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The Sylow Theorems Chapter 10: The Sylow Theorems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Spring 2014 M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 1 / 29 Overview This chapter about one question: What groups are there? In Chapter 5, we saw five families of groups: cyclic, dihedral, abelian, symmetric, alternating. In Chapter 8, we classified all (finitely generated) abelian groups. But what other groups are there, and what do they look like? For example, for a fixed order jGj, we may ask the following questions about G: 1. How big are its subgroups? 2. How are those subgroups related? 3. How many subgroups are there? 4. Are any of them normal? There is no one general method to answer this for any given order. However, the Sylow Theorems, developed by Norwegian mathematician Peter Sylow (1832{1918), are powerful tools that help us attack this question. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 2 / 29 The Sylow Theorems Definition A p-group is a group whose order is a power of a prime p. A p-group that is a subgroup of a group G is a p-subgroup of G. Notational convention n n Throughout, G will be a group of order jGj = p · m, with p - m. That is, p is the highest power of p dividing jGj. There are three Sylow theorems, and loosely speaking, they describe the following about a group's p-subgroups: 1. Existence: In every group, p-subgroups of all possible sizes exist. 2. Relationship: All maximal p-subgroups are conjugate. 3. Number: There are strong restrictions on the number of p-subgroups a group can have. Together, these place strong restrictions on the structure of a group G with a fixed order. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 3 / 29 Our unknown group of order 200 Throughout this chapter, we will have a running example, a \mystery group" M of order 200. jMj=200 ? ? ? ? ? e ? ? • ? ? ? ? ? Using only the fact that jMj = 200, we will unconver as much about the structure of M as we can. We actually already know a little bit. Recall Cauchy's theorem: Cauchy's theorem If p is a prime number dividing jGj, then G has an element g of order p. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 4 / 29 Our mystery group of order 200 Since our mystery group M has order jMj = 23 · 52 = 200, Cauchy's theorem tells us that: M has an element a of order2; M has an element b of order5; Also, by Lagrange's theorem, hai \ hbi = feg. jMj=200 ? ? ? b ? • 2 jbj = 5 jaj = 2 •b e ? ? a• • • 3 • b ? 4 ? b ? ? M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 5 / 29 p-groups Before we introduce the Sylow theorems, we need to better understand p-groups. n Recall that a p-group is any group of order p . For example, C1, C4, V4, D4 and Q4 are all 2-groups. p-group Lemma If a p-group G acts on a set S via φ: G ! Perm(S), then j Fix(φ)j ≡p jSj : Proof (sketch) Fix(φ) non-fixed points all in size-pk orbits Suppose jGj = pn. p elts pi elts p6 elts · · By the Orbit-Stabilizer theorem, the · · · · only possible orbit sizes are 3 2 n p elts p elts 1; p; p ;:::; p . · · · M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 6 / 29 p-groups Normalizer lemma, Part 1 If H is a p-subgroup of G, then [NG (H): H] ≡p [G : H] : Proof Let S = G=H = fHx j x 2 Gg. The group H acts on S by right-multiplication, via φ: H ! Perm(S), where φ(h) = the permutation sending each Hx to Hxh. The fixed points of φ are the cosets Hx in the normalizer NG (H): Hxh = Hx; 8h 2 H () Hxhx −1 = H; 8h 2 H () xhx −1 2 H; 8h 2 H () x 2 NG (H) : Therefore, j Fix(φ)j = [NG (H): H], and jSj = [G : H]. By our p-group Lemma, j Fix(φ)j ≡p jSj =) [NG (H): H] ≡p [G : H] : M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 7 / 29 p-groups Here is a picture of the action of the p-subgroup H on the set S = G=H, from the proof of the Normalizer Lemma. S = G=H = set of right cosets of H in G NG (H) Hg1 Hg4 Hg Hg H 3 2 Hg6 Hg5 Hg7 Ha1 Hg1 Hg8 Hg14 Hg9 Ha2 Hg13 Hg10 Ha3 Hg12 Hg11 The fixed points are precisely Orbits of size > 1 are of various sizes the cosets in NG (H) dividing jHj, but all lie outside NG (H) M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 8 / 29 p-subgroups The following result will be useful in proving the first Sylow theorem. The Normalizer lemma, Part 2 n i n Suppose jGj = p m, and H ≤ G with jHj = p < p . Then H NG (H), and the index [NG (H): H] is a multiple of p. [G : H] cosets of H (a multiple of p) Hy2 ... H NG (H) ≤ G H Hx2 Hxk Hy1 Hy3 . [NG (H): H] > 1 cosets of H Conclusions: H = NG (H) is impossible! i+1 p divides jNG (H)j. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 9 / 29 Proof of the normalizer lemma The Normalizer lemma, Part 2 n i n Suppose jGj = p m, and H ≤ G with jHj = p < p . Then H NG (H), and the index [NG (H): H] is a multiple of p. Proof Since H C NG (H), we can create the quotient map q : NG (H) −! NG (H)=H ; q : g 7−! gH : The size of the quotient group is [NG (H): H], the number of cosets of H in NG (H). By The Normalizer lemma Part 1, [NG (H): H] ≡p [G : H]. By Lagrange's theorem, jGj pnm [N (H): H] ≡ [G : H] = = = pn−i m ≡ 0 : G p jHj pi p Therefore, [NG (H): H] is a multiple of p, so NG (H) must be strictly larger than H. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 10 / 29 The 1st Sylow Theorem: Existence of p-subgroups First Sylow Theorem G has a subgroup of order pk , for each pk dividing jGj. Also, every p-subgroup with fewer than pn elements sits inside one of the larger p-subgroups. The First Sylow Theorem is in a sense, a generalization of Cauchy's theorem. Here is a comparison: Cauchy's Theorem First Sylow Theorem If p divides jGj, then ... If pk divides jGj, then ... There is a subgroup of order p There is a subgroup of order pk which is cyclic and has no non-trivial proper subgroups. which has subgroups of order 1; p; p2 ::: pk . G contains an element of order p G might not contain an element of order pk . M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 11 / 29 The 1st Sylow Theorem: Existence of p-subgroups Proof The trivial subgroup feg has order p0 = 1. Big idea: Suppose we're given a subgroup H < G of order pi < pn. We will construct a subgroup H0 of order pi+1. By the normalizer lemma, H NG (H), and the order of the quotient group NG (H)=H is a multiple of p. By Cauchy's Theorem, NG (H)=H contains an element (a coset!) of order p. Call this element aH. Note that haHi is cyclic of order p. Claim: The preimage of haHi under the quotient q : NG (H) ! NG (H)=H is the subgroup H0 we seek. The preimages q−1(H), q−1(aH), q−1(a2H), . , q−1(ap−1H) are all distinct cosets i of H in NG (H), each of size p . 0 −1 i+1 Thus, the preimage H = q (haHi) contains p · jHj = p elements. M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 12 / 29 The 1st Sylow Theorem: Existence of p-subgroups Here is a picture of how we found the group H0 = q−1(haHi). q NG (H) NG (H) H H • • • aH H H0 • aH • ·· haHi a3H a2H a3H a2H · • • q−1 • g1 • g3 g1H g3H · · · · • g2 • g4 · g2H g4H · p−1 0 [ k Since jHj = pi , the subgroup H = a H contains p · jHj = pi+1 elements. k=0 M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 13 / 29 Our unknown group of order 200 We now know a little bit more about the structure of our mystery group of order jMj = 23 · 52: 3 M has a 2-subgroup P2 of order2 = 8; 2 M has a 5-subgroup P5 of order 25 = 5 ; Each of these subgroups contains a nested chain of p-subgroups, down to the trivial group, feg. jMj=200 ? ? b 8 • b2 jbj = 5 4 • jaj = 2 e a• • 25 • 3 • b ? b4 ? ? M. Macauley (Clemson) Chapter 10: The Sylow Theorems Math 4120, Spring 2014 14 / 29 The 2nd Sylow Theorem: Relationship among p-subgroups Definition n n A subgroup H < G of order p , where jGj = p · m with p - m is called a Sylow p-subgroup of G. Let Sylp(G) denote the set of Sylow p-subgroups of G.
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