1 Introduction
In 1872 while studying Galois Theory, Ludwig Sylow wrote a paper that contained some theorems that are now known as Sylow’s Theorems. The content of these theorems focuses on how many subgroups of fixed order a given finite group holds [18, p. 280]. Why are these theorems important? For starters, given the proof of Sylow’s we can establish Cauchy’s theorem via induction. On the reverse, Cauchy’s Theorem is often used as a method to prove Sylow’s Theorems [18, p. 281]. In addition to this, these theorems play a large role in finite group theory as they aid greatly in classifying finite, simple groups.
For the purposes of this paper, let p be a prime number, G be a finite group, e be the identity of G, Z(G) be the center of G, and |G| be the order of group G.
2 Definitions
2.1 Sylow p-Group
A group is a Sylow p-group if the order of the group is a power of p.
2.2 Sylow p-Subgroup
If pk is the highest power of a prime p dividing the order of finite group G, then a subgroup of G of order pk is called a Sylow p-subgroup of G.
Note
Sometimes this is referred to as a p-Sylow subgroup of G. Example
Let us look at D6. The elements of D6 are {e, r, r2, r3, r4, r5, s, rs, r2s, r3s, r4s, r5s}. The only 3-Sylow subgroup is the set {e, r2, r4} but there are three 2-Sylow sub- groups: {e, r3, s, r3s}, {e, r3, rs, r4s}, {e, r3, r2s, r5s}.
2.3 np (G)
Let the number of Sylow p-subgroups of G be denoted np(G) = |Sylp(G)|.
3 Lagrange’s Theorem
The order of any subgroup of a group must divide the order of the group.
In 1872 while studying Galois Theory, Ludwig Sylow wrote a paper that contained some theorems that are now known as Sylow’s Theorems. The content of these theorems focuses on how many subgroups of fixed order a given finite group holds [18, p. 280]. Why are these theorems important? For starters, given the proof of Sylow’s we can establish Cauchy’s theorem via induction. On the reverse, Cauchy’s Theorem is often used as a method to prove Sylow’s Theorems [18, p. 281]. In addition to this, these theorems play a large role in finite group theory as they aid greatly in classifying finite, simple groups.
For the purposes of this paper, let p be a prime number, G be a finite group, e be the identity of G, Z(G) be the center of G, and |G| be the order of group G.
2 Definitions
2.1 Sylow p-Group
A group is a Sylow p-group if the order of the group is a power of p.
2.2 Sylow p-Subgroup
If pk is the highest power of a prime p dividing the order of finite group G, then a subgroup of G of order pk is called a Sylow p-subgroup of G.
Note
Sometimes this is referred to as a p-Sylow subgroup of G. Example
Let us look at D6. The elements of D6 are {e, r, r2, r3, r4, r5, s, rs, r2s, r3s, r4s, r5s}. The only 3-Sylow subgroup is the set {e, r2, r4} but there are three 2-Sylow sub- groups: {e, r3, s, r3s}, {e, r3, rs, r4s}, {e, r3, r2s, r5s}.
2.3 np (G)
Let the number of Sylow p-subgroups of G be denoted np(G) = |Sylp(G)|.
3 Lagrange’s Theorem
The order of any subgroup of a group must divide the order of the group.
3.1 Proof
Let H be a subgroup of G. We want to show that |H| divides |G|. We know that the right cosets of H form a partition in G so every element of G belongs to at least one right coset of H in G. We also know that there is no element that belongs to two distinct right cosets of H in G. Thus, every element of G belongs to exactly one right coset of H. We know that every right coset of H contains |H| elements so |G| = u|H| with u being the number of right cosets of H in G. Therefore |H|||G|.
QED
Note
The converse is not true. For example, A4 = {1, 2, 3, 4} has order 12 but there does not exist a subgroup of order 6. Sylow’s Theorems provides more information as to what the properties of subgroups of a certain group are which is why Sylow’s Theorems are sometimes considered a partial converse to Lagrange [8, p. 1].
4 Cauchy’s Theorem
Let p divide |G|. Then G has an element of order p.
4.1 Proof
Let S = {a1,a2,a3,...,ap} such that every ai is in G and a1a2a3 ...ap = e for |G| = n. We know S has np−1 elements. Let us define a relation on S where two sets with p elements are equivalent if one set is a cyclic permutation of the other set.
Case 1
If all components of a set with p elements are equal then its equivalence class has one member.
Case 2
If two components of a set with p elements are distinct then the equivalence class has p members.
Let the number of solutions to xp = e be called r. Then r is the number of equivalence classes with just one member. Let s denote the number of equiv- alence class with p members so np−1 = r + sp thus p|r.
QED
Note
The proof for Cauchy’s Theorem and Sylow’s 1st Theorem are very similar in formatting. Oftentimes, Cauchy’s Theorem is considered to be a special case of Sylow’s 1st Theorem [2, p. 1]. While originally Cauchy was used for a finite, abelian group, it has since been shown that Cauchy holds for any finite group regardless of whether or not it is abelian [16, p. 3].
Let H be a subgroup of G. We want to show that |H| divides |G|. We know that the right cosets of H form a partition in G so every element of G belongs to at least one right coset of H in G. We also know that there is no element that belongs to two distinct right cosets of H in G. Thus, every element of G belongs to exactly one right coset of H. We know that every right coset of H contains |H| elements so |G| = u|H| with u being the number of right cosets of H in G. Therefore |H|||G|.
QED
Note
The converse is not true. For example, A4 = {1, 2, 3, 4} has order 12 but there does not exist a subgroup of order 6. Sylow’s Theorems provides more information as to what the properties of subgroups of a certain group are which is why Sylow’s Theorems are sometimes considered a partial converse to Lagrange [8, p. 1].
4 Cauchy’s Theorem
Let p divide |G|. Then G has an element of order p.
4.1 Proof
Let S = {a1,a2,a3,...,ap} such that every ai is in G and a1a2a3 ...ap = e for |G| = n. We know S has np−1 elements. Let us define a relation on S where two sets with p elements are equivalent if one set is a cyclic permutation of the other set.
Case 1
If all components of a set with p elements are equal then its equivalence class has one member.
Case 2
If two components of a set with p elements are distinct then the equivalence class has p members.
Let the number of solutions to xp = e be called r. Then r is the number of equivalence classes with just one member. Let s denote the number of equiv- alence class with p members so np−1 = r + sp thus p|r.
QED
Note
The proof for Cauchy’s Theorem and Sylow’s 1st Theorem are very similar in formatting. Oftentimes, Cauchy’s Theorem is considered to be a special case of Sylow’s 1st Theorem [2, p. 1]. While originally Cauchy was used for a finite, abelian group, it has since been shown that Cauchy holds for any finite group regardless of whether or not it is abelian [16, p. 3].
5 Sylow’s 1st Theorem
If pk divides |G| then G has a Sylow p-subgroup.
5.1 Proof
Let pk divide |G|. We want to show G has a subgroup of order pk.
Case 1:
Suppose p divides the order of Z(G). By Cauchy’s Theorem for abelian groups, Z(G) must have an element of order p, denote it m. By induction, the quotient group G/ < m > for abelian G has a subgroup R of order pk−1 so the preimage of R in Z(G) is a subgroup of order pk.
Case 2:
Suppose p does not divide the order of Z(G). Let C1, C2, . . . , Cn be the conju- gacy classes in G with more than one element. Then |G| = |Z(G)| + |C1| + · · · + |Cn|.
Since p by definition does not divide the order of the center, there exists at least one conjugacy class whose order is not divisible by p. Let us denote this conjugacy class as cong(c). We know that
|cong(c)| = [G : CG(c)] = |G|/|CG(c)|.
Note that pk must divide the order of subgroup CG(c) so by induction, G has a Sylow p-subgroup.
QED
6 Sylow’s 2nd Theorem
In a finite group with p dividing |G|, all the Sylow p-subgroups are conjugate for some fixed p.
6.1 Proof
View [1, 2, 9, 12] to examine how this proof can be done using orbits.
QED
7 Sylow’s 3rd Theorem
The number of Sylow p-subgroups for a fixed p is congruent to 1 ≡ mod p and divides |G|.
7.1 Proof
Let P be a Sylow p-subgroup that acts on the set S = {P = P1, P2, ..., Pr} via conjugation. We know from Sylow’s 2nd Theorem that the only P conjugate of P is itself and the the other conjugacy classes have orders of some power of p. This will be shown in more detail in the note below. Let |S| be the sum of positive powers of p so |S| ≡ 1 mod p. Suppose G acts on |S| via conjugation. Since all p-Sylow subgroups are conjugate, there exists only one orbit so for P in S, |S| = |P| = [G : N(P)]. However, [G : N(P)] must divide |G| so np(G) must divide |G|.
QED
7.2 Note
There are a multiple of ways to solve Sylow’s second and third theorems includ- ing separating the statements into parts and solving them in varying orders. To show an example of this, let us look at part of theorem three:
The number of Sylow p-groups for a fixed p is congruent to 1 ≡ mod p.
Proof
We want to show that np(G) ≡ 1 mod p. Let H be a Sylow p-subgroup that acts on the set S = {H = H1,H2,...,Hr} via conjugation where all Hi are Sylow p-subgroups. We know G acts on S by conjugation. Since H < G, H also acts on S by conjugation.
Consider the orbits of H as shown in [1, 2, 9]. The order of an orbit is the index (P : Ps) where (H : Hs)||H|. Since the order of H is a power of p, the order of each orbit is also a power of p. Then the order of an orbit is divisible by p except in the case p0 =1 where hHih−1 =Hi for some i and all h∈H so h ∈ Hi. Since Hi is a Sylow p-subgroup, H = Hi. This means that if an orbit consists of a single element, it must be {H}. Conversely, for h ∈ H, hHh−1 = H. Then the orders of the orbits of S under conjugation by H are all divisible by p except for {H}. It follows that |S| is congruent to 1 mod p where |S| is the number of conjugates of H.
QED
8 Example
Let G be a group of order pq with p<q where p and q are both prime. Then G has exactly one subgroup of order q, which is a normal subgroup of G.
8.1 Proof
We know nq has to divide pq/q=p so nq =1 or p. We also know nq ≡1mod
q. Since p>1 and p<q then p<q+1 so nq =1. Thus the Sylow q-subgroup
with order q is the only one, making it a normal subgroup of G.
QED
9 Corollary
Let P be a Sylow p-subgroup of G. The following are equivalent.
P is the unique Sylow p-subgroup of G so np(G) = 1.
P is normal in G.
All subgroups generated by elements of the p-power order are p-groups. In other words, if X is a subset of G such that |x| is a power of p for all x in X then < X > is a p-group.
10 Theorem
The groups A5 and S5 each have ten subgroups of size 3 and six subgroups of size 5.
10.1 Proof
Any element with odd order in a symmetric group is an even permutation thus subgroups 3-Sylow and 5-Sylow are in A5. We know |A5| = 60 = 22 ∗3∗5 so the 3-Sylow subgroups have size 3 and the 5-Sylow subgroups have size 5. Let us denote these numbers n3 and n5 respectively. By Sylow’s 3rd Theorem, n3|20 and n3 ≡ 1 mod 3 thus n3 = 1,4, or 10. There are twenty 3-cycles in A5 so there are ten subgroups of size 3 since the 3-cyles come in inverse pairs, making n3 =10. For the 5-Sylow subgroup, n5|12 and n5 ≡1mod5 thus n5 =1, or 6. There are at least two subgroups of size 5 in A5 so n5 >1 and n5 = 6.
QED
11 Normal Sylow Subgroups Theorem
The condition np = 1 means that a p-Sylow subgroup is a normal subgroup.
11.1 Proof
We know by Sylow’s 2nd Theorem that all p-Sylow subgroups are conjugate. Then np = 1 when a p-Sylow subgroup of G is self conjugate thus a normal subgroup of G.
QED
12 Theorem
QED
9 Corollary
Let P be a Sylow p-subgroup of G. The following are equivalent.
P is the unique Sylow p-subgroup of G so np(G) = 1.
P is normal in G.
All subgroups generated by elements of the p-power order are p-groups. In other words, if X is a subset of G such that |x| is a power of p for all x in X then < X > is a p-group.
10 Theorem
The groups A5 and S5 each have ten subgroups of size 3 and six subgroups of size 5.
10.1 Proof
Any element with odd order in a symmetric group is an even permutation thus subgroups 3-Sylow and 5-Sylow are in A5. We know |A5| = 60 = 22 ∗3∗5 so the 3-Sylow subgroups have size 3 and the 5-Sylow subgroups have size 5. Let us denote these numbers n3 and n5 respectively. By Sylow’s 3rd Theorem, n3|20 and n3 ≡ 1 mod 3 thus n3 = 1,4, or 10. There are twenty 3-cycles in A5 so there are ten subgroups of size 3 since the 3-cyles come in inverse pairs, making n3 =10. For the 5-Sylow subgroup, n5|12 and n5 ≡1mod5 thus n5 =1, or 6. There are at least two subgroups of size 5 in A5 so n5 >1 and n5 = 6.
QED
11 Normal Sylow Subgroups Theorem
The condition np = 1 means that a p-Sylow subgroup is a normal subgroup.
11.1 Proof
We know by Sylow’s 2nd Theorem that all p-Sylow subgroups are conjugate. Then np = 1 when a p-Sylow subgroup of G is self conjugate thus a normal subgroup of G.
QED
12 Theorem
If p and q are different prime factors of |G|, np = 1, nq = 1 then the elements
of the p-Sylow subgroup commute with the elements of the q-Sylow subgroup.
12.1 Proof
Let P be the p-Sylow subgroup, Q be the q-Sylow subgroup. Since P and Q have relatively prime sizes, P∩Q = {e} by Lagrange’s Theorem. We know that the subgroups P and Q are normal in G so for a in P and b in Q, aba−1 = (aba−1)b−1 = a(ba−1b−1) in P∩Q = {e} so ab = ba.
QED
13 Example
Prove that if |G| = 30 then the group is not simple.
13.1 Proof
Assume |G| = 30. We know 30 = 2∗3∗5 and that there exists a number of Sylow p-subgroups for each p such that p||G|. For p = 2, n2|(3 ∗ 5) and n2 ≡ 1 mod 2. The factors of p = 2 would then be 1, 3, 5, 15 with all factors congruent. For p=3, n3|(2∗5) and n3 ≡1mod3. The factors of p = 3 would then be 1, 2, 5, 10 with 1, 10 congruent. For p = 5, n5|(2 ∗ 3) and n5 ≡ 1 mod 5. The factors of p = 5 would then be 1, 2, 3, 6 with 1, 6 congruent.
Since there are several numbers that could possibly work for the primes, suppose to show a contradiction that n3 = 10 and n5 = 6. Then n3(G) = 10 so |n3(G)| = 3 elements, two of which are non identity, giving a total of 21 elements. Repeat for n5 = 6 to get 24 elements total. We know 21+24 = 45 > |G| = 30, which is a contradiction. Therefore, n3 = 1 or n5 = 1 so n3(G) or n5(G) is characteristically normal in G thus G is not simple.
QED
14 Conclusion
Sylow’s Theorems may appear to be a simple list on finite group properties but the applications of Sylow’s Theorems are rather diverse including classifying groups with order pq with p < q for p and q both prime, determining if a finite group is nilpotent, determining normality of a Sylow subgroup, determining if a group of a certain size is abelian, determining if a group of a certain order is simple, and so much more.
12.1 Proof
Let P be the p-Sylow subgroup, Q be the q-Sylow subgroup. Since P and Q have relatively prime sizes, P∩Q = {e} by Lagrange’s Theorem. We know that the subgroups P and Q are normal in G so for a in P and b in Q, aba−1 = (aba−1)b−1 = a(ba−1b−1) in P∩Q = {e} so ab = ba.
QED
13 Example
Prove that if |G| = 30 then the group is not simple.
13.1 Proof
Assume |G| = 30. We know 30 = 2∗3∗5 and that there exists a number of Sylow p-subgroups for each p such that p||G|. For p = 2, n2|(3 ∗ 5) and n2 ≡ 1 mod 2. The factors of p = 2 would then be 1, 3, 5, 15 with all factors congruent. For p=3, n3|(2∗5) and n3 ≡1mod3. The factors of p = 3 would then be 1, 2, 5, 10 with 1, 10 congruent. For p = 5, n5|(2 ∗ 3) and n5 ≡ 1 mod 5. The factors of p = 5 would then be 1, 2, 3, 6 with 1, 6 congruent.
Since there are several numbers that could possibly work for the primes, suppose to show a contradiction that n3 = 10 and n5 = 6. Then n3(G) = 10 so |n3(G)| = 3 elements, two of which are non identity, giving a total of 21 elements. Repeat for n5 = 6 to get 24 elements total. We know 21+24 = 45 > |G| = 30, which is a contradiction. Therefore, n3 = 1 or n5 = 1 so n3(G) or n5(G) is characteristically normal in G thus G is not simple.
QED
14 Conclusion
Sylow’s Theorems may appear to be a simple list on finite group properties but the applications of Sylow’s Theorems are rather diverse including classifying groups with order pq with p < q for p and q both prime, determining if a finite group is nilpotent, determining normality of a Sylow subgroup, determining if a group of a certain size is abelian, determining if a group of a certain order is simple, and so much more.
Bibliography
-
[1] A. Bohmann The Sylow Theorems. (Massachusetts Institute of Technology,
Massachusetts).
-
[2] Boyer Notes on the Proof of the Sylow Theorems. (Drexel University, Penn-
sylvania).
-
[3] K. Conrad Consequences of the Sylow Theorems (Univeristy of Connecticut,
Connecticut).
-
[4] K. Conrad The Sylow Theorems. (University of Connecticut, Connecticut).
-
[5] Sylow’s Theorems. (Dartmouth University, New Hampshire).
-
[6] P Hall Theorems Like Sylow’s (Proceedings of the London Mathematical
Society).
-
[7] T. Hungerford. Abstract Algebra: An Introduction (Saint Louis University,
Missouri, 2014).
-
[8] A. Ideljaj The Sylow Theorems and Their Applications (University of
Chicago, Chicago, 2016).
-
[9] B. Ikenaga The Sylow Theorems (Millersville University, Pennsylvania,
2012).
-
[10] W. Kantor Sylow’s Theorem in Polynomial Time (University of Oregon,
Oregon, 1985).
-
[11] The Sylow Theorems (Lawrence University, Wisconsin).
-
[12] KP Mann Notes on Sylow’s Theorems (Berkeley University, California,
2011).
-
[13] J. McKernan Sylow Theorems and Applications (Massachusetts Institute
of Technology, Massachusetts).
-
[14] J. Pakianathan Sylow Theory (University of Rochester, New York, 2003).
-
[15] F. Kammuller, L. Paulson A Formal Proof of Sylow’s Theorem (University
of Cambridge, UK, 2000).
[16] H. Poutiainen On Sylow’s Theorems (University of Oulu, Hawaii, 2015). - [17] G. Singh, A. Dar The Sylow Theorem and its Consequences (Barkatuallah University Institute of Technology Bhopal, Inida, 2014).
- [18] W. Waterhouse The Early Proofs of Sylow’s Theorem (Pennsylvania State University, Pennsylvania, 1979).
- [19] E. Weisstein Sylow Theorems (MathWorld, Wolfram Web Resource).[20] E. Weisstein Sylow p-Subgroup (MathWorld, Wolfram Web Resource).
No comments:
Post a Comment