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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1 1. Group Theory II In this section we consider groups operating on sets. This is not particularly new. For example, the permutation group S n acts on the subset N n = {1, 2,...,n} of N. Also the group D n acts on the vertices of a regular n-gon. We see immediately that both actions satisfy the conditions of the subsequent definition. These conditions are essential for establishing a suitable equivalence relation on S with respect to the action of G. The latter is then used for showing the existence of certain subgroups of a given group. The concept is also applied in other areas of mathematics. Definition 1.1. Let G be a group and S be a non-empty set. We say that G acts on S if there is a map G S S : (g,s) g s satisfying the following conditions: (1) (gh) s = g (h s) g,h G, s S, (2) e s = s s S, e the unit element of G. Examples (1) Let U be a subgroup of a group G, S G and U S S : (g,h) ghg 1. The 2 conditions of the definition are easily checked: g 1 (g 2 h) = g 1 (g 2 hg2 1 ) = g 1 g 2 hg2 1 g1 1 = g 1 g 2 h (g 1 g 2 ) 1 = (g 1 g 2 ) h for all g 1,g 2 U, h S as well as e h = h for the unit element e of U. (2) Let G be a group and m a natural number satisfying m G. Let S be a subset of the power set of G consisting of all subsets of G with exactly m elements. The action of G on S is given by G S S : (g,t) gt = {gt t T }. We leave it to the reader to verify the conditions of the definition. If a group G acts on a set S then S decomposes into equivalence classes with respect to the following relation on S. Two elements s,t of S are said to be equivalent (with respect to G), iff there exists g G 1

2 2 with g s = t. We show that this is indeed an equivalence relation. The relation is clearly reflexive because of the second condition in??. It is symmetric because g s = t implies g 1 t = s. For this we need both conditions. Eventually, g s = t and h t = u (g,h G, s,t,u S) yield (hg) s = h (g s) = h t = u, hence the transitivity of that relation. The equivalence classes are also called orbits. The orbit containing s S is denoted by O(s). We will see that the length of an orbit can be interpreted as a group index of G. For this we introduce the notion of an inertia group. Definition 1.2. An action of the group G on the set S is said to be transitive if there is exactly one orbit in S, namely S itself. For s S the subset of G fixing s, i.e. Stab(s) := {g G g s = s}, is called stabilizer (or inertia group, fix group) of the element s. It is easy to see that Stab(s) is actually a subgroup of G for each s S. Clearly, the unit element e of G belongs to Stab(s). If g,h are in Stab(s), then also h 1 and gh 1 are in Stab(s). In order to establish a connection between the length of an orbit O(s) and the number of elements of G, Stab(s) we need to consider how the fix groups of elements of an orbit are related. Let s,t S belong to the same orbit, say O(s). Hence, there exists g G with g s = t. We obtain the following chain of equivalences: 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. Here, the second but last equation is a consequence of g 1 hg = k gkg 1 = h. Hence, the fix groups of any two elements of an orbit are conjugate subgroups of G. Lemma 1.3. The length O(s) of the orbit O(s) equals the group index (G : Stab(s)). If G is finite then the length of every orbit is a divisor of the group order. Proof We consider the map ϕ : O(s) {g Stab(s) g G} : g s g Stab(s).

3 ϕ is clearly surjective. It is also well defined and injective because of the following chain of equivalences: g Stab(s) = h Stab(s) h 1 g Stab(s) h 1 g s = s g s = h s. Hence, we have established a bijection between the elements g s of the orbit O(s) and the left cosets g Stab(s) of Stab(s) in G. If G acts on the set S then S decomposes into orbits. Hence, there is a set R S of representatives such that. S = O(r). (1) r R Since that union is disjoint the number of elements in S is obtained as a sum of group indices: S = r R(G : Stab(r)). (2) Those elements s S whose orbits O(s) consist of just one element, O(s) = {s}, are called fix points under the action of G. They play a distinguished role. If we denote the set of them by F(S) then the decomposition of S into orbits yields the following relation between the lengths of those orbits. S = F(S) + r R\F(S) (G : Stab(r)). (3) The last equation becomes even more important if we consider the action of a group G on subsets of G by conjugation. Let T be a fixed non-empty subset of G. We put S := {gtg 1 g G} and define the action of G on S via G S S : (h, gtg 1 ) hgt(hg) 1. Obviously, G operates transitively on S, there exists only one orbit, the set S itself. In this special situation we have Stab(T) = {g G gtg 1 = T } = N T, that is the fix group of T equals the normalizer N T of T in G. If T is even a subgroup of G then T has exactly (G : N T ) conjugate subgroups. In a slightly different situation let G act on the set of its own elements by conjugation. In that case, the set of fixed points F(G) coincides with 3

4 4 the center Z(G) of G. The orbits of the elements are called classes of conjugate elements. The stabilizer of an element coincides with its normalizer. The class equation (??) becomes G = Z(G) + (G : N r ) (4) r R\Z(G) if R again denotes a full set of representatives of the orbits. These results will now be used to exhibit the existence of subgroups of prime power order in any finite group G. Definition 1.4. A finite group G is called a p-group if its order (G : 1) is a power of a prime number p, say p r with p P, r N. If G is a finite group its subgroups of prime power order, say p m, are called p-subgroups. A p-subgroup H of maximal p-power, i.e. (H : 1) = p m and p does not divide (G : H), is called a p-sylow-subgroup. Remark The center of a p-group is non-trivial, i.e. it contains more than one element. This is an immediate consequence of (??) where the left-hand side and also all terms of the second summand of the righthand side are divisible by p with the consequence that also Z(G) must be divisible by p. Example Let G = V 4 be the Klein Four Group. G is its own p- Sylow-subgroup for the only prime number p = 2 dividing the order of G, and G has 3 p-subgroups of order 2. We want to show the existence of p-subgroups for all finite groups G whose order is divisible by p. For this we start in a slightly more general context. We assume that the order of the given finite group G is n = p m q for a prime number p not dividing q N. From Lagrange s theorem we know that the order of a subgroup of G divides (G : 1). Hence, we assume that k is a positive integer subject to k > 1 and k (G : 1). If there is a subgroup of G of order k then it has to be one of the ( n k) subsets of G of k elements. As in the second example in this section let S be the set of all subsets of G of k elements. The group G acts on S by multiplication. For T S, say T = {t 1,...,t k }, we have G S S : g T {gt 1,...,gt k }. What can we say about the stabilizer of T? If gt = T for some g G then gt 1 = t j for some j {1,...,k}, and the group element g is uniquely determined by that index j: g = t j t 1 1. Hence, the order of the stabilizer of T is bounded by k.

5 On the other hand, if U S is indeed a subgroup of G then the stabilizer of U equals U because of g U = U g U. From this we conclude that an element U S is a subgroup of G if and only if U equals its stabilizer. In that case the length of the orbit O(U) of U is G /k. Now let us assume that the length of the orbit O(T) of T S is G /k. We want to prove that O(T) contains exactly one subgroup U of G. The stabilizer Stab(T) of T satisfies Stab(T)x T for all x T and therefore Stab(T)x = T because of Stab(T) = k. Hence, x 1 Stab(T)x = x 1 T is a subgroup in the orbit of T. Lagrange s theorem tells us that O(T) can contain at most one subgroup of G. For detecting subgroups of order k we therefore just need to check whether S contains an orbit of length G /k. However, this can require lengthy computations as the following example demonstrates. Exercise We recommend that the reader generates the subsets of k {3, 4, 6} elements of the alternating group A 4 to check computationally that A 4 has subgroups of orders 3,4, respectively, but has no subgroup of order 6. The results of that exercise show that we cannot expect the existence of subgroups of arbitrary order dividing the group order n. It is the merit of the group theoretician L. Sylow ( ) from Norway to have recognized and proved the existence of subgroups of prime power order p a (1 a m) for prime numbers p with p m G. The special case a = 1 is due to Cauchy and readers not familiar with the subject are advised to assume a = 1 at first reading of the following, though the arguments are the same for a > 1. We recall that all we need to prove is the existence of an orbit O(T) of S of length equal to p m a q, the stabilizer of that orbit being a subgroup we are looking for. (If we additionally require e T we already have Stab(T) = Stab(T)e = T.) We already know that O(T) p m a q. Since the length of any orbit divides G it therefore suffices to exhibit the existence of an orbit whose orbit length is not divisible by p m a+1. The latter is an easy consequence of the following lemma from elementary number theory and the class equation of the action of G on S. 5 Lemma 1.5. Let n = p m q for a prime number ( p not dividing q N. n Then for 1 a m the binomial coefficient is divisible by p m a p a )

6 6 ( but not divisible by p m a+1 n. The quotient )/p m a is congruent to p a 1 modulo p. Proof ( n with x = p a ) ( = n 1 p a 1 p a 1 i=0 p n i a 1 p a i = p m q i pm a q p a i i=1 = p m a qx ). In the product for x we write every index i {1,...,p a 1} in the form i = p l i x i with 0 l i < a and x i not divisible by p. Dividing the numerator and the denominator of the i-th factor of that product by p l i we obtain for the numerator: p a 1 i=1 p ( p m l i ) a 1 q x i = up + y (u,y N, y = ( x i ), p y), and for the denominator: and therefore p a 1 i=1 i=1 ( p a l i x i ) = v p + y (v N), x(vp + y) = up + y. Because of p y this yields the result x 1 mod p. The following theorem contains the most important results on the existence of p-subgroups. Theorem 1.6 (Sylow). Let G be a group of order n = p m q with p m and let 1 a m (a N). (1) The number of subgroups of G of order p a is congruent to 1 modulo p, i.e. these subgroups do always exist. (2) Let H be a p-sylow-subgroup of G and U an arbitrary p-subgroup of G. Then one of the conjugates of U is contained in H. (3) All p-sylow-subgroups of G are conjugate. The number of p- Sylow-subgroups of G divides q. Proof

7 (1) We have seen that subgroups of order p a are in 1 1-correspondence to the orbits of length p m a q of the set S of subsets of G with p a elements under the action of G. Because of the preceding Lemma S is not divisible by p m a+1. Hence, the class equation (??) tells us that there are orbits of length p m a q and it follows that their number is congruent to 1 modulo p. (2) Let H be a fixed p-sylow-subgroup of G. We consider the action of G on the set S H := {ghg 1 g G} by conjugation. Obviously, G acts transitively on S H. The length S H equals the group index (G : N H ) for the normalizer N H of H. Since N H contains H that group index is not divisible by p. Next let U be an arbitrary p-subgroup of G, say of order p a. We let U act on S H by conjugation. S H decomposes into orbits whose lengths are divisors of p a. Because of p S H there must exist orbits of length 1. Hence, there exists a p-sylow-subgroup K which is conjugate to H and for which U is contained in the normalizer N K. Then U and K are both contained in the normalizer N K of K in G. Also, K is always a normal subgroup of N K. Hence, we can apply the first isomorphism theorem for groups and get UK/K = U/U K. Therefore UK is a supergroup of K for which the index (U K : K) equals (U : U K), a p-power. Then also the order of UK is a p-power divisible by K. K being a p-sylow-subgroup of G we must necessarily have UK = K and therefore U K. Since K = ghg 1 for a suitable element g of G we obtain g 1 Ug g 1 Kg = H. 7 (3) If we choose U to be a p-sylow-subgroup, too, then the same considerations as in the previous part of the proof yield that any two p-sylow-subgroups of G are conjugate, i.e. the orbit S H already coincides with the set of all p-sylow-subgroups of G. We already noted that the orbit length S H = (G : N H ) is not divisible by p. Being a divisor of G it must therefore divide q. Remark A p-sylow-subgroup H of G is a normal subgroup of G if and only if S H = 1.

8 8 Corollary 1.7 (Cauchy). If the order of the finite group G is divisible by the prime number p then G contains a subgroup and therefore an element of order p. Proof According to Sylow s theorem G contains a subgroup U with (U : 1) = p. U is necessarily cyclic, and all its elements except the unit element have order p. Corollary 1.8. A finite group G is a p-group if and only if the order of each of its elements is a p-power. Example We want to determine all non-abelian groups of order 8. G cannot contain an element of order 8 (in that case G would be cyclic) nor can all group elements of G have 2 as exponent (in which case G would be abelian). Hence, G must contain at least one element, say b, of order 4. Then U = b has index 2 in G and is therefore a normal subgroup. G decomposes into 2 equivalence classes with respect to U, say G = U Ua with Ua = au. Since a is not contained in U the element a 2 is not contained in Ua, therefore it must belong to U. Because of ord(a) 4 we obtain a 2 = e or a 2 = b 2. We shall see that both options yield a group of order 8 which is unique up to isomorphism. Since we are looking for non-abelian groups we need to know what aba 1 is. Because of au = Ua we must have aba 1 U. The option aba 1 = b is to be excluded since G would be abelian in that case. The choice aba 1 = e is impossible since it yields b = e. Finally, a choice aba 1 = b 2 implies ab 2 a 1 = (aba 1 ) 2 = b 4 = e, hence b 2 = e which is in contradiction with ord(b) = 4. The only remaining possibility is therefore aba 1 = b 1 yielding ab m = b m a for all m N. We now distinguish the two cases a 2 = e and a 2 = b 2. (1) a 2 = e yields G = D 4. The generators a,b of G satisfy the relations defining D 4 : a 2 = e = b 4 and aba 1 = b 1. For example, these are fulfilled by the matrices ( ) ( ) cos(2π/4) sin(2π/4) 1 0 b :=,a :=. sin(2π/4) cos(2π/4) 0 1 (2) a 2 = b 2 yields G = Q 8. The relations b 4 = e, a 2 = b 2, aba 1 = b 1 are easily seen to be satisfied by the matrices ( ) ( ) i b :=,a := 0 i 1 0

9 from C 2 2, where i 2 = 1 denotes the imaginary unit. Hence, those 2 matrices generate a group of 8 elements which is isomorphic to the quaternion group. We note that the special representation of the group elements by matrices in both cases makes it superfluous to test whether the composition (here: matrix multiplication) is associative. From Sylow s theorem we can easily conclude that p-groups do not only contain subgroups for every p-power dividing the group order but also normal subgroups. Proposition 1.9. Let G be a p-group of order p m and let a {1, 2,...,m}. Then G contains a normal subgroup of order p a, moreover, the number of these normal subgroups of G is congruent to 1 modulo p. Proof We denote by S the (non-empty) set of subgroups U of G of order p a. According to the Sylow theorem the number of elements in S is congruent to 1 modulo p. We let G act on S by conjugation. If R denotes a full set of representatives of the corresponding orbits then the class equation reads S = U R(G : N U ). 9 Denoting the set of fix points by F(S) we get S F(S) mod p, and in?? we proved S 1 mod p. For any V F(S) its normalizer N V coincides with G. As a consequence, V is a normal subgroup of G. Finally, we apply Sylow s theorem to finite abelian groups. Theorem Let G be a finite abelian group. Then G is the direct sum of its p-sylow-subgroups. Proof Let the prime factor decomposition of the order n of G be r n = p m i i. i=1 Since G is commutative it has precisely one p i -Sylow-subgroup U i for each p i (1 i r). Writing the composition in G additively we obtain that U := U U r is a subgroup of G. We show that this sum of subgroups is direct. Then its order equals the order of G and it must therefore coincide with G.

10 10 For this we put Ũ i := j=1 j i and demonstrate U i Ũi = {0} for 1 i r. Every element x of that intersection has an order dividing p m i i as well as G /p m i i. Hence, we must have x = 0. U j

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