Sylow theorems - Group Theory, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


THE SYLOW THEOREMS
1. Introduction
The converse of Lagrange's theorem is false: if G is a nite group and djjGj, then there
may not be a subgroup of G with orderd. The simplest example of this is the group A
4
, of
order 12, which has no subgroup of order 6. The Norwegian mathematician Peter Ludwig
Sylow [1] discovered that a converse result is true when d is a prime power: if p is a prime
number and p
k
jjGj then G must contain a subgroup of order p
k
. Sylow also discovered
important relations among the subgroups with order the largest power of p dividingjGj,
such as the fact that all subgroups of that order are conjugate to each other.
For example, a group of order 100 = 2
2

 5
2
must contain subgroups of order 1, 2, 4, 5,
and 25, the subgroups of order 4 are conjugate to each other, and the subgroups of order
25 are conjugate to each other. It is not necessarily the case that the subgroups of order 2
are conjugate or that the subgroups of order 5 are conjugate.
Denition 1.1. LetG be a nite group andp be a prime. Any subgroup ofG whose order
is the highest power ofp dividingjGj is called a p-Sylow subgroup ofG. Ap-Sylow subgroup
for some p is called a Sylow subgroup.
In a group of order 100, a 2-Sylow subgroup has order 4, a 5-Sylow subgroup has order
25, and a p-Sylow subgroup is trivial if p6= 2 or 5.
In a group of order 12, a 2-Sylow subgroup has order 4, a 3-Sylow subgroup has order 3,
and ap-Sylow subgroup is trivial if p> 3. Let's look at a few examples of Sylow subgroups
in groups of order 12.
Example 1.2. In Z=(12), the only 2-Sylow subgroup is f0; 3; 6; 9g =h3i and the only
3-Sylow subgroup isf0; 4; 8g =h4i.
Example 1.3. In A
4
there is one subgroup of order 4, so the only 2-Sylow subgroup is
f(1); (12)(34); (13)(24); (14)(23)g =h(12)(34); (14)(23)i:
There are four 3-Sylow subgroups:
f(1); (123); (132)g =h(123)i; f(1); (124); (142)g =h(124)i;
f(1); (134); (143)g =h(134)i; f(1); (234); (243)g =h(234)i:
Example 1.4. In D
6
there are three 2-Sylow subgroups:
f1; r
3
; s; r
3
sg =hr
3
;si; f1; r
3
; rs; r
4
sg =hr
3
;rsi; f1; r
3
; r
2
s; r
5
sg =hr
3
;r
2
si:
The only 3-Sylow subgroup of D
6
isf1;r
2
;r
4
g =hr
2
i.
In a group of order 24, a 2-Sylow subgroup has order 8 and a 3-Sylow subgroup has order
3. Let's look at two examples.
Example 1.5. In S
4
, the 3-Sylow subgroups are the 3-Sylow subgroups of A
4
(an element
of 3-power order inS
4
must be a 3-cycle, and they all lie inA
4
). We determined the 3-Sylow
subgroups of A
4
in Example 1.3; there are four of them.
Page 2


THE SYLOW THEOREMS
1. Introduction
The converse of Lagrange's theorem is false: if G is a nite group and djjGj, then there
may not be a subgroup of G with orderd. The simplest example of this is the group A
4
, of
order 12, which has no subgroup of order 6. The Norwegian mathematician Peter Ludwig
Sylow [1] discovered that a converse result is true when d is a prime power: if p is a prime
number and p
k
jjGj then G must contain a subgroup of order p
k
. Sylow also discovered
important relations among the subgroups with order the largest power of p dividingjGj,
such as the fact that all subgroups of that order are conjugate to each other.
For example, a group of order 100 = 2
2

 5
2
must contain subgroups of order 1, 2, 4, 5,
and 25, the subgroups of order 4 are conjugate to each other, and the subgroups of order
25 are conjugate to each other. It is not necessarily the case that the subgroups of order 2
are conjugate or that the subgroups of order 5 are conjugate.
Denition 1.1. LetG be a nite group andp be a prime. Any subgroup ofG whose order
is the highest power ofp dividingjGj is called a p-Sylow subgroup ofG. Ap-Sylow subgroup
for some p is called a Sylow subgroup.
In a group of order 100, a 2-Sylow subgroup has order 4, a 5-Sylow subgroup has order
25, and a p-Sylow subgroup is trivial if p6= 2 or 5.
In a group of order 12, a 2-Sylow subgroup has order 4, a 3-Sylow subgroup has order 3,
and ap-Sylow subgroup is trivial if p> 3. Let's look at a few examples of Sylow subgroups
in groups of order 12.
Example 1.2. In Z=(12), the only 2-Sylow subgroup is f0; 3; 6; 9g =h3i and the only
3-Sylow subgroup isf0; 4; 8g =h4i.
Example 1.3. In A
4
there is one subgroup of order 4, so the only 2-Sylow subgroup is
f(1); (12)(34); (13)(24); (14)(23)g =h(12)(34); (14)(23)i:
There are four 3-Sylow subgroups:
f(1); (123); (132)g =h(123)i; f(1); (124); (142)g =h(124)i;
f(1); (134); (143)g =h(134)i; f(1); (234); (243)g =h(234)i:
Example 1.4. In D
6
there are three 2-Sylow subgroups:
f1; r
3
; s; r
3
sg =hr
3
;si; f1; r
3
; rs; r
4
sg =hr
3
;rsi; f1; r
3
; r
2
s; r
5
sg =hr
3
;r
2
si:
The only 3-Sylow subgroup of D
6
isf1;r
2
;r
4
g =hr
2
i.
In a group of order 24, a 2-Sylow subgroup has order 8 and a 3-Sylow subgroup has order
3. Let's look at two examples.
Example 1.5. In S
4
, the 3-Sylow subgroups are the 3-Sylow subgroups of A
4
(an element
of 3-power order inS
4
must be a 3-cycle, and they all lie inA
4
). We determined the 3-Sylow
subgroups of A
4
in Example 1.3; there are four of them.
There are three 2-Sylow subgroups of S
4
, and they are interesting to work out since they
can be understood as copies of D
4
inside S
4
. The number of ways to label the four vertices
of a square as 1, 2, 3, and 4 is 4! = 24, but up to rotations and re
ections of the square
there are really just three dierent ways of carrying out the labeling, as follows.
1
2 3
4 1
2 4
3 1
3 2
4
Any other labeling of the square is a rotated or re
ected version of one of these three squares.
For example, the square below is obtained from the middle square above by re
ecting across
a horizontal line through the middle of the square.
2
1 3
4
WhenD
4
acts on a square with labeled vertices, each motion of D
4
creates a permutation
of the four vertices, and this permutation is an element of S
4
. For example, a 90 degree
rotation of the square is a 4-cycle on the vertices. In this way we obtain a copy of D
4
inside
S
4
. The three essentially dierent labelings of the vertices of the square above embed D
4
into S
4
as three dierent subgroups of order 8:
f1; (1234); (1432); (12)(34); (13)(24); (14)(23); (13); (24)g =h(1234); (13)i;
f1; (1243); (1342); (12)(34); (13)(24); (14)(23); (14); (23)g =h(1243); (14)i;
f1; (1324); (1423); (12)(34); (13)(24); (14)(23); (12); (34)g =h(1324); (12)i:
These are the 2-Sylow subgroups of S
4
.
Example 1.6. The group SL
2
(Z=(3)) has order 24. An explicit tabulation of the elements
of this group reveals that there are only 8 elements in the group with 2-power order:

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

;

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

:
These form the only 2-Sylow subgroup, which is isomorphic to Q
8
by labeling the matrices
in the rst row as 1;i;j;k and the matrices in the second row as1;i;j;k.
There are four 3-Sylow subgroups:h(
1 1
0 1
)i,h(
1 0
1 1
)i,h(
0 1
2 2
)i, andh(
0 2
1 2
)i.
Here are the Sylow theorems. They are often given in three parts. The result we call
Sylow III* is not always stated explicitly as part of the Sylow theorems.
Page 3


THE SYLOW THEOREMS
1. Introduction
The converse of Lagrange's theorem is false: if G is a nite group and djjGj, then there
may not be a subgroup of G with orderd. The simplest example of this is the group A
4
, of
order 12, which has no subgroup of order 6. The Norwegian mathematician Peter Ludwig
Sylow [1] discovered that a converse result is true when d is a prime power: if p is a prime
number and p
k
jjGj then G must contain a subgroup of order p
k
. Sylow also discovered
important relations among the subgroups with order the largest power of p dividingjGj,
such as the fact that all subgroups of that order are conjugate to each other.
For example, a group of order 100 = 2
2

 5
2
must contain subgroups of order 1, 2, 4, 5,
and 25, the subgroups of order 4 are conjugate to each other, and the subgroups of order
25 are conjugate to each other. It is not necessarily the case that the subgroups of order 2
are conjugate or that the subgroups of order 5 are conjugate.
Denition 1.1. LetG be a nite group andp be a prime. Any subgroup ofG whose order
is the highest power ofp dividingjGj is called a p-Sylow subgroup ofG. Ap-Sylow subgroup
for some p is called a Sylow subgroup.
In a group of order 100, a 2-Sylow subgroup has order 4, a 5-Sylow subgroup has order
25, and a p-Sylow subgroup is trivial if p6= 2 or 5.
In a group of order 12, a 2-Sylow subgroup has order 4, a 3-Sylow subgroup has order 3,
and ap-Sylow subgroup is trivial if p> 3. Let's look at a few examples of Sylow subgroups
in groups of order 12.
Example 1.2. In Z=(12), the only 2-Sylow subgroup is f0; 3; 6; 9g =h3i and the only
3-Sylow subgroup isf0; 4; 8g =h4i.
Example 1.3. In A
4
there is one subgroup of order 4, so the only 2-Sylow subgroup is
f(1); (12)(34); (13)(24); (14)(23)g =h(12)(34); (14)(23)i:
There are four 3-Sylow subgroups:
f(1); (123); (132)g =h(123)i; f(1); (124); (142)g =h(124)i;
f(1); (134); (143)g =h(134)i; f(1); (234); (243)g =h(234)i:
Example 1.4. In D
6
there are three 2-Sylow subgroups:
f1; r
3
; s; r
3
sg =hr
3
;si; f1; r
3
; rs; r
4
sg =hr
3
;rsi; f1; r
3
; r
2
s; r
5
sg =hr
3
;r
2
si:
The only 3-Sylow subgroup of D
6
isf1;r
2
;r
4
g =hr
2
i.
In a group of order 24, a 2-Sylow subgroup has order 8 and a 3-Sylow subgroup has order
3. Let's look at two examples.
Example 1.5. In S
4
, the 3-Sylow subgroups are the 3-Sylow subgroups of A
4
(an element
of 3-power order inS
4
must be a 3-cycle, and they all lie inA
4
). We determined the 3-Sylow
subgroups of A
4
in Example 1.3; there are four of them.
There are three 2-Sylow subgroups of S
4
, and they are interesting to work out since they
can be understood as copies of D
4
inside S
4
. The number of ways to label the four vertices
of a square as 1, 2, 3, and 4 is 4! = 24, but up to rotations and re
ections of the square
there are really just three dierent ways of carrying out the labeling, as follows.
1
2 3
4 1
2 4
3 1
3 2
4
Any other labeling of the square is a rotated or re
ected version of one of these three squares.
For example, the square below is obtained from the middle square above by re
ecting across
a horizontal line through the middle of the square.
2
1 3
4
WhenD
4
acts on a square with labeled vertices, each motion of D
4
creates a permutation
of the four vertices, and this permutation is an element of S
4
. For example, a 90 degree
rotation of the square is a 4-cycle on the vertices. In this way we obtain a copy of D
4
inside
S
4
. The three essentially dierent labelings of the vertices of the square above embed D
4
into S
4
as three dierent subgroups of order 8:
f1; (1234); (1432); (12)(34); (13)(24); (14)(23); (13); (24)g =h(1234); (13)i;
f1; (1243); (1342); (12)(34); (13)(24); (14)(23); (14); (23)g =h(1243); (14)i;
f1; (1324); (1423); (12)(34); (13)(24); (14)(23); (12); (34)g =h(1324); (12)i:
These are the 2-Sylow subgroups of S
4
.
Example 1.6. The group SL
2
(Z=(3)) has order 24. An explicit tabulation of the elements
of this group reveals that there are only 8 elements in the group with 2-power order:

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

;

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

:
These form the only 2-Sylow subgroup, which is isomorphic to Q
8
by labeling the matrices
in the rst row as 1;i;j;k and the matrices in the second row as1;i;j;k.
There are four 3-Sylow subgroups:h(
1 1
0 1
)i,h(
1 0
1 1
)i,h(
0 1
2 2
)i, andh(
0 2
1 2
)i.
Here are the Sylow theorems. They are often given in three parts. The result we call
Sylow III* is not always stated explicitly as part of the Sylow theorems.
Theorem 1.7 (Sylow I). A nite group G has a p-Sylow subgroup for every prime p and
any p-subgroup of G lies in a p-Sylow subgroup of G.
Theorem 1.8 (Sylow II). For each prime p, the p-Sylow subgroups of G are conjugate.
Theorem 1.9 (Sylow III). For each prime p, let n
p
be the number of p-Sylow subgroups of
G. WritejGj =p
k
m, where p doesn't divide m. Then
n
p
 1 modp and n
p
jm:
Theorem 1.10 (Sylow III*). For each prime p, let n
p
be the number of p-Sylow subgroups
of G. Then n
p
= [G : N(P )], where P is any p-Sylow subgroup and N(P ) is its normalizer.
Sylow II says for two p-Sylow subgroups H and K of G that there is some g2 G such
that gHg
1
=K. This is illustrated in the table below.
Example Group Size p H K g
1.3 A
4
12 3 h(123)i h(124)i (243)
1.4 D
6
12 2 hr
3
;si hr
3
;rsi r
2
1.5 S
4
24 2 h(1234); (13)i h(1243); (14)i (34)
1.6 SL
2
(Z=(3)) 24 3 h(
1 1
0 1
)i h(
1 0
1 1
)i (
0 1
2 1
)
When trying to conjugate one cyclic subgroup to another cyclic subgroup, be careful: not
all generators of the two groups have to be conjugate. For example, in A
4
the subgroups
h(123)i =f(1); (123); (132)g andh(124)i =f(1); (124); (142)g are conjugate, but the
conjugacy class of (123) in A
4
isf(123); (142); (134); (243)g, so there's no way to conjugate
(123) to (124) by an element of A
4
; we must conjugate (123) to (142). The 3-cycles (123)
and (124) are conjugate in S
4
, but not in A
4
. Similarly, (
1 1
0 1
) and (
1 0
1 1
) are conjugate in
GL
2
(Z=(3)) but not in SL
2
(Z=(3)), so when Sylow II says the subgroupsh(
1 1
0 1
)i andh(
1 0
1 1
)i
are conjugate in SL
2
(Z=(3)) a conjugating matrix must send (
1 1
0 1
) to (
1 0
1 1
)
2
= (
1 0
2 1
).
Let's see what Sylow III tells us about the number of 2-Sylow and 3-Sylow subgroups of a
group of order 12. Forp = 2 andp = 3 in Sylow III, the divisibility conditions aren
2
j 3 and
n
3
j 4 and the congruence conditions are n
2
 1 mod 2 and n
3
 1 mod 3. The divisibility
conditions imply n
2
is 1 or 3 and n
3
is 1, 2, or 4. The congruence n
2
 1 mod 2 tells us
nothing new (1 and 3 are both odd), but the congruence n
3
 1 mod 3 rules out the option
n
3
= 2. Therefore n
2
is 1 or 3 and n
3
is 1 or 4 whenjGj = 12. IfjGj = 24 we again nd n
2
is 1 or 3 whilen
3
is 1 or 4. (For instance, from n
3
j 8 andn
3
 1 mod 3 the only choices are
n
3
= 1 and n
3
= 4.) Therefore as soon as we nd more than one 2-Sylow subgroup there
must be three of them, and as soon as we nd more than one 3-Sylow subgroup there must
be four of them. The table below shows the values of n
2
and n
3
in the examples above.
Group Size n
2
n
3
Z=(12) 12 1 1
A
4
12 1 4
D
6
12 3 1
S
4
24 3 4
SL
2
(Z=(3)) 24 1 4
2. Proof of the Sylow Theorems
Our proof of the Sylow theorems will use group actions. The table below is a summary.
For each theorem the table lists a group, a set it acts on, and the action. We write Syl
p
(G)
for the set of p-Sylow subgroups of G, so n
p
=j Syl
p
(G)j.
Page 4


THE SYLOW THEOREMS
1. Introduction
The converse of Lagrange's theorem is false: if G is a nite group and djjGj, then there
may not be a subgroup of G with orderd. The simplest example of this is the group A
4
, of
order 12, which has no subgroup of order 6. The Norwegian mathematician Peter Ludwig
Sylow [1] discovered that a converse result is true when d is a prime power: if p is a prime
number and p
k
jjGj then G must contain a subgroup of order p
k
. Sylow also discovered
important relations among the subgroups with order the largest power of p dividingjGj,
such as the fact that all subgroups of that order are conjugate to each other.
For example, a group of order 100 = 2
2

 5
2
must contain subgroups of order 1, 2, 4, 5,
and 25, the subgroups of order 4 are conjugate to each other, and the subgroups of order
25 are conjugate to each other. It is not necessarily the case that the subgroups of order 2
are conjugate or that the subgroups of order 5 are conjugate.
Denition 1.1. LetG be a nite group andp be a prime. Any subgroup ofG whose order
is the highest power ofp dividingjGj is called a p-Sylow subgroup ofG. Ap-Sylow subgroup
for some p is called a Sylow subgroup.
In a group of order 100, a 2-Sylow subgroup has order 4, a 5-Sylow subgroup has order
25, and a p-Sylow subgroup is trivial if p6= 2 or 5.
In a group of order 12, a 2-Sylow subgroup has order 4, a 3-Sylow subgroup has order 3,
and ap-Sylow subgroup is trivial if p> 3. Let's look at a few examples of Sylow subgroups
in groups of order 12.
Example 1.2. In Z=(12), the only 2-Sylow subgroup is f0; 3; 6; 9g =h3i and the only
3-Sylow subgroup isf0; 4; 8g =h4i.
Example 1.3. In A
4
there is one subgroup of order 4, so the only 2-Sylow subgroup is
f(1); (12)(34); (13)(24); (14)(23)g =h(12)(34); (14)(23)i:
There are four 3-Sylow subgroups:
f(1); (123); (132)g =h(123)i; f(1); (124); (142)g =h(124)i;
f(1); (134); (143)g =h(134)i; f(1); (234); (243)g =h(234)i:
Example 1.4. In D
6
there are three 2-Sylow subgroups:
f1; r
3
; s; r
3
sg =hr
3
;si; f1; r
3
; rs; r
4
sg =hr
3
;rsi; f1; r
3
; r
2
s; r
5
sg =hr
3
;r
2
si:
The only 3-Sylow subgroup of D
6
isf1;r
2
;r
4
g =hr
2
i.
In a group of order 24, a 2-Sylow subgroup has order 8 and a 3-Sylow subgroup has order
3. Let's look at two examples.
Example 1.5. In S
4
, the 3-Sylow subgroups are the 3-Sylow subgroups of A
4
(an element
of 3-power order inS
4
must be a 3-cycle, and they all lie inA
4
). We determined the 3-Sylow
subgroups of A
4
in Example 1.3; there are four of them.
There are three 2-Sylow subgroups of S
4
, and they are interesting to work out since they
can be understood as copies of D
4
inside S
4
. The number of ways to label the four vertices
of a square as 1, 2, 3, and 4 is 4! = 24, but up to rotations and re
ections of the square
there are really just three dierent ways of carrying out the labeling, as follows.
1
2 3
4 1
2 4
3 1
3 2
4
Any other labeling of the square is a rotated or re
ected version of one of these three squares.
For example, the square below is obtained from the middle square above by re
ecting across
a horizontal line through the middle of the square.
2
1 3
4
WhenD
4
acts on a square with labeled vertices, each motion of D
4
creates a permutation
of the four vertices, and this permutation is an element of S
4
. For example, a 90 degree
rotation of the square is a 4-cycle on the vertices. In this way we obtain a copy of D
4
inside
S
4
. The three essentially dierent labelings of the vertices of the square above embed D
4
into S
4
as three dierent subgroups of order 8:
f1; (1234); (1432); (12)(34); (13)(24); (14)(23); (13); (24)g =h(1234); (13)i;
f1; (1243); (1342); (12)(34); (13)(24); (14)(23); (14); (23)g =h(1243); (14)i;
f1; (1324); (1423); (12)(34); (13)(24); (14)(23); (12); (34)g =h(1324); (12)i:
These are the 2-Sylow subgroups of S
4
.
Example 1.6. The group SL
2
(Z=(3)) has order 24. An explicit tabulation of the elements
of this group reveals that there are only 8 elements in the group with 2-power order:

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

;

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

:
These form the only 2-Sylow subgroup, which is isomorphic to Q
8
by labeling the matrices
in the rst row as 1;i;j;k and the matrices in the second row as1;i;j;k.
There are four 3-Sylow subgroups:h(
1 1
0 1
)i,h(
1 0
1 1
)i,h(
0 1
2 2
)i, andh(
0 2
1 2
)i.
Here are the Sylow theorems. They are often given in three parts. The result we call
Sylow III* is not always stated explicitly as part of the Sylow theorems.
Theorem 1.7 (Sylow I). A nite group G has a p-Sylow subgroup for every prime p and
any p-subgroup of G lies in a p-Sylow subgroup of G.
Theorem 1.8 (Sylow II). For each prime p, the p-Sylow subgroups of G are conjugate.
Theorem 1.9 (Sylow III). For each prime p, let n
p
be the number of p-Sylow subgroups of
G. WritejGj =p
k
m, where p doesn't divide m. Then
n
p
 1 modp and n
p
jm:
Theorem 1.10 (Sylow III*). For each prime p, let n
p
be the number of p-Sylow subgroups
of G. Then n
p
= [G : N(P )], where P is any p-Sylow subgroup and N(P ) is its normalizer.
Sylow II says for two p-Sylow subgroups H and K of G that there is some g2 G such
that gHg
1
=K. This is illustrated in the table below.
Example Group Size p H K g
1.3 A
4
12 3 h(123)i h(124)i (243)
1.4 D
6
12 2 hr
3
;si hr
3
;rsi r
2
1.5 S
4
24 2 h(1234); (13)i h(1243); (14)i (34)
1.6 SL
2
(Z=(3)) 24 3 h(
1 1
0 1
)i h(
1 0
1 1
)i (
0 1
2 1
)
When trying to conjugate one cyclic subgroup to another cyclic subgroup, be careful: not
all generators of the two groups have to be conjugate. For example, in A
4
the subgroups
h(123)i =f(1); (123); (132)g andh(124)i =f(1); (124); (142)g are conjugate, but the
conjugacy class of (123) in A
4
isf(123); (142); (134); (243)g, so there's no way to conjugate
(123) to (124) by an element of A
4
; we must conjugate (123) to (142). The 3-cycles (123)
and (124) are conjugate in S
4
, but not in A
4
. Similarly, (
1 1
0 1
) and (
1 0
1 1
) are conjugate in
GL
2
(Z=(3)) but not in SL
2
(Z=(3)), so when Sylow II says the subgroupsh(
1 1
0 1
)i andh(
1 0
1 1
)i
are conjugate in SL
2
(Z=(3)) a conjugating matrix must send (
1 1
0 1
) to (
1 0
1 1
)
2
= (
1 0
2 1
).
Let's see what Sylow III tells us about the number of 2-Sylow and 3-Sylow subgroups of a
group of order 12. Forp = 2 andp = 3 in Sylow III, the divisibility conditions aren
2
j 3 and
n
3
j 4 and the congruence conditions are n
2
 1 mod 2 and n
3
 1 mod 3. The divisibility
conditions imply n
2
is 1 or 3 and n
3
is 1, 2, or 4. The congruence n
2
 1 mod 2 tells us
nothing new (1 and 3 are both odd), but the congruence n
3
 1 mod 3 rules out the option
n
3
= 2. Therefore n
2
is 1 or 3 and n
3
is 1 or 4 whenjGj = 12. IfjGj = 24 we again nd n
2
is 1 or 3 whilen
3
is 1 or 4. (For instance, from n
3
j 8 andn
3
 1 mod 3 the only choices are
n
3
= 1 and n
3
= 4.) Therefore as soon as we nd more than one 2-Sylow subgroup there
must be three of them, and as soon as we nd more than one 3-Sylow subgroup there must
be four of them. The table below shows the values of n
2
and n
3
in the examples above.
Group Size n
2
n
3
Z=(12) 12 1 1
A
4
12 1 4
D
6
12 3 1
S
4
24 3 4
SL
2
(Z=(3)) 24 1 4
2. Proof of the Sylow Theorems
Our proof of the Sylow theorems will use group actions. The table below is a summary.
For each theorem the table lists a group, a set it acts on, and the action. We write Syl
p
(G)
for the set of p-Sylow subgroups of G, so n
p
=j Syl
p
(G)j.
Theorem Group Set Action
Sylow I p-subgroup H G=H left mult.
Sylow II p-Sylow subgroup Q G=P left mult.
Sylow III (n
p
 1 modp) P2 Syl
p
(G) Syl
p
(G) conjugation
Sylow III (n
p
jm) G Syl
p
(G) conjugation
Sylow III

G Syl
p
(G) conjugation
The two conclusions of Sylow III are listed separately in the table since they are proved
using dierent group actions.
Our proofs will usually involve the action of a p-group on a set and use the xed-point
congruence for such actions:jXjj Fix

(X)j modp, whereX is a nite set being acted on
by a nite p-group .
Proof of Sylow I: Letp
k
be the highest power ofp injGj. The result is obvious ifk = 0,
since the trivial subgroup is a p-Sylow subgroup, so we can take k 1, hence pjjGj.
Our strategy for proving Sylow I is to prove a stronger result: there is a subgroup
of order p
i
for 0 i k. More specically, ifjHj = p
i
and i < k, we will show there is
a p-subgroup H
0
 H with [H
0
: H] = p, sojH
0
j = p
i+1
. Then, starting with H as the
trivial subgroup, we can repeat this process with H
0
in place of H to create a rising tower
of subgroups
feg =H
0
H
1
H
2




wherejH
i
j =p
i
, and after k steps we reach H
k
, which is a p-Sylow subgroup of G.
Consider the left multiplication action of H on the left cosets G=H (this need not be
a group). This is an action of a nite p-group H on the set G=H, so by the xed-point
congruence for actions of nontrivial p-groups,
(2.1) jG=Hjj Fix
H
(G=H)j modp:
Let's unravel what it means for a coset gH in G=H to be a xed point by the group H
under left multiplication:
hgH =gH for all h2H () hg2gH for all h2H
() g
1
hg2H for all h2H
() g
1
HgH
() g
1
Hg =H becausejg
1
Hgj =jHj
() g2 N(H):
Thus Fix
H
(G=H) =fgH :g2 N(H)g = N(H)=H, so (2.1) becomes
(2.2) [G :H] [N(H) :H] modp:
Because HC N(H), N(H)=H is a group.
When jHj = p
i
and i < k, the index [G : H] is divisible by p, so the congruence
(2.2) implies [N(H) : H] is divisible by p, so N(H)=H is a group with order divisible by
p. Thus N(H)=H has a subgroup of order p by Cauchy's theorem. All subgroups of the
quotient group N(H)=H have the form H
0
=H, where H
0
is a subgroup between H and
N(H). Therefore a subgroup of order p in N(H)=H is H
0
=H such that [H
0
: H] = p, so
jH
0
j =pjHj =p
i+1
.
Proof of Sylow II: Pick two p-Sylow subgroups P and Q. We want to show they are
conjugate.
Page 5


THE SYLOW THEOREMS
1. Introduction
The converse of Lagrange's theorem is false: if G is a nite group and djjGj, then there
may not be a subgroup of G with orderd. The simplest example of this is the group A
4
, of
order 12, which has no subgroup of order 6. The Norwegian mathematician Peter Ludwig
Sylow [1] discovered that a converse result is true when d is a prime power: if p is a prime
number and p
k
jjGj then G must contain a subgroup of order p
k
. Sylow also discovered
important relations among the subgroups with order the largest power of p dividingjGj,
such as the fact that all subgroups of that order are conjugate to each other.
For example, a group of order 100 = 2
2

 5
2
must contain subgroups of order 1, 2, 4, 5,
and 25, the subgroups of order 4 are conjugate to each other, and the subgroups of order
25 are conjugate to each other. It is not necessarily the case that the subgroups of order 2
are conjugate or that the subgroups of order 5 are conjugate.
Denition 1.1. LetG be a nite group andp be a prime. Any subgroup ofG whose order
is the highest power ofp dividingjGj is called a p-Sylow subgroup ofG. Ap-Sylow subgroup
for some p is called a Sylow subgroup.
In a group of order 100, a 2-Sylow subgroup has order 4, a 5-Sylow subgroup has order
25, and a p-Sylow subgroup is trivial if p6= 2 or 5.
In a group of order 12, a 2-Sylow subgroup has order 4, a 3-Sylow subgroup has order 3,
and ap-Sylow subgroup is trivial if p> 3. Let's look at a few examples of Sylow subgroups
in groups of order 12.
Example 1.2. In Z=(12), the only 2-Sylow subgroup is f0; 3; 6; 9g =h3i and the only
3-Sylow subgroup isf0; 4; 8g =h4i.
Example 1.3. In A
4
there is one subgroup of order 4, so the only 2-Sylow subgroup is
f(1); (12)(34); (13)(24); (14)(23)g =h(12)(34); (14)(23)i:
There are four 3-Sylow subgroups:
f(1); (123); (132)g =h(123)i; f(1); (124); (142)g =h(124)i;
f(1); (134); (143)g =h(134)i; f(1); (234); (243)g =h(234)i:
Example 1.4. In D
6
there are three 2-Sylow subgroups:
f1; r
3
; s; r
3
sg =hr
3
;si; f1; r
3
; rs; r
4
sg =hr
3
;rsi; f1; r
3
; r
2
s; r
5
sg =hr
3
;r
2
si:
The only 3-Sylow subgroup of D
6
isf1;r
2
;r
4
g =hr
2
i.
In a group of order 24, a 2-Sylow subgroup has order 8 and a 3-Sylow subgroup has order
3. Let's look at two examples.
Example 1.5. In S
4
, the 3-Sylow subgroups are the 3-Sylow subgroups of A
4
(an element
of 3-power order inS
4
must be a 3-cycle, and they all lie inA
4
). We determined the 3-Sylow
subgroups of A
4
in Example 1.3; there are four of them.
There are three 2-Sylow subgroups of S
4
, and they are interesting to work out since they
can be understood as copies of D
4
inside S
4
. The number of ways to label the four vertices
of a square as 1, 2, 3, and 4 is 4! = 24, but up to rotations and re
ections of the square
there are really just three dierent ways of carrying out the labeling, as follows.
1
2 3
4 1
2 4
3 1
3 2
4
Any other labeling of the square is a rotated or re
ected version of one of these three squares.
For example, the square below is obtained from the middle square above by re
ecting across
a horizontal line through the middle of the square.
2
1 3
4
WhenD
4
acts on a square with labeled vertices, each motion of D
4
creates a permutation
of the four vertices, and this permutation is an element of S
4
. For example, a 90 degree
rotation of the square is a 4-cycle on the vertices. In this way we obtain a copy of D
4
inside
S
4
. The three essentially dierent labelings of the vertices of the square above embed D
4
into S
4
as three dierent subgroups of order 8:
f1; (1234); (1432); (12)(34); (13)(24); (14)(23); (13); (24)g =h(1234); (13)i;
f1; (1243); (1342); (12)(34); (13)(24); (14)(23); (14); (23)g =h(1243); (14)i;
f1; (1324); (1423); (12)(34); (13)(24); (14)(23); (12); (34)g =h(1324); (12)i:
These are the 2-Sylow subgroups of S
4
.
Example 1.6. The group SL
2
(Z=(3)) has order 24. An explicit tabulation of the elements
of this group reveals that there are only 8 elements in the group with 2-power order:

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

;

1 0
0 1

;

0 1
1 0

;

1 1
1 1

;

1 1
1 1

:
These form the only 2-Sylow subgroup, which is isomorphic to Q
8
by labeling the matrices
in the rst row as 1;i;j;k and the matrices in the second row as1;i;j;k.
There are four 3-Sylow subgroups:h(
1 1
0 1
)i,h(
1 0
1 1
)i,h(
0 1
2 2
)i, andh(
0 2
1 2
)i.
Here are the Sylow theorems. They are often given in three parts. The result we call
Sylow III* is not always stated explicitly as part of the Sylow theorems.
Theorem 1.7 (Sylow I). A nite group G has a p-Sylow subgroup for every prime p and
any p-subgroup of G lies in a p-Sylow subgroup of G.
Theorem 1.8 (Sylow II). For each prime p, the p-Sylow subgroups of G are conjugate.
Theorem 1.9 (Sylow III). For each prime p, let n
p
be the number of p-Sylow subgroups of
G. WritejGj =p
k
m, where p doesn't divide m. Then
n
p
 1 modp and n
p
jm:
Theorem 1.10 (Sylow III*). For each prime p, let n
p
be the number of p-Sylow subgroups
of G. Then n
p
= [G : N(P )], where P is any p-Sylow subgroup and N(P ) is its normalizer.
Sylow II says for two p-Sylow subgroups H and K of G that there is some g2 G such
that gHg
1
=K. This is illustrated in the table below.
Example Group Size p H K g
1.3 A
4
12 3 h(123)i h(124)i (243)
1.4 D
6
12 2 hr
3
;si hr
3
;rsi r
2
1.5 S
4
24 2 h(1234); (13)i h(1243); (14)i (34)
1.6 SL
2
(Z=(3)) 24 3 h(
1 1
0 1
)i h(
1 0
1 1
)i (
0 1
2 1
)
When trying to conjugate one cyclic subgroup to another cyclic subgroup, be careful: not
all generators of the two groups have to be conjugate. For example, in A
4
the subgroups
h(123)i =f(1); (123); (132)g andh(124)i =f(1); (124); (142)g are conjugate, but the
conjugacy class of (123) in A
4
isf(123); (142); (134); (243)g, so there's no way to conjugate
(123) to (124) by an element of A
4
; we must conjugate (123) to (142). The 3-cycles (123)
and (124) are conjugate in S
4
, but not in A
4
. Similarly, (
1 1
0 1
) and (
1 0
1 1
) are conjugate in
GL
2
(Z=(3)) but not in SL
2
(Z=(3)), so when Sylow II says the subgroupsh(
1 1
0 1
)i andh(
1 0
1 1
)i
are conjugate in SL
2
(Z=(3)) a conjugating matrix must send (
1 1
0 1
) to (
1 0
1 1
)
2
= (
1 0
2 1
).
Let's see what Sylow III tells us about the number of 2-Sylow and 3-Sylow subgroups of a
group of order 12. Forp = 2 andp = 3 in Sylow III, the divisibility conditions aren
2
j 3 and
n
3
j 4 and the congruence conditions are n
2
 1 mod 2 and n
3
 1 mod 3. The divisibility
conditions imply n
2
is 1 or 3 and n
3
is 1, 2, or 4. The congruence n
2
 1 mod 2 tells us
nothing new (1 and 3 are both odd), but the congruence n
3
 1 mod 3 rules out the option
n
3
= 2. Therefore n
2
is 1 or 3 and n
3
is 1 or 4 whenjGj = 12. IfjGj = 24 we again nd n
2
is 1 or 3 whilen
3
is 1 or 4. (For instance, from n
3
j 8 andn
3
 1 mod 3 the only choices are
n
3
= 1 and n
3
= 4.) Therefore as soon as we nd more than one 2-Sylow subgroup there
must be three of them, and as soon as we nd more than one 3-Sylow subgroup there must
be four of them. The table below shows the values of n
2
and n
3
in the examples above.
Group Size n
2
n
3
Z=(12) 12 1 1
A
4
12 1 4
D
6
12 3 1
S
4
24 3 4
SL
2
(Z=(3)) 24 1 4
2. Proof of the Sylow Theorems
Our proof of the Sylow theorems will use group actions. The table below is a summary.
For each theorem the table lists a group, a set it acts on, and the action. We write Syl
p
(G)
for the set of p-Sylow subgroups of G, so n
p
=j Syl
p
(G)j.
Theorem Group Set Action
Sylow I p-subgroup H G=H left mult.
Sylow II p-Sylow subgroup Q G=P left mult.
Sylow III (n
p
 1 modp) P2 Syl
p
(G) Syl
p
(G) conjugation
Sylow III (n
p
jm) G Syl
p
(G) conjugation
Sylow III

G Syl
p
(G) conjugation
The two conclusions of Sylow III are listed separately in the table since they are proved
using dierent group actions.
Our proofs will usually involve the action of a p-group on a set and use the xed-point
congruence for such actions:jXjj Fix

(X)j modp, whereX is a nite set being acted on
by a nite p-group .
Proof of Sylow I: Letp
k
be the highest power ofp injGj. The result is obvious ifk = 0,
since the trivial subgroup is a p-Sylow subgroup, so we can take k 1, hence pjjGj.
Our strategy for proving Sylow I is to prove a stronger result: there is a subgroup
of order p
i
for 0 i k. More specically, ifjHj = p
i
and i < k, we will show there is
a p-subgroup H
0
 H with [H
0
: H] = p, sojH
0
j = p
i+1
. Then, starting with H as the
trivial subgroup, we can repeat this process with H
0
in place of H to create a rising tower
of subgroups
feg =H
0
H
1
H
2




wherejH
i
j =p
i
, and after k steps we reach H
k
, which is a p-Sylow subgroup of G.
Consider the left multiplication action of H on the left cosets G=H (this need not be
a group). This is an action of a nite p-group H on the set G=H, so by the xed-point
congruence for actions of nontrivial p-groups,
(2.1) jG=Hjj Fix
H
(G=H)j modp:
Let's unravel what it means for a coset gH in G=H to be a xed point by the group H
under left multiplication:
hgH =gH for all h2H () hg2gH for all h2H
() g
1
hg2H for all h2H
() g
1
HgH
() g
1
Hg =H becausejg
1
Hgj =jHj
() g2 N(H):
Thus Fix
H
(G=H) =fgH :g2 N(H)g = N(H)=H, so (2.1) becomes
(2.2) [G :H] [N(H) :H] modp:
Because HC N(H), N(H)=H is a group.
When jHj = p
i
and i < k, the index [G : H] is divisible by p, so the congruence
(2.2) implies [N(H) : H] is divisible by p, so N(H)=H is a group with order divisible by
p. Thus N(H)=H has a subgroup of order p by Cauchy's theorem. All subgroups of the
quotient group N(H)=H have the form H
0
=H, where H
0
is a subgroup between H and
N(H). Therefore a subgroup of order p in N(H)=H is H
0
=H such that [H
0
: H] = p, so
jH
0
j =pjHj =p
i+1
.
Proof of Sylow II: Pick two p-Sylow subgroups P and Q. We want to show they are
conjugate.
Consider the action of Q on G=P by left multiplication. Since Q is a nite p-group,
jG=Pjj Fix
Q
(G=P )j modp:
The left side is [G : P ], which is nonzero modulo p since P is a p-Sylow subgroup. Thus
j Fix
Q
(G=P )j can't be 0, so there is a xed point in G=P . Call it gP . That is, qgP = gP
for all q2 Q. Equivalently, qg2 gP for all q2 Q, so Q gPg
1
. Therefore Q = gPg
1
,
since Q and gPg
1
have the same size.
Proof of Sylow III: We will prove n
p
 1 modp and then n
p
jm.
To show n
p
 1 modp, consider the action of P on the set Syl
p
(G) by conjugation. The
size of Syl
p
(G) is n
p
. Since P is a nite p-group,
n
p
jfxed pointsgj modp:
Fixed points forP acting by conjugation on Syl
p
(G) areQ2 Syl
p
(G) such thatgQg
1
=Q
for all g2 P . One choice for Q is P . For any such Q, P N(Q). Also Q N(Q), so P
andQ arep-Sylow subgroups in N(Q). Applying Sylow II to the group N(Q),P andQ are
conjugate in N(Q). Since QC N(Q), the only subgroup of N(Q) conjugate to Q is Q, so
P =Q. Thus P is the only xed point when P acts on Syl
p
(G), so n
p
 1 modp.
To show n
p
jm, consider the action of G by conjugation on Syl
p
(G). Since the p-Sylow
subgroups are conjugate to each other (Sylow II), there is one orbit. A set on which a group
acts with one orbit has size dividing the size of the group, so n
p
jjGj. From n
p
 1 modp,
the number n
p
is relatively prime to p, so n
p
jm.
Proof of Sylow III

: Let P be a p-Sylow subgroup of G and let G act on Syl
p
(G) by
conjugation. By the orbit-stabilizer formula,
n
p
=j Syl
p
(G)j = [G : Stab
fPg
]:
The stabilizer Stab
fPg
is
Stab
fPg
=fg :gPg
1
=Pg = N(P ):
Thus n
p
= [G : N(P )].