Solutions for Assignment 4 Math 402

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1 Solutions for Assignment 4 Math 402 Page 74, problem 6. Assume that φ : G G is a group homomorphism. Let H = φ(g). We will prove that H is a subgroup of G. Let e and e denote the identity elements of G and G, respectively. We will use the properties of group homomorphisms proved in class. Since φ(e) = e, it follows that e H. To show that H is closed under the group operation for G, suppose that a,b H. Since H = φ(g), there exists elements a,b G such that φ(a) = a, φ(b) = b. Then a b = φ(a)φ(b) = φ(ab) φ(g) since ab G. Hence, if a,b H, then a b H. Finally, suppose that a H. Then a = φ(a) for some a G. Now a 1 G and φ(a 1 ) = φ(a) 1 = (a ) 1. Therefore (a ) 1 H. We have proved that H is a subgroup of G. Page 74, problem 8. The answer to problem 25 on page 15 is an answer to this question too. Let G be the group of real numbers under addition and let G be the group of positive real numbers under multiplication. Define φ : G G by φ(x) = e x for all x G = R. Then φ is a bijection, φ(x + y) = φ(x)φ(y) for all x,y G, and hence φ is an isomorphism of G to G. Page 74, problem 14. We assume that G is abelian and that φ : G G is a surjective homomorphism. Suppose that a,b G. Then there exists elements a,b G such that φ(a) = a and φ(b) = b. Since G is abelian, we have ab = ba. Using this and the assumption that φ is a homomorphism, it follows that a b = φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a) = b a Therefore, for any a,b G, we have a b = b a. Therefore G is abelian. Page 75, problem 26. If a G, the definition of σ a : G G is as follows: For any g G, σ a (g) = aga 1. We will use the fact that σ a is an automorphism of G. (This is proved in the solution to question 2 on the sample midterm.) In particular, σ a A(G). (a) Suppose that a,b G. If g G, then we have σ ab (g) = (ab)g(ab) 1 = (ab)g(b 1 a 1 ) = a ( bgb 1) a 1 = a ( σ b (g) ) a 1 = σ a ( σb (g) ). Thus σ ab (g) = σ a ( σb (g) ) for all g G. Therefore, we have σ ab = σ a σ b. This is true for all a,b G.

2 Define ψ : G A(G) by ψ(a) = σ a for all a G. We have and therefore ψ is a group homomorphism. ψ(ab) = σ ab = σ a σ b = ψ(a) ψ(b) (b) Let i G denote the identity map from G to G. Suppose that a G. Then Now, for a,g G, we have Therefore, That is, Ker(ψ) = Z(G). a Ker(ψ) σ a = i G σ a (g) = g for all g G. σ a (g) = g aga 1 = g ag = ga. a Ker(ψ) ag = ga for all g G a Z(G). Page 83, problem 7. Assume that G is cyclic and that N is a subgroup of G. Since G is abelian, N is a normal subgroup of G. Hence we can consider the quotient group G/N. Let a be a generator for G. This element a will be fixed in the rest of this proof. Since a is a generator of G, we have G = {a k k Z } Therefore, every element of G/N will have the form a k N, for some k Z. Furthermore, we have a k N = (an) k. One way to explain that equality is the following. Consider the map φ : G G/N defined by φ(g) = gn for all g G. The map φ is a group homomorphism (as explained in class one day). According to proposition 1 on the Group Homomorphism handout, we have φ(g k ) = φ(g) k for all g G and k Z. Taking g = a gives the above equality. It follows that every element of G/N has the form (an) k. Therefore, every element of G/N is a power of the element an. Therefore, G/N is indeed a cyclic group. Page 83, problem 8. Suppose that G is an abelian group and that N is a subgroup of G. Since G is abelian, N is a normal subgroup of G. If a,b G, we have ab = ba. Therefore, (an)(bn) = abn = ban = (bn)(an)

3 for all a,b G. Thus, for any elements an, bn G/N, we have and therefore the group G/N is abelian. (an)(bn) = (bn)(an) Page 87, problem 2. Let G be the group of real valued functions on the interval [0, 1] under the operation of addition of functions: (f + g)(x) = f(x) + g(x) for all x [0, 1]. We won t bother verifying that G is a group. It is rather straightforward to do that. Define a map φ : G R as follows. For any f G, define φ(f) = f( 1 4 ) The map φ is a homomorphism from G to R. One might call φ the evaluation homomorphism at x = 1. To verify that φ is a group homomorphism, we have 4 φ(f + g) = (f + g)( 1) = 4 f(1) + 4 g(1 ) = φ(f) + φ(g) 4 for all f,g G. Therefore, φ : G R is a homomorphism. The fact that φ is surjective is easy to see. For if r R, let f be the constant function defined by f(x) = r for all x [0, 1]. Then φ(f) = f( 1 4 ) = r. Therefore, by the first homomorphism theorem, we have G/N = R, where N = Ker(φ). To finish the solution, note that N = Ker(φ) = {f G φ(f) = 0 } = {f G f( 1 4 ) = 0 } and so N is precisely the given subgroup of G described in the problem. Page 87, problem 3. Let G be the group of nonzero real numbers under the operation of multiplication. Let G be the group of positive real numbers under multiplication. One knows that every positive real number y is of the form y = x 2, where x is a real number. Define φ : G G by defining φ(x) = x 2 for all x G. Note that if x G, then φ(x) G. The fact that φ is a homomorphism is rather obvious: If x 1,x 2 G, then we have φ(x 1 x 2 ) = (x 1 x 2 ) 2 = x 2 1x 2 2 = φ(x 1 )φ(x 2 ). Also, as noted above, φ is surjective. We have Ker(φ) = {x G φ(x) = 1 } = {x G x 2 = 1 } = {1, 1} which is the subgroup N specified in the problem. Therefore, by the first homomorphism theorem, we have G/N = G, as stated.

4 Page 87, problem 4. This problem concerns the direct product G = G 1 G 2 of two groups G 1 and G 2. The elements of G have the form (a,b), where a G 1 and b G 2. Let e 1 and e 2 denote the identity elements of G 1 and G 2, respectively. Define a map π : G G 2 by π ( (a,b) ) = b for all (a,b) G. We verify that π is a homomorphism as follows. Suppose that g = (a,b), g = (a,b ) are elements of G. Then, by definition, gg = (aa,bb ). We have π(gg ) = π ( (aa,bb ) ) = bb = π(g)π(g ) and so π is indeed a homomorphism. Also, π is surjective because, if b G 2, then (e 1,b) G and π ( (e 1,b) ) = b. The kernel of π is Ker(π) = {(a,b) G π ( (a,b) ) = e 2 } = {(a,b) G b = e 2 } = {(a,e 2 ) a G 1 } which is precisely the subset N defined in this problem. Thus, N = Ker(π) and therefore N is a normal subgroup of G. The first homomorphism theorem implies that G/N = G 2 since π is a surjective homomorphism. Finally, we prove that N = G 1. To see this, define ε : N G by ε ( (a,e 2 ) ) = a for all a G 1. It is clear that ε is a bijection and that ε is a homomorphism. Hence ε is an isomorphism. A. This problem concerns the group G = S 4. For each j {1, 2, 3, 4}, let H j = {f f S 4, f(j) = j} which is easily seen to be a subgroup of S 4 of order 6. We will denote H 4 by H. Suppose that j {1, 2, 3, 4} and that y S 4 satisfies y(4) = j. We will prove that (1) yhy 1 = H j To prove (1), suppose that h H. This means that h(4) = 4. Note that y 1 (j) = 4. It follows that yhy 1 (j) = yh ( y 1 (j) ) = yh(4) = y ( h(4) ) = y(4) = j

5 and so, by definition, we have yhy 1 H j. We have proved the inclusion (2) yhy 1 H j. Thus, we have proved that yhy 1 is a subset of H j. We mentioned before that H j = 6. We also have H = 6. We can define a map f : H yhy 1 as follows: f(h) = yhy 1 for all h H. Since every element of yhy 1 is of the form yhy 1 for some h H, the map f is surjective. the map f is also injective for the following reason: If a,b H (or even just in G), then yay 1 = yby 1 = y 1( yay 1) y = y 1( yby 1) y = a = b. Since f is surjective and injective, it follows that f is a bijection. Therefore, yhy 1 has the same cardinality as H. Thus, like H, yhy 1 has 6 elements. The inclusion (2) is an equality since both sets have 6 elements. Hence (1) is true. Alternatively, we can also argue as follows. Suppose f H j. Then f(j) = j. If y satisfies y(4) = j, then we have y 1 fy(4) = y 1 f(j) = y 1 (j) = 4 and so y 1 fy H. Denote y 1 fy by h. Thus, h = y 1 fy is in H and therefore f = yhy 1 yhy 1 We have proved that H j yhy 1. Combining this with the inclusion (2), we can now conclude (again) that (1) is true. The fact that these four subgroups are different can be verified as follows. Suppose that j,j {1, 2, 3, 4} and j j. Suppose that k,k are the remaining two elements of {1, 2, 3, 4}. Then consider the 3-cycle f = (j k k ) Since f(j) = j, we have f H j. But f(j ) = k j and so f H j. Hence H j H j. Finally, H 1 H 4 = {f S 4 f(1) = 1, f(4) = 4} is a subgroup of S 4 with two elements: i and the 2-cycle (23). That is, H 1 H 4 = {i, (23)}, a subgroup of S 4 of order 2.