ A.But
since 〈A
B
0
0
〉 is a p-group we get 〈A
B
0
0
〉
b
is a p-group. Therefore, 〈(xgA
0
(xg)
−1
)
B
0
〉 is a
p-group. Since P
0
is a p-group we get H = P
0
〈(xgA
0
(xg)
−1
)
B
0
〉 is a p-group, a con-
tradiction to the maximality of P
0
.So,P
0
triangleleftequalP. Now since G = AB and P ∈ Syl
p
(G)
we get G = PB so then since B
0
triangleleftequal B, B
0
≤ N
P
(P
0
), and P ≤ N
G
(P
0
)weget
1 negationslash= B
0
≤
intersectiontext
b∈B
bN
G
(P
0
)b
−1
=
intersectiontext
g∈G
gN
G
(P
0
)g
−1
triangleleftequalG.If
intersectiontext
g∈G
gN
G
(P
0
)g
−1
negationslash= G we get G
46
is not simple. If G =
intersectiontext
g∈G
gN
G
(P
0
)g
−1
then G = N
G
(P
0
). But then 1 negationslash= P
0
triangleleftequalG and
P
0
= G since P
0
is a p-group. Hence, G is not simple.
Definition 4.6. Let G be a group and p be a prime. Define
Ω
1
(G)=〈x ∈ G | x
p
=1〉.
Definition 4.7. Let G be group and P ≤ G be a p-group. Define
J(P)=〈A | A ≤ P is abelian and |A| is maximal〉.
Then J(P) is called the Thompson Subgroup.
Theorem 4.13 (Baer). Let G be a group and H ≤ G such that 〈H,gHg
−1
〉 is a
p-group for all g ∈ G. Then H ≤ O
p
(G).
Theorem 4.14. Let G be a group and x negationslash∈ O
2
(G) such that x
2
=1. Then there exists
y ∈ G such that |y| is odd and xyx
−1
= y
−1
.
Theorem 4.15. Let G be a group such that |G| = p
a
q
b
for odd primes p and q and
P ∈ Syl
p
(G) such that C
G
(Ω
1
(Z(P))) = P. Then J(P)triangleleftequalG.
Definition 4.8. Let G be a group and H ≤ G. Then H is a p-central subgroup
of G is there exists P ∈ Syl
p
(G) such that H ≤ Z(P). We write Hp-central ≤ G.
5 Burnsides p
a
q
b
Theorem
We now have all the group theoretical tools needed to begin our proof of
Burnside’s Theorem.
47
Theorem 5.1 (Burnside’s Theorem). Let G be a group such that |G| = p
a
q
b
. Then
G is solvable.
Proof. Assume the theorem is false and let G be a minimal counterexample. We to
prove the following about G,
1. G is simple.
Assume not. There exists 1 negationslash= N triangleleftequalG and N negationslash= G.Then
G
N
is a group such
that
vextendsingle
vextendsingle
vextendsingle
G
N
vextendsingle
vextendsingle
vextendsingle < |G| and N is a group such that |N| < |G| and they are both pq
groups. Hence, by the minimality of G we get
G
N
and N are solvable and so by
Theorem 2.3, G is solvable, a contradiction. Thus, G is simple.
2. If M is a maximal subgroup of G then F(M)isap or q group.
Suppose p ||F(M)| and q ||F(M)|.LetZ = Z
p
Z
q
where Z
p
=Ω
1
(Z(O
p
(M)))
and Z
q
=Ω
1
(Z(O
q
(M))). Then Z
p
triangleleftequalM and so M ≤ N
G
(Z
p
) ≤ G.Bythe
maximality of M we get M = N
G
(Z
p
)orG = N
G
(Z
p
). If G = N
G
(Z
p
) then
we get Z
p
triangleleftequalG, a contradiction since G is simple. Therefore, M = N
G
(Z
p
)and
similarly M = N
G
(Z
q
). We claim that M is the unique maximal subgroup of G
such that Z ≤ M. Suppose Z ≤ H and H is a maximal subgroup of G.Then
O
p
(M)∩H triangleleftequalM ∩H is a q-group. But then,
O
p
(M)∩H ≤ O
p
(M ∩H)
= O
p
(N
G
(Z
q
)∩H)
= O
p
(N
H
(Z
q
))
≤ O
p
(H).
48
Similarly, using M = M
G
(Z
p
)wegetO
q
(M)∩H ≤ O
q
(H). Hence,
F(M)∩H = O
p
(M)O
q
(M)∩H
=(O
p
(M)∩H)(O
q
(M)∩H)
≤ O
p
(H)O
q
(H)
= F(H).
Thus, F(M) ∩ H ≤ F(H). Now, Z = Z
q
Z
q
≤ F(M) ∩ H ≤ F(H). Hence,
by Sylow’s Theorem Z
p
≤ O
p
(H)andZ
q
≤ O
q
(H). Now, [Z
p
,O
q
(H)] ≤
[O
p
(H),O
q
(H)] ≤ O
p
(H)∩O
q
(H)=1andso[Z
p
,O
q
(H)] = 1. Thus, O
q
(H) ≤
C
G
(Z
p
) ≤ N
G
(Z
p
)=M. Similarly,O
p
(H) ≤ M andsoF(M)=O
p
(H)O
q
(H) ≤
M. Since Z
p
≤ O
p
(H)andZ
q
≤ O
q
(H)wegetp ||F(H)| and q ||F(H)|.
Similarly, since H is maximal subgroup, using Z
∗
p
=Ω
1
(Z(O
p
(H))) and Z
∗
q
=
Ω
1
(Z(O
q
(H))) we get F(M)∩M ≤ F(M)andF(M) ≤ H.ButthenF(M)=
F(M)∩H ≤ F(H)=F(H)∩M ≤ F(M). Thus, F(M)=F(H). Now since M
and H are maximal and G is simple we get M = N
G
(F(M)) = N
G
(F(H)) = H.
We claim M does not contain a Sylow p-subgroup of G.LetM
p
∈ Syl
p
(M).
If M
p
∈ Syl
p
(G) then by Sylow’s Theorem there exists G
q
∈ Syl
q
(G)such
that O
q
(M) ≤ G
q
.ThenG = M
p
G
q
. Now since O
q
(M) triangleleftequal M we get 1 negationslash=
O
q
(M) ≤
intersectiontext
x∈M
p
xG
q
x
−1
=
intersectiontext
x∈G
xG
q
x
−1
triangleleftequal G.Thus,1negationslash=
intersectiontext
x∈G
xG
q
x
−1
triangleleftequal G but
intersectiontext
x∈G
xG
q
x
−1
≤ G
q
<G, a contradiction since G is simple. Hence, M does
not contain a Sylow p-subgroup of G and similarly M does not contain a Sy-
low q-subgroup of G.LetM
p
∈ Syl
p
(M). Then there exists G
p
∈ Syl
p
(G)
such that M
p
<G
p
. Since G is nilpotent, by Theorem 3.4, M
p
<N
G
p
(M
p
).
Let x ∈ N
G
p
(M
p
) \ M
p
. Since Z
p
triangleleftequal M is a p-group by Sylow’s Theorem
49
Z
p
≤ M
p
. Hence, Z
p
≤ M
p
= xM
p
x
−1
≤ xMx
−1
. Now since Z
q
triangleleftequalM we get
xZ
q
x
−1
triangleleftequalxMx
−1
. Since Z(O
p
(M)) is abelian we know Z
p
=Ω
1
(Z(O
p
(M))) is
an elementary abelian p-group. From the action of Z
p
and xZ
q
x
−1
,byTheorem
4.11, we get xZ
q
x
−1
= 〈C
xZ
q
x
−1(U) | U ≤ Z
p
,
|Z
p
|
|U|
= p〉.LetU ≤ Z
p
such that
|Z
p
|
|U|
= p. Since Z is abelian we get Z
p
≤ C
G
(U) <G. By the uniqueness of M
we have C
G
(U) ≤ M. Thus, since MC
xZ
q
x
−1(U) ≤ C
G
(U)wegetxZ
q
x
−1
≤ M.
But then Z
q
≤ x
−1
Mx. Hence, Z = Z
p
Z
q
≤ x
−1
Mx. Again by the unique-
ness of M we get M = x
−1
Mx. Hence, x ∈ N
G
(M). But since M is maximal
and G is simple we have M = N
G
(M). Thus, x ∈ M and so x ∈ M ∩ G
p
.
Now by Sylow’s Theorem there exists m ∈ M such that G
p
∩M ≤ mM
p
m
−1
.
Hence we get M
p
≤ G
p
∩ M ≤ mM
p
m
−1
.andsoM
p
= G
p
∩ M.Thuswe
get x ∈ G
p
∩ M = M
p
, a contradiction. Hence, Z
p
∼
= Z
p
and {1} is the only
subgroup of Z
p
with index p thus Z
p
is cyclic. Similarly, Z
q
∼
= Z
q
is cyclic.
Since Z
p
≤ xMx
−1
and xZ
q
x
−1
triangleleftequalxMx
−1
we know H = Z
p
xZ
q
x
−1
is a sub-
group. Without loss of generality, p>q.Thenn
p
=1andn
q
= 1. Hence,
Z
p
triangleleftequalH and xZ
q
x
−1
triangleleftequalH.Butthen[Z
p
,xZ
q
x
−1
] ≤ Z
p
∩ xZ
q
x
−1
=1. Thus,
xZ
q
x
−1
≤ C
G
(Z
p
) ≤ N
G
(Z
p
)=M and so x ∈ M
p
.Also,Z
p
≤ M
p
≤ M
so xZx
−1
= xZ
p
x
−1
xZ
q
x
−1
≤ M or Z = x
−1
Mx and M = x
−1
Mx.Thus,
x ∈ N
G
(M)=M, a contradiction.
3. Let M beamaximalsubgroup of G then M cannotcontaina p-centralsubgroup
of G and a q-central subgroup of G.
By (2) we may assume F(M)isap-group. By Sylow’s Theorem there exists
M
p
∈ Syl
p
(M) such that F(M) ≤ M
p
and there exists G
p
∈ Syl
p
(G) such that
M
p
≤ G
p
.Thus,F(M) ≤ G
p
.IfC
G
(F(M)) negationslash≤ M then M<〈C
G
(F(M)),M〉≤
50
G. Hence, by the maximality of M we get G = 〈C
G
(F(M)),M〉.Butthen
F(M)triangleleftequal〈C
G
(F(M)),M〉 = G,acontradictionsinceGissimple. Thus,C
G
(F(M)) ≤
M andsoC
G
(F(M)) = C
M
(F(M)). Now,Z(G
p
) ≤ C
G
(F(M)) = C
M
(F(M)) ≤
F(M) by Theorem 3.9 since M is solvable. Hence, Z(G
p
) ≤ M and Z(G
p
)isa
p-central subgroup of G. Suppose H ≤ M such that H is a q-central subgroup
of G. Then there exists G
q
∈ Syl
q
(G) such that H ≤ Z(G
q
). Then G = G
p
G
q
,
Z(G
p
)triangleleftequalG
p
, HtriangleleftequalG
q
,and〈Z(G
p
)
H
〉≤〈F(M)
H
〉≤F(M) since F(M)triangleleftequalM and
F(M)isap-group. Hence 〈Z(G
p
)
H
〉 is a p-group. But then by Theorem 4.12
we get G is not simple, a contradiction.
4. A p-central subgroup of G cannot normalize a q-subgroup of G.
Suppose H is a p-central subgroup of G and Q ≤ G is a q-group such that
H ≤ N
G
(Q). By Sylow’s Theorem, there exists G
q
∈ Syl
q
(G) such that Q ≤
G
q
. Since N
G
(Q) <Gthere exists a maximal subgroup M of G such that
N
G
(Q) ≤ M.NowZ(G
q
) ≤ C
G
(Q) ≤ N
G
(Q) ≤ M.Also,H ≤ N
G
(Q) ≤ M.
But Z(G
q
)isaq-central subgroup of G and H is a p-central subgroup of G,
which contradicts (3). Thus, a p-central subgroup of G cannot normalize a
q-subgroup of G.
5. |G| is odd.
Suppose not. Then, 2 ||G| and so by Sylow’s Theorem there exists G
2
∈
Syl
2
(G). Then by Theorem 3.2, Z(G
2
) negationslash= 1. Hence, by Theorem 1.10, there
exists 1 negationslash= x ∈ Z(G
2
) such that x
2
= 1. Since G is simple we get O
2
(G)=1.
Hence x negationslash∈ O
2
(G). By Theorem 4.14, there exists y ∈ G such that xyx
−1
= y
−1
and |y| is odd. Hence, we get 〈x〉≤N
G
(〈y〉). But 〈x〉 is a 2-central subgroup
of G and 〈y〉 is a q-group, which contradicts (4). Therefore, |G| is odd.
51
6. Let M be a maximal subgroup of G such that F(M)isap-group and M
p
∈
Syl
p
(M) such that F(M) ≤ M
p
.ThenJ(M
p
)triangleleftequalM and M
p
∈ Syl
p
(G).
We want to show that J(M
p
) triangleleftequal M. By Theorem 4.15 it is enough to show
C
M
(Ω
1
(Z(M
p
))) = M
p
. SinceΩ
1
(Z(M
p
)) ≤ Z(M
p
)wegetM
p
≤ C
M
(Ω
1
(Z(M
p
))).
Let G
p
∈ Syl
p
(G) such that M
p
≤ G
p
.ThenF(M) ≤ M
p
≤ G
p
.Thus,
Z(G
p
) ≤ C
G
(F(M))
= C
M
(F(M))
≤ F(M) since M is solvable
≤ M
p
.
Thus, Z(G
p
) ≤ M
p
and so Z(G
p
) ≤ Z(M
p
). Hence, Ω
1
(Z(G
p
)) ≤ Ω
1
(Z(M
p
))
and so C
M
(Ω
1
(Z(M
p
))) ≤ C
G
(Ω
1
(Z(G
p
))). But, by (4) C
G
(Ω
1
(Z(G
p
))) has
no q-subgroups and so C
G
(Ω
1
(Z(G
p
))) is a p-group. Hence, C
M
(Ω
1
(Z(M
p
)))
is a p-group. But M
p
≤ C
M
(Ω
1
(Z(M
p
))) and M
p
∈ Syl
p
(M). Hence, M
p
=
C
M
(Ω
1
(Z(M
p
))) and so Theorem 4.15 J(M
p
)triangleleftequalM.IfM
p
<G
p
then since G
p
is
nilpotent, by Theorem 3.4, M
p
<N
G
p
(M
p
)=H.Now,M
p
triangleleftequalH.IfH ≤ M then
H ≤ G
p
∩M.ButG
p
∩M = M
p
and so we get H ≤ M
p
, a contradiction. Thus,
H negationslash≤ M.ButhenM<〈M,H〉≤G and so G = 〈M,H〉 by the maximality of
M.Now,J(M
p
)triangleleftequalM.alsoM
p
triangleleftequalH we get J(M
p
)triangleleftequalH. Hence, J(M
p
)triangleleftequal〈M,H〉,
a contradiction, since G is simple. Thus, M
p
= G
p
∈ Syl
p
(G).
52
Let
C
p
= {M | M is a maximal subgroup of G and F(M)isap-group}
and
C
q
= {M | M is a maximal subgroup of G and F(M)isaq-group}
Let M
1
,M
2
∈ C
p
and P
1
∈ Syl
p
(M
1
)andP
2
∈ Syl
p
(M
2
). By (6) P
1
,P
2
∈ Syl
p
(G).
By Sylow’s Theorem there exists g ∈ G such that gP
1
g
−1
= P
2
.IfgM
1
g
−1
negationslash= M
2
then
M
2
< 〈gM
1
g
−1
,M
2
〉≤G. Hence we get G = 〈gM
1
g
−1
,M
2
〉 since M
2
is maximal.
By (6) we know J(P
2
)=J(gP
1
g
−1
)triangleleftequal〈gM
1
g
−1
,M
2
〉 = G, a contradiction since G
is simple. Thus, gM
1
g
−1
= M
2
and so G acts transitively on C
p
by conjugation.
Similarly, G acts transitively on C
q
by conjugation. Let M
1
,M
2
∈ C
p
such that
|M
1
∩M
2
|
p
ismaximal. If|M
1
∩M
2
|
p
negationslash=1thenletP ∈ Syl
p
(M
1
∩M
2
). IfP ∈ Syl
p
(M
1
)
then, by (6), we get P ∈ Syl
p
(G). Hence, since P ≤ M
2
we get P ∈ Syl
p
(M
2
). Now,
by (6), we get J(P) triangleleftequal 〈M
1
,M
2
〉 = G, a contradiction since G is simple. Hence,
P negationslash∈ Syl
p
(M
1
) and similarly P negationslash∈ Syl
p
(M
2
). Therefore, P<N
M
1
(P) ≤ N
G
(P)and
P<N
M
2
(P) ≤ N
G
(P). Since N
G
(P) <G, there exists a maximal subgroup R of
G such that N
G
(P) ≤ R.IfF(R)isaq-group let G
p
∈ Syl
p
(G) such that P ≤ G
p
.
Then, Z(G
p
) ≤ C
G
(P) ≤ N
G
(P) ≤ R and F(R)triangleleftequalR. Hence Z(G
p
) ≤ N
G
(F(R)),
but Z(G
p
)isap-central subgroup of G and F(R)isaq-group, a contradiction of (4).
53
Thus, F(R)isap-group and R ∈ C
p
.Now
|M
1
∩R|
p
≥|M
1
∩N
G
(P)|
p
= |N
M
1
(P)|
p
> |P|
= |M
1
∩M
2
|
p
.
Hence, by the maximality of |M
1
∩ M
2
|
p
we get R = M
1
. Also, similarly R = M
2
.
Thus, M
1
= R = M
2
, a contradiction since M
1
and M
2
are distinct. Therefore,
|M
1
∩M
2
|
p
= 1 and similarly |H
1
∩H
2
|
q
=1forallH
1
,H
2
∈ C
q
. Suppose p
a
>q
b
.
Let M
1
,M
2
∈ C
p
be distinct and P
1
∈ Syl
p
(M
1
)andP
2
∈ Syl
p
(M
2
). Then P
1
∩P
2
≤
M
1
∩M
2
is a p-group and |M
1
∩M
2
|
p
= 1. Hence |P
1
∩P
2
| =1.Butthenweget
p
a
q
b
= |G|
≥|P
1
P
2
|
=
|P
1
||P
2
|
|P
1
∩P
2
|
=
p
a
p
a
1
= p
2a
>p
a
q
b
a contradiction. Similarly, we get a contradiction if q
b
>p
a
. Therefore, G is solvable.
54
References
[1] Bender, Helmut: A Group Theoretic Proof of Burnside’s p
a
q
b
Theorem. Math. Z.
126, (1972), 327 - 338.
[2] Burnside, William: On Groups of Order p
a
q
b
. Proc. London Math Soc. (1904), 2,
388 - 391.
[3] Goldschmidt, David: A Group Theoretic Proof of the p
a
q
b
Theorem for Odd
Primes. Math. Z. 113, (1970), 373 - 375.
[4] Matsuyama, Hiroshi: Solvability of Groups of Order 2
a
p
b
. Osaka J. Math. 10,
(1973), 375 - 378.
[5] Papantonopoulou, Aligli: Algebra Pure and Applied. Prentice-Hall, (2002).
55