15 Permutation representations and G-sets

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1 15 Permutation representations and G-sets Recall. If C is a category and c C then Aut(c) =the group of automorphisms of c 15.1 Definition. A representation of a group G in a category C is a homomorphism : G Aut(c) Special types of representations: linear representations = representations in the category of vector spaces permutation representations = representations in the category of sets Note. Let S be a set and let : G Aut(S) be a permutation representation of G. Thehomomorphism defines a map G S S, (a, x) ρ(a)(x) Denote a x := (a)(x). Wehave: 1) (ab) x = a (b x) for a, b G, x S 2) e x = x for all x S 15.3 Definition. An action of a group G on a set S is a map G S S, (a, x) a x satisfying conditions 1) - 2) above. A G-set is a set S equipped with an action of the group G. 54

2 15.4 Note. ApermutationrepresentationG Aut(S) determines a G-set structure on the set S. Conversely, the structure of a G-set on S determines a permutation representation G Aut(S) Examples. 1) Take S = G. Themap G G G, (a, b) ab defines a G-set structure on G. Wesaythatthe group G acts on itself by left translations. 2) Let H be a subgroup of G. TakeS = G/H. Themap G G/H G/H, (a, bh) (ab)h defines a G-set structure on G/H. 3) Take again S = G. Themap G G G, (a, b) aba 1 defines another G-set structure on G. We say that the group G acts on itself by conjugations. 4) R is a Z/2Z-set with the action given by 0 x = x, 1 x = x. Z/2Z R R 15.6 Definition. If S, T are G-sets then a G-equivariant map is a function f : S T such that f(a x) =a f(x) for all a G, x S. 55

3 15.7 Note. For a given group G the collection of G-sets and G-equivariant map forms a category Set G Proposition. A G-equivariant map f : S T is an isomorphism of G-sets iff f is a bijection Proposition/Definition. Let S be a G-set and let x G. Theset G x := {a G a x = x} is a subgroup of G. Itiscalledthestabilizer of x (or the isotropy group of x) Definition. Let S be a G-set. The action of the group G on S is transitive if for any x, y S there is a G such that a x = y Proposition. For any transitive G-set S we have an isomorphism of G- sets: S = G/G x where x is any element of S. Proof. We have a map f : G/G x S, f(ag x )=a x Check that f is well defined f is G-equivariant f is a bijection. 56

4 It follows that f an isomorphism of G-sets Proposition. If S is a transitive G-set and x, y S then where y = a x G x = a 1 G y a Proof. For b G y we have (a 1 ba) x =(a 1 b) y = a 1 y = x It follows that a 1 G y a G x.ontheotherhand,ifc G x then (aca 1 ) y =(ac) x = a x = y Thus aca 1 G y and so c = a 1 (aca 1 )a a 1 G y a. This shows that G x a 1 G y a Definition. If S is a G-set then the orbit of an element x S is the set Orb(x) :={y S y = a x for some a G} Proposition. Let S be a G-set. 1) If x S then Orb(x) is the unique subset of S such that x Orb(x) G acts transitively on Orb(x). 2) If x, y S then either Orb(x) Orb(y) = or Orb(x) =Orb(y). 3) If x, y S then Orb(x) =Orb(y) iff x = a y for some a G. Proof. Exercise. 57

5 15.15 Corollary. Every G-set S is a disjoint union of its orbits. For an element x S we have an isomorphism of G-sets Orb(x) = G/G x Proof. This follows directly from Proposition and Proposition Corollary. Let G be a finite group and let S be a finite G-set. 1) For any x S we have Orb(x) G x = G 2) If Orb(x 1 ),...Orb(x n ) are all distinct orbits of S then S = n Orb(x i ) i=1 Proof. 1) By Corollary we have Orb(x) =[G : G x ],sobylagrange stheorem (7.4) G =[G : G x ] G x = Orb(x) G x. 58

6 16 Some applications of G-sets Recall. If G is a finite group and a G then a divides G Theorem (Cauchy). Let G be a finite group. If G is divisible by a prime number p then there exists a G such that a = p. Proof. Let S = {(a 1,...,a p ) a i G, a 1... a p = e} Notice that (a 1,...,a p ) S iff a 1,...,a p 1 are arbitrary elements of G and a p =(a 1... a p 1 ) 1.Iffollowsthat S = G p 1 In particular p divides S. Define an action of Z/pZ on S by cyclic permutations: for k Z/pZ. Notice that: k (a 1,...,a p ):=(a k+1,...,a p,a 1,...,a k ) 1) The number of elements of each orbit of this action divides Z/pZ = p. Therefore for any (a 1,...,a p ) S we have Orb((a 1,...,a p )) = p or Orb((a 1,...,a p )) =1 2) Orb((a 1,...,a p )) =1iff a 1 = = a p.wehavethen 3) Orb((e,..., e)) =1 a p 1 = a 1... a p = e Assume that Orb((e,..., e)) is the only orbit consisting of only one element. Then every other orbit of S has p elements. From part 2) of Corollary we obtain then S = kp +1 for some k. This is however impossible since p divides S. Therefore there is some element (a 1,...,a p ) S such that (a 1,...,a p ) = (e,..., e) and Orb((a 1,...,a p )) =1.Itfollowsthata 1 = e and a p 1 = e. Thus a 1 = p. 59

7 16.2 Definition. Let p be a prime number. A group G is a p-group if order of every element of G is a power of p Proposition. If G is a finite group then G is a p-group iff G = p n for some n. Proof. If G = p n and a G then by (7.7) a divides p n,so a = p r for some r 0. Conversely, assume that G = qm for some prime q = p. Then by Cauchy s Theorem (16.1) thereisanelementa G such that a = q. Therefore G is not a p-group. Recall. If G is a group then the center of G is the subgroup Z(G) ={a G ab = ba for all b G} Note. In general it may happen that Z(G) ={e} (take e.g. G = G T ) Theorem. If G is a finite p-group then Z(G) = {e}. Proof. Consider the action of G on itself by conjugations: G G G, a b := aba 1 Let G = p n.noticethat: 1) If b G then Orb(b) divides p n,so Orb(b) = p r for some r 0. In particular either p divides Orb(b) or Orb(b) =1. 2) Orb(b) =1iff b Z(G). 60

8 As a consequence if Z(G) ={e} then Orb(e) =1and p divides the number of elements in every other orbit. From part 2) of Corollary we obtain then p n = G = kp +1 which is impossible. Therefore Z(G) = {e} Note. Not every p-group is abelian. For example take the group of quaternions Q 8 (see Hungerford p. 33) and the group of symmetries of a square D 4 (see Hungerford p. 25). These group are non-abelian, but Q 8 = D 4 =8,so they are 2-groups Definition. Let S be a G-set. A element x S is a fixed point of the action of G on S if a x = x for all a G. Note. If S is a G-set then x is a fixed point of the action of G iff Orb(x) =1. 61

9 17 The Sylow theorems Recall. If G is a finite group and H G is a subgroup then H divides G. Note. 1) If G is an abelian group and n divides G then there is a subgroup H G such that H = n (homework). 2) This is not true in general when G is a non-abelian group (homework) Definition. Let G be a finite group such that G = p r m where p is a prime and p m. WesaythatasubgroupP G is a Sylow p-subgroup of G if P = p r First Sylow Theorem. If G is a finite group and p is a prime number dividing G then G contains a Sylow p-subgroup. Proof. Assume that G = p r m where p m. Let S be the set of all subsets A G such that A = p r. Define an action of G on S as follows. If A S, A = {a 1,...,a p r} and b G then Notice that b A := {ba 1,...,ba p r} p r m S = = p r p r j=1 p r (m 1) + j j so p does not divide S. AaconsequencethereisanorbitOrb(A 0 ) in S such that p Orb(A 0 ). TakethestabilizerG A0.Wewanttoshowthat G A0 = p r. We have p r m = G = Orb(A 0 ) G A0 62

10 so p r divides G A0.Ontheotherhand,noticethatforanya Gthere is some element b A 0 Orb(A 0 ) such that a b A 0. Ineed, if A 0 = {a 1,...,a p r} then (aa 1 1 ) A 0 = {a,... } Since G has p r m elements and each set b A 0 contains p r elements, thus we must have Orb(A 0 ) m, andconsequently G A0 p r.itfollowsthat G A0 = p r Second Sylow Theorem. If P is a Sylow p-subgroup of a finite group G then for any p-subgroup H G we have for some a G. aha 1 P In particular if P, P are two Sylow p-subgroups of G then P = ap a 1 for some a G. Proof. Assume that G = p r m where p m and let H = p s. Take G/P,thesetofleftcosetsofP. translations: H G/P G/P, The group H acts on G/P by left a (bp ):=(ab)p Since H = p s,foreverybp G/P we have Orb(bP ) = p k for some k s. So, either p divides Orb(bP ) or Orb(bP ) =1. On the other hand G/P =[G : P ]=m, andp m, sotheremustbean orbit whose number of elements is not divisible by p. It follows that there is b 0 P G/P such that Orb(b 0 P ) =1 As a consequence ab 0 P = b 0 P for all a H, i.e. Hb 0 P b 0 P. Since Hb 0 Hb 0 P we obtain Hb 0 b 0 P or equivalently: b 1 0 Hb 0 P. 63

11 17.4 Example. Take the group G T.Wehave: G T =6=3 2. Sylow 3-subgroup of G T : {I,R 1,R 2 } Sylow 2-subgroups of G T : P = {I,S 1 }, P = {I,S 2 }, P = {I,S 3 }. We have: P = R 1 P R 1 1 and P = R 2 P R Third Sylow Theorem. Let G be a finite group such that G = p r m where r>0 and p m. Ifs is the number of Sylow p-subgroups of G then 1) s m 2) s 1(modp) 17.6 Definition. Let G be a group and H be a subgroup of G. Thenormalizer of H in G is the subgroup N G (H) G given by N G (H) ={a G aha 1 = H} 17.7 Note. 1) H N G (H). Infact,N G (H) is the biggest subgroup of G that contains H aitsnormalsubgroup. 2) H G iff N G (H) =G. 3) Let S be the set of all subgroups of G. The group G acts on S by conjugations: G S S, a H := aha 1 If H is a subgroup of G then Orb(H) ={ all subgroups of G conjugate to H } N G (H) =stabilizer of H By (15.16) thisgives: (number of subgroups conjugate to H) = Orb(H) =[G : N G (H)] 64

12 Proof of Theorem Let P be a Sylow p-subgroup of G. By(17.3) allother Sylow p-subgroups of G are conjugate to P,soweget s =(number of subgroups conjugate to P )= [G : N G (P )] Since P N G by Lagrange s Theorem (7.4) wehave G p r m It follows that s m. =[G : N G (P )] N G (P ) =[G : N G (P )] s [N G (P ):P ] P p r Next, let S = {P 1,...,P s } be the set of all Sylow p-subgroups of G. Thegroup P 1 acts on S by conjugations: P 1 S S, a P i := ap i a 1 Notice that: Orb(P i ) divides P 1 = p r,sofori =1,...,s either p divides Orb(P i ) or Orb(P i ) =1. Orb(P 1 ) =1 If i>1 then Orb(P i ) > 1. Indeed, if Orb(P i ) =1for some i>1 then ap i a 1 = P i for all a P 1. Check: in this case P 1 P i is a p-subgroup of G. Since P 1 = P i we would also have P 1 P i >p r which is impossible. As a consequence we get s = S = Orb(P 1 ) + ( all other orbits ) 1 multiples of p Therefore s 1(modp). 65

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