1 Chapter 6 - Exercise 1.8.cf

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1 1 CHAPTER 6 - EXERCISE 1.8.CF 1 1 Chapter 6 - Exercise 1.8.cf Determine 1 The Class Equation of the dihedral group D 5. Note first that D 5 = 10 = 5 2. Hence every conjugacy class will have order 1, 2 or 5. At first we have that C 1 = 1 and that we are left with 9 to assign. Now we look at the centralizer of D 5. If you consider x j D 5 then you have that it commutes with any other power of x but that yx j y = x 5 j. It follows that we require for the centralizer j 5 j(mod5) that is 5 2j(mod5) which is impossible. Similarly if you take y D 5, then it won t work with any x j as it will require that 5 2j(mod5) again. So the centralizer of D 5 is C 1. Thus all the other conjugacy classes are of order over 2. And since we need them to divide 10 and add up to 9 we are left finally to 10 = Note that C x C x 2 and that C x = C x 2 = 2. On the other hand C y = 5. Determine 1 The Class Equation of the group of upper triangular matrices in GL 2 (F 3 ). GL 2 (F 3 ) is the set of 2 2 invertible upper triangular matrices with elements in {0, 1, 2}. Its order is then = 12 as we have to consider that 0 can not be an element of the diagonal. Remark first that half of the elements are simply twice the other half as in F 3, 2 2 = 1. Then we can reduce the work on one element out of two and just say that the pattern repeat twice. C 1 = 1 represents the identity matrix, and thus twice the identity will give another element of the same kind. a b Now if you consider a random element of the group, say x = then its 0 c c b a abc inverse is given by x 1 = (ac) 1 =. 0 a 0 c ac Applying this on the element we get which could be 1 2 either the original one or, just by noticing in the calculation that 1 2ac a 2 = c 2 = 1 in F 3. Doing similarly on the obtained element we get

2 2 CHAPTER 6 - EXERCISE 1.10.D 2 which turns out to be the same answer as ( before. ) Hence ( this gives ) another conjugacy class of order 2. Similarly will do and This gives as a total = 6 and we are just left with 6 which is either as already ( obtained ) ( or in ) order to keep our symmetry ac But noticing that x 1 x = we see that there should be 2 0 at least three elements associated with to complete the conjugacy class. Hence we have 12 = Chapter 6 - Exercise 1.10.d Determine 1 The centralizer of in GL 3(R). Note first that the identity matrix commutes with every element. study could be reduced to the following. Thus our Write a b c d e f g h j = a b c d e f g h j This reduces to two matrices that give a veryeasy to solve system from which we deduce that the centralizer of 1 in GL 3(R) is 0 a b c Z = 0 a b : a, b, c R, a a 3 Chapter 6 - Exercise 2.12 Prove or disprove 1 A nonabelian simple group can not operate non trivially on a set containing less than 5 elements.

3 4 CHAPTER 6 - EXERCISE Consider such a group G acting on a set S containing less than 5 elements. We know that there is a homomorphism ψ : G P erm(s) describing the action of G. Then, since G is simple, we have that ker(ψ) is either G or {1}. If ker(ψ) = G we are done because it would mean that G acts trivially. So let us consider the case ker(ψ) = {1}. In this case, our homomorphism becomes an injective homomorphism. Hence we can say that G = Im(G) P erm(s). Thus we can write G < S 4. Now we can use the map φ : S 4 {±1} known as the sign function. We get that G ker(φ) G. But as G is simple we have only two cases: G ker(φ) = {e}: This means that every permutation in G is odd and thus that g G, g 2 = e as it would be an even permutation. As a consequence every element in G has order 2 and hence G is abelian. CONTRADIC- TION. G ker(φ) = G: It follows that G ker(φ) = A 4. But we know that A 4 has no simple subgroup from the study of groups of order 12. Hence this is a contradiction. So the action should be trivial. 4 Chapter 6 - Exercise 3.13 H is a normal subgroup of G of order 2. Prove 1 H is in the center of G. We have H = {e, h} where e is the identity element and h 2 = e. Then note first that e is an element of the center of G. We are left to deal with h. Since H is normal in G, we have that g G, ghg 1 H. Then two cases are possible. Either ghg 1 = h and it is an element of the center or ghg 1 = e from which we conclude h = e, contradicting the order of H. Hence only the first case is true and H is an element of the center of G. Let now H be a normal subgroup of G finite of prime order p such that p is the smallest prime dividing G. Prove 1 H is in the center Z(G).

4 5 CHAPTER 6 - EXERCISE Since H is of order p, then it is a cyclic group H =< h h p = e >. Consider the action of G on H by conjugation. Then since H is simple, we have that ghg 1 = H. The orbit of e H is simply itself as the identity commutes with everything. Then, since the orbits are disjoint, the other orbits contain at most p 1 elements. We hence conclude that h H, O h < p but we still require the order of orbit to divide the total order of the group whose smallest prime factor is p. As a consequence we must impose that O h = 1 from which we conclude that G h = G. So every element in G stabilize h H. This means that H is part of the center Z(G). 5 Chapter 6 - Exercise 4.17 Prove 1 There are at most 5 isomorphism classes of groups of order 20. Let us denote the number of Sylow n-subgroups by s n. Also note that 20 = First we consider the Sylow 5-subgroups. Then s 5 divides 4 which gives as possibilities 1, 2 and 4, but also s 5 1(mod5), from which we get finally s 5 = 1. Hence the only Sylow 5-subgroup H is normal in G. On the other hand for the Sylow 2-subgroups we have that s 2 divides 5, which gives as possibilities 1 and 5, and also it verifies s 2 1(mod2), from which we can not conclude anything. If we first consider the case where there is only one Sylow 3-subgroup K, this one should be normal. As a consequence we have that G has two normal subgroups of order 4 and 5. Then there intersection has to be trivial as its order should divide both the order of H and the order of K, as they both are normal. Hence the order of the intersection divides both 4 and 5, which leads to 1. Moreover we have that HK = G. So altogether we get that G = H K, where H = C 5 as it is a normal group of order 5. Similarly K = C 4 or K = C 2 C 2 that are known to be different. So that gives two equivalence classes of G, C 5 C 4 and C 5 C 2 C 2. If there are 5 Sylow 2-subgroups, then those are cyclic of order 4 or the direct product of two of order 2. First consider the cyclic group of order 4. Take two elements x H and y K i so that x 5 = e and y 4 = e. Then we have that, since H is normal in G, yxy 1 = x j for some j < 5. It gives x = y 4 xy 4 = y 3 (yxy 1 )y 3 = y 3 x j y 3 =... = x j4 from which we conclude that x = x j4 and thus j 4 1(mod5) and thus j can take the following values: 1, 2, 3 or 4. This

5 5 CHAPTER 6 - EXERCISE gives 4 cases to analyze. i = 1: This gives yxy 1 = x that is yx = xy and then K is abelian and thus normal in G, case already treated above. i = 2: This gives yxy 1 = x 2 which leads to G =< x, y x 5 = e, y 4 = e, yx = x 2 y >. i = 3: This gives yxy 1 = x 3 which leads to G =< x, y x 5 = e, y 4 = e, yx = x 3 y > which is equivalent to the previous one as y 3 xy 3 = x 33 = x 2 and that y and y 1 could be arbitrarily interchanged. i = 4: This gives yxy 1 = x 4 which leads to G =< x, y x 5 = e, y 4 = e, xyx = y >, which is different from the two previous as y 3 xy 3 = x 43 = x 64 = x 4. If we consider S 2 =< u u 2 = e > < v v 2 = e >, we then have uxu 1 = x i and vxv 1 = x j. It leads in both cases to i 2 1(mod5) and j 2 1(mod5) that is i, j {1, 4}. We then have four cases: i = 1, j = 1: u and x commute and so do v with x. This then reduces to the second abelian group described in the beginning. i = 1, j = 4: this defines a different group with an operation that is gives vx = x 4 v. i = 4, j = 1: this is simply the previous case replacing u for v. As they are arbitrary the isomorphic class is the same. i = 4, j = 4. Then we can write (uv)x(uv) 1 = u(vxv 1 )u 1 = ux 4 u 1 = (uxu 1 ) 4 = x 4 4 = x 16 = x. This is then the same form as the second of this list. Altogether this defines 5 isomorphic classes, two of them abelian as described in the first part, and three non abelian from the previous lists.