Algebra homework 6 Homomorphisms, isomorphisms


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1 MATHUA T.A. Louis Guigo Algebra homework 6 Homomorphisms, isomorphisms Exercise 1. Show that the following maps are group homomorphisms and compute their kernels. (a f : (R, (GL 2 (R, given by For x, y R, x, y 0, we have: f(x f(y ( x 0 Therefore, f is a homomorphism. f(x ( x 0 ( y 0. ( xy 0 f(xy Ker(f f 1 (I 2 {1} since f(x I 2 implies x 1 by identification. (b g : (R, + (GL 2 (R, given by g(x ( 1 x. For x, y R, we have: g(x g(y ( 1 x ( 1 y ( 1 x + y g(x + y Therefore, f is a homomorphism. Ker(g g 1 (I 2 {0} since g(x I 2 implies x 0 by identification.
2 (c h : (R 2, + (R, + given by h(x, y x. For x, y, x, y R, we have: h(x, y + h(x, y x + x h(x + x Therefore, h is a homomorphism. Ker(h h 1 (0 {(0, y y R} since h(x, y 0 implies x 0 by identification. (d The complex conjugation map j : (C, + (C, +, given by j(x + iy x iy. For x, y, x, y R, we have: j(x + iy + j(x + iy x iy + x iy x + x i(y + y j((x + iy + (x + iy Therefore, j is a homomorphism. Ker(j j 1 (0 {0} since j(x + iy x iy 0 + i0 implies x, y 0 by identification. (e k : G G given by k(x x n if G is an abelian group (written in multiplicative notation. What if G is not abelian? For x, y R, we have: k(xk(y x n y n (xy n k(xy by commutativity of the operation on G. Therefore, k is a homomorphism. Ker(G k 1 (1 {x G x n 1} That is, all the elements of order n in G. If G is not supposed to be abelian, then x n y n homomorphism. (xy n in general, thus k is not a
3 Exercise 2. Let φ : G H be a group homomorphism. 1. Show that if G is abelian, then Im(φ is also abelian. Let us suppose that G is abelian, and show that Im(φ is also abelian. Let x, y Im(φ. There exists a, b G such that x φ(a, y φ(b. Then: xy φ(aφ(b φ(ab φ(ba φ(bφ(a yx. by property of φ and commutativity on G. Therefore Im(φ is abelian. 2. Show that if G is cyclic, then Im(φ is also cyclic. Let us suppose that G is cyclic, and denote by g a generator of G. Let s show that Im(φ is also cyclic. Let x Im(φ. There exists a G such that x φ(a. Because G is cyclic, there exists m Z such that a g m. Then: x φ(a φ(g m φ(g m. Therefore Im(φ is cyclic, generated by φ(g. Exercise 3. Let T denote the group of invertible upper triangular 2 2 matrices ( a b A, a, b, d R, ad Show that T is a subgroup of GL 2 (R. 1. For A T, A T GL 2 (R. 2. I 2 T, with parameters a, d 1 and b For a, b, d, e, g, h R, ad 0, eh 0, we have: since ae dh 0., det(a ad 0. Therefore A GL 2 (R. So ( e g 0 h ( ae h T 4. Finally, T is stable by inversion, since for a, b, d R, ad 0, 1 ( a 1 db a 1 T
4 2. Let φ : T R be the map given by sending a matrix A as above to a 2. Show that φ is a homomorphism, and give its kernel and image. For a, b, d, e, g, h R, ad 0, eh 0, we have: φ ( e g φ 0 h Therefore, φ is a homomorphism. We have: since φ Ker(φ φ 1 (1 ( ae a 2 e 2 (ae 2 φ h {( 1 b ( 1 b, ( φ } b, d R, d 0 1 implies a 2 1 by identification, so a ±1. ({ Im(φ φ ( e g 0 h } a, b, d R, ad 0 { a 2 a 0 } R + \{0} since every positive real number can be expressed as the square of a real number. Exercise 4. Show that all the nontrivial subgroups of Z are isomorphic to Z. As mentioned in class, every subgroup of Z is cyclic. That is, for G a subgroup of Z different from {0}, there exists n Z, n such that G n nz. Let us define φ : nz Z by φ(a n 1 a for any a nz. φ is a homomorphism. Indeed, for a, b nz, there exist p, q Z such that a np, b nq. φ(a + φ(b p + q φ(a + b It is bijective since for any c Z, φ(x c if and only if x nc. Therefore, φ is an isomorphism from G to Z. As a consequence, any subgroup G of Z, different from {0} is isomorphic to Z.
5 Exercise 5. Recall that the group (Z/8Z is of order 4. Is it isomorphic to Z/4Z? If not, find another group of order 4 it is isomorphic to. We have: (Z/8Z {1, 3, 5, 7}. Here is its multiplication table: We see that the elements 3, 5 and 7 are all of order 2, whereas Z/4Z is cyclic. If (Z/8Z were isomorphic to Z/4Z, it would have an element of order 4. So these two groups are not isomorphic. However, (Z/8Z is isomorphic to a group of order 4 with multiplication table is given by: 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 An example of such a group is Z/2Z Z/2Z. + (0, 0 (1, 0 (0, 1 (1, 1 (0, 0 (0, 0 (1, 0 (0, 1 (1, 1 (1, 0 (1, 0 (0, 0 (1, 1 (0, 1 (0, 1 (0, 1 (1, 1 (0, 0 (1, 0 (1, 1 (1, 1 (0, 1 (1, 0 (0, 0 You can check (with the multiplication tables for instance that φ : (Z/8Z (Z/2Z (Z/2Z defined as follows is an isomorphism. φ(1 (0, 0. φ(3 (1, 0. φ(5 (0, 1. φ(7 (1, 1. Exercise Show that for any a Z, the map φ : Z Z defined by φ(n an is a group homomorphism. Give its kernel and image. For n, n Z, φ(n + φ(n an + an a(n + n φ(n + n. Therefore, φ is a homomorphism. And we have φ(n 0 iff an 0 iff n 0, if we suppose a 0. So Ker(φ Z if a 0 and Ker(φ {0} if a 0. If a 0 then clearly Im(φ {0}. Otherwise Im(φ az, the multiples of a.
6 2. Conversely, show that a homomorphism φ : Z Z is of the form φ(n an for some a Z. Thus, the homomorphisms Z Z are exactly the maps n an. If φ is such a homomorphism, then for any n Z, we have: φ(n φ( φ(1n an (by induction, with a φ(1 Z. Therefore, every homomorphism from Z to Z can be written n an, for a certain a Z. 3. Determine all the automorphisms of Z, that is, the isomorphisms Z Z. Let φ : Z Z be a automorphism. An automorphism of Z is in particular a homomorphism from Z to Z. As a consequence, there exists a Z such that φ n an. Since φ is a automorphism, Im(φ Z. But we have seen that Im(n an az. Therefore az Z, which implies a ±1. Conversely, n n and n n are automorphisms of Z. Therefore, all the automorphisms of Z are n n and n n. Exercise 7. Let G and H be two groups. Show that G H is isomorphic to H G. Let φ : G H H G be defined as follows: For any (x, y G H, φ(x, y (y, x. φ is a homomorphism since for (x, y, (x, y G H, φ((x, y (x, y (y y, x x (y, x (y, x φ((x, y φ((x, y. φ is bijective since φ(a, b (x, y if and only if a y and b x. Therefore it is an isomorphism from G H to H G. Exercise 8. Are the groups Z/6Z and Z/2Z Z/3Z isomorphic? Justify your answer by either constructing an isomorphism or explaining why it does not exist. Those two group are isomorphic, since they are both cyclic generated by one element (of order 6: Z/6Z 1. Z/2Z Z/3Z (1, 1. You can check that φ : Z/6Z Z/2Z Z/3Z, defined by φ(1 (1, 1 (it is sufficient to know φ completely since 1 is a generator is an isomorphism.
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