LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS


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1 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS Recall Lagrange s theorem says that for any finite group G, if H G, then H divides G. In these lectures we will be interested in establishing certain partial converses to this result. It should be noted that a true converse is not possible: for instance, the alternating group A 4 has order 12 but no subgroup of 6 elements. However, one can still deduce weaker statements about the existence of subgroups of a group of a given order. As an example we state the following fundamental theorem. Theorem (Cauchy s theorem). Let p be a prime and suppose G is a finite group for which p G. Then G contains an element of order p and, consequently, a cyclic subgroup of order p. Thus, Cauchy s theorem is a partial converse to Lagrange s theorem, in which one restricts one s attention to prime order subgroups. We will prove this result, and far reaching generalisations known as the Sylow theorems, below. The key tool in the proofs will be the theory of group actions, is reviewed presently. Group actions. Definition. Let X be a set and G be a group. An action of G on X is a map : G X X such that i) ex = x for all x X; ii) (g 1 g 2 )x = g 1 (g 2 x) for all x X and g 1, g 2 G. In this situation, we say X is a Gset. Definition. On any Gset X one may define an equivalence relation x 1 x 2 if and only if there exists some g G such that gx 1 = x 2. The equivalence classes of this relation are called the orbits in X under G. If x X, the equivalence class containing x is called the orbit of x and is denoted by orb G (x) = Gx := {gx : g G}. Definition. If X is a Gset and x X, then one may define the stabilizer stab G (x) = G x of x in G to be the subgroup stab G (x) = {g G : gx = x}. Theorem (OrbitStabilizer theorem). Let X be a Gset and x X. Then orb G (x) = (G : stab G (x)). Thus, if G is finite, then orb G (x) divides G. Definition. For any Gset X we define the set of fixed points to be X G := {x X : gx = x for all g G}; that is, the union of all the 1element orbits of X. Suppose X is a finite Gset and let x 1,..., x r X be a set of representatives for the orbits of the action of G on X so that r X = orb G (x i ). i=1 1
2 2 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS Thus, if X G = s, then continuing with the above setup r X = X G + orb G (x i ). (0.1) Henceforth, p denotes a prime. i=s+1 Theorem. Let G be a group of order p n and X a finite Gset. Then X X G mod p. By the OrbitStabilizer theorem, p divides orb G (x i ) for each i = s + 1,..., r. Cauchy s theorem. Theorem (Cauchy s theorem). Suppose G is a finite group for which p G. Then G contains an element of order p and, consequently, a cyclic subgroup of order p. Consider the set X := { (g 1,..., g p ) G p : g 1... g p = e }. Any arbitrary choice of (g 1,..., g p 1 ) G p 1 determines a unique element of X: indeed, if we define g p := (g 1... g p 1 ) 1, then (g 1,..., g p ) X. Thus, X = G p 1 and since p G, it follows that p X. Let σ S p by the cycle (1, 2,..., p) and define an action of the group σ S p on X by permuting the coordinates. By the theorem, X X σ mod p and so p X σ. Since (g 1,..., g p ) X σ if and only if g 1 = = g p, and X σ > 1, it follows that there exists some a G such that a e and a p = e, as required. As an application of Cauchy s theorem we prove a simple classification theorem. Corollary. Let p be a prime and G be a finite group. The following are equivalent: i) Every element of G has order given by a power of p. ii) G is a power of p. If either of these (equivalent) conditions hold, we say G is a pgroup. Definition. We say H G is a psubgroup of G if H is a pgroup. Proof (of the Corollary). If G = p n for some n N {0}, then G is clearly a pgroup by Lagrange s theorem. For the converse, suppose that G is a finite group and G p n for all n N {0}. Then there exists a prime q p such that q G. By Cauchy s theorem there exists an element of order q, so G is not a pgroup. The First Sylow theorem. In this section we prove a farreaching generalisation of Cauchy s theorem. Theorem (First Sylow theorem). Let G be a finite group and suppose p n G but p n+1 G for some prime p and n N. 1) G contains a subgroup of order p i for each 1 i n. 2) Every H G of order p i is a normal subgroup of a subgroup of order p i+1 for 1 i n 1.
3 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS 3 Note part 1) with i = 1 recovers Cauchy s theorem. The proof of the First Sylow theorem will require the notion of a normaliser of a group. Let G be a group and S denote the set of all subgroups of G. We can define an action G S S of G on S by conjugation: (g, H) ghg 1. Recall, for any H G it follows that stab G (H) := {g G : ghg 1 = H} is a subgroup of G and, furthermore, it follows by definition that H stab G (H). In fact, clearly stab G (H) is the largest subgroup of G for which H is a normal subgroup. Definition. If G is a group and H G, then we define the normaliser of H to be the subgroup N[H] := stab G (H), where the stabilizer is as defined above. Lemma. If G is a finite group and H G, then g N[H] if and only if ghg 1 H for all h H. Proof. If g N[H], then by definition ghg 1 = H and so ghg 1 H for all h H. Conversely, suppose ghg 1 H for all h H so that the conjugation map ι g : H H given by ι g (h) := ghg 1 is welldefined and clearly injective. Since H is a finite set, it must also be surjective so that ghg 1 = H, as required. Lemma. If H is a psubgroup of a finite group G, then (N[H] : H) (G : H) mod p. In particular, if p (G : H), then N[H] H. Let L denote the set of left cosets of H and let H act of L by left translation: H L L, (h, xh) hxh. Note that L = (G : H) by definition. By the previous lemma, the set L H of fixed points of the action satisfies L H = {xh L : hxh = xh for all h H} = {xh L : xh N[H]} and so L H = (N[H] : H) Since H is a psubgroup, it has order given by a power of p and so L L H mod p. For the final part of the lemma, if p (G : H), then p (N[H] : H) and so (N[H] : H) > 1. We are now in a position to prove the First Sylow theorem. Proof (of First Sylow theorem). The key ideas of the proof of part 1) are: Proceed by induction on i. The case i = 1 is Cauchy s theorem. Suppose, by way of induction hypothesis that H G has order p i for some 1 i n 1. Since p (G : H), it follows that p (N[H] : H) so that p N[H]/H. By Cauchy s theorem, there exists K N[H]/H with K = p. If π : N[H] N[H]/H is the canonical homomorphism, then π 1 K G satisfies H π 1 K. Since π 1 K/H = K by the first isomorphism theorem, it follows that π 1 K = p i+1. The key ideas of the proof of part 2) are: Arguing as before, H π 1 K N[H] where π 1 K = p i+1. Since H N[H], it follows that H π 1 K.
4 4 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS We consider some simple consequences of the First Sylow Theorem. Theorem. i) Every group of primepower order is solvable. ii) No group of order p r for some r 2 is simple. The Second and Third Sylow theorems. Definition. A Sylow psubgroup of a finite group G is a maximal psubgroup. Thus, by the First Sylow theorem if p n G but p n+1 G, then the Sylow psubgroups are precisely the subgroups of G of order p n. The remaining Sylow theorems give insight into the structure of the Sylow p subgroups. Lemma. If G is a finite group and P a Sylow psubgroup then for any g G the conjugation gp g 1 is also a Sylow psubgroup. Corollary. Suppose G is a finite group and p, q are distict prime divisors of G. If G has only one Sylow psubgroup, then this group is normal and so G is not simple. Proof. If G has only one Sylow psubgroup, then this group is a nontrivial proper subgroup of G which is fixed under conjugation, and therefore normal. The converse of the lemma also holds. Theorem (Second Sylow theorem). If P 1, P 2 are Sylow psubgroups of a finite group G, then they are conjugate. That is, there exists some g G such that P 2 = gp 1 g 1. Let L denote the left cosets of P 1 and let P 2 act of L by left translation. Note that L P2 L mod p and p L = (G : P 1 ) so that L P2 0. If xp 1 L P2, then P 1 = x 1 P 2 x. Theorem (Third Sylow theorem). If G is a finite group and p G, then the number of Sylow psubgroups is congruent to 1 modulo p and divides G. Let S denote the set of all Sylow psubgroups and fix some P S. Then P acts on S by conjugation: (x, Q) xqx 1 for all (x, Q) P S. If Q S P, then P, Q N[Q] are Sylow psubgroups of N[Q]. By Sylow II, it follows P, Q are conjugate in N[Q]. By the definition of N[Q], it follows that P = Q, so that S P = {P }. S S P = 1 mod p. Let G act on S by conjugation. By Sylow II there is only one orbit and so, by the OrbitStabilizer theorem, S = (G : stab G (P )) for any P S. In particular, S G. We conclude with some examples which demonstrate the utility of the Sylow theorems. Example. No group of order 20 is simple. Indeed, by Sylow III the number of Sylow 5subgroups is congruent to 1 modulo 5 and divides 20. Thus there is precisely 1 Sylow 5subgroup which must be normal.
5 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS 5 Example. No group of order 30 is simple. By Sylow III the number of Sylow 5subgroups is congruent to 1 modulo 5 and divides 30, so there are either 1 or 6 such subgroups. Similarly, the number of Sylow 3subgroups is congruent to 1 modulo 3 and divides 30, so there are either 1 or 10 such subgroups. If in either case there is only 1 subgroup, then we are done so we may suppose there are 6 Sylow 5subgroup and 10 Sylow 3subgroups. Each nonidentity element of a given Sylow 5subgroup is a generator and thus has order 5 and is distinct from all the nonidentity elements of the other Sylow 5subgroups. Hence G contains precisely 24 = 4 6 elements of order 5. Similarly, G contains 20 = 2 10 elements of order 3. But in this case, the number of elements of G is 45, a contradiction. Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, address:
Stab(t) = {h G h t = t} = {h G h (g s) = g s} = {h G (g 1 hg) s = s} = g{k G k s = s} g 1 = g Stab(s)g 1.
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