D-MATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups

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1 D-MATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition being composition of functions, that is isomorphic to the symmetric group S 3. Solution The function f has order 2 in the law of composition of functions, in fact f(f(x)) = 1/f(x) = 1/(1/x) = x, similarly the function g has order 3: g 2 (x) = g(x) 1 g(x) = ((x 1)/x) 1 (x 1)/x = 1 x 1 g 3 (x) = g 2 (g(x)) = 1 g(x) 1 = 1 (x 1)/x 1 = 1 Moreover the equality f g 2 = g f holds: 1 x = f g 2 = g f = 1/x 1 1/x 1/x = x = x(1 x)/x = 1 x. This implies that the group generated by f and g has the same multiplication table as S 3, hence the homomorphism defined by sending f to (12) and g to (123) gives an isomorphism of the two groups. 2. In the additive group R m of vectors, let W be the set of solutions of a system of homogeneous linear equations AX = 0. Show that the set of solutions of an inhomogeneous system AX = B is either empty, or is an additive coset of W. Solution A vector space V is an abelian group that is endowed with the additional structure of scalar multiplication. Any linear map between vector spaces is a group homomorphism, that is also compatible with the additional structure. Let us now fix the matrix A representing an endomorphism of the vector space V. And let W < V denote the image of A, K its kernel. The First Isomorphism Theorem says that a surjective map A : V W whose kernel is K, induces a group isomorphism V/K W. Since this is the case, the space W corresponds bijectively to the left cosets of K. Indeed since the group V is abelian, left cosets and right cosets are equal and any subgroup of V is normal. The inhomogeneous equation AX = B admits solution if and only if the vector B belongs to the image W, and in that case the set of solutions consist of the vectors that can be written as X 0 + v where v is in the kernel K of A. This shows that the solutions of AX = B are left cosets of K and any left coset of K correspond to the solution of AX = B for an element B in W. 1

2 3. Prove that every subgroup of index 2 is a normal subgroup. Find an example of a subgroup of index 3 that is not normal. Solution Let N be a subgroup of index two in G. Since the left cosets of N form a partition of G consisting of exactly two elements, the coset of N distinct from N consists of all the elements of G that are not in N. The same is true for the right cosets, that, again, are N and N c. We want to show that for any element g G, gng 1 = N. We distinguish two cases. Let us assume first that g belongs to N. In that case for any element n N the element gng 1 is still an element of N since N is a group, hence gng 1 = N. Let us now assume that g doesn t belong to N, then the left coset gn is not N since it contains the element g that, by assumption, doesn t belong to N and hence consists of all the element of G that are not in N. The same is true for the right coset, in particular we get that gn = Ng. This implies that for any n in N there is n N such that gn = n g multiplying on the right g 1 we get gng 1 = n and this implies that gng 1 = N. In order to give an example of a subgroup of index 3 that is not normal, let us consider the symmetric group S 3 and the subgroup H generated by the permutation h = (12). The group S 3 has order 6, and the group H has order 2, since the transposition (12) has order 2. In order to show that H is not normal let us consider the permutation g = (13): ghg 1 = (13)(12)(13) = (132). Since ghg 1 do not belong to H, the subgroup H is not normal. It is interesting to compute the left and the right cosets of the group H and see that they give two different partitions of G: Left cosets Right cosets eh = {e, (12)} He = {e, (12)} (13)H = {(13), (123)} H(13) = {(13), (132)} (23)H = {(23), (132)} H(23) = {(23), (123)}. 4. Let H < G be a finite index subgroup, then there exist a subgroup N of H that is normal in G and has finite index. Solution Let us denote by n the index of H in G and let eh = g 1 H, g 2 H,... g n H be the left cosets of H in G. Let us now consider the intersection n N = g i Hg 1 i. i=1 N is a subgroup of H, since the intersection of groups is a group and since g 1 H = H hence g 1 belongs to H and g 1 Hg 1 1 = H. 2

3 Let us now check that N is normal in G. Let g be an element of G. The left multiplication by g induces a permutation of the left cosets of H, in particular there is a permutation σ S n and elements h i in H such that gg i = g σ(i) h i. This gives that g(g i Hg 1 i )g 1 = g σ(i) Hg 1 σ(i), hence gng 1 = N. To show that N is finite index we will prove the following Lemma: Lemma: If A and B are subgroups of index a and b in G, then A B is a subgroup of index at most ab. Proof: Indeed the cosets of A B give a partition of G, but since A B is a subgroup of A the left cosets of A B that are contained in A give a partition of A. We claim that this partition has at most b elements: indeed the cosets of B in G give a partition of G consisting of b elements and (gb) A is either empty or a coset of A B in A. Since the same is true for any coset ha of A in G, we have that for any coset of A there are at most b cosets of A B covering it, and since there are a cosets of A in G there are at most ab cosets of A B in G. End of the proof. Notice that the index of A B can be smaller then ab, this happens for example when A = B. By induction this implies that the intersection H = n i=1 H i of the finite index subgroups H i of index h i has index at most h 1... h n, in particular has finite index. This applies in particular to N that is hence a normal subgroup of G of finite index that is contained in H. 5. Determine the integers n for which the pair of congruences 2x y 1 and 4x+3y 2 modulo n has a solution. Solution If the system of equations { 2x y 1 4x + 3y 2 are satisfied modulo n, we get, subtracting two times the first equation to the second one that 5y 0 modulo n. This implies that n must be odd. Indeed let us assume by contradiction that n is even and there is a solution for both equations, we get from the first equation that y must be odd, but since 5y must be zero, y must be even that gives a contradiction. We will now show that if n is odd there is always at least a solution. Indeed, since n is odd, 2 is invertible modulo n, this implies that there exists a number x such that 2x = 1 + n. The pair (x, n) solves both equations modulo n. 6. Prove the Chinese Remainder Theorem: let a, b, u, v be integers, and assume that the greatest common divisor of a and b is 1. Then there is an integer x such that 3

4 x u modulo a and x v modulo b. Find a counterexample if a and b are not coprime. Solution Let us consider first the case u = 1, v = 0. In this case we look for a number x that satisfies x = 1 am and x = bn. Finding x amounts to find integers m and n such that 1 = am + bn. This problem is known as Bezout s Lemma. Bezout s Lemma: if a and b are coprime there exist integers m and n such that an + bm = 1. Proof of Bezout s Lemma: let us consider the smallest positive integer d that can be written as an+bm for a pair of integers n, m. We want to show that d = 1. Assume by contradiction that d > 1. Let n and m be numbers such that d = na+mb. Since a and b are coprime, and d is bigger than 1, there is at least one between a and b that is not exactly divisible by d, let us assume that a is such number and let us consider the reminder c of the division of a by d. The number c is strictly smaller than d, but positive moreover c = a kd = a k(na + mb) = (1 kn)a + kmb. And this gives a contradiction. End of the proof. Let us now consider the element x = ubm + van, where m and n are integers given by Bezout s Lemma. We claim that x is a solution for the problem. Indeed x = u(1 an) + v(an) = u + a(vn un) u modulo a. Similarly x = v(1 bm) + u(bm) = v + bm(u v) v modulo b. 7. In each of the following cases determine whether or not G is isomorphic to the product group H K (a) G = R, H = {±1}, K = {positive real numbers}. Solution Yes: indeed R is abelian, both K and H are normal subgroups, obviously they intersect only in 1, that is the identity of R, and any real number can be written as hk with h H, k K. (b) G = {invertible upper triangular 2 2 matrices}, H = {invertible diagonal matrices}, K = {upper triangular matrices with diagonal entries 1}. Solution No: we will first show that the natural map α : H K G (h, k) hk doesn t give a group isomorphism. Indeed if that was true, for any element h in H and k in K, since by the group law in the product group we have (0, k)(h, 0) = (h, k), we would get that kh = α(0, k)α(h, 0) = α(h, k) = hk. But this is not true, indeed, for example [ 2 0 ] [ ] 1 2 = [ 2 4 ] [ 2 2 ] = [ 1 2 ] [ ] 2 0. Anyway it would still be possible that H K was isomorphic to G under a different isomorphism. For example, if G = Z 2 xz 2 and H = K = Z 2 x0 then 4

5 HxK and G are isomorphic but not under the natural isomorphism given by inclusion of H and K in G, since in that case the inclusions induce an homomorphism that is not injective nor surjective. We will now show that this is not the case in the problem 7b): the map β : R R t imes [ H ] a 0 (a, b) 0 b [ ] a 0 gives a group isomorphism: the subgroups H 0 = { a R } and [ ] 1 0 H 1 { a R 0 a } are normal subgroups of H whose product is the whole H and intersect in the trivial group. The group K, instead is isomorphic to R + : it is easy to check that the map β : R + [ K ] 1 a a gives a group isomorphism. Let us now assume by contradiction that there exists an abstract isomorphism H K = G. Then we would also get an isomorphism G = H 0 H 1 K = R R R +. Since both R and BR + are abelian and the product of abelian groups is abelian we would get that the group G would be abelian. But this is not true and gives a contradiction. (c) G = C, H = {unit circle}, K = {positive real numbers}. Solution Yes: since C is abelian, both K and H are normal subgroups. Moreover the two subgroups intersect only in the identity of C. The fact that the product KH is the whole C follows from the fact that every complex number has an expression in polar coordinates, that is as a product of a positive real number and a number in the unit circle. 8. Let H = {±1, ±i} be the subgroup of C of fourth roots of unity. Describe the cosets of H in G explicitly. Is G/H isomorphic to G? 5