Download Center and Commutator Subgroups of Z3 × S3 - Prof. Huajun Huang and more Assignments Abstract Algebra in PDF only on Docsity! 15.14 Find both the center and the commutator subgroup of Z3 × S3. Solution: The center subgroup of G := Z3 × S3 is Z(G) = {g ∈ G | gx = xg for all x ∈ G} = Z3 × {e}. The commutator subgroup [G,G] is generated by all g1g2g−11 g −1 2 for all g1, g2 ∈ G. Suppose g1 = (a1, σ1) and g2 = (a2, σ2). Then g1g2g −1 1 g −1 2 = (a1, σ1)(a2, σ2)(a1, σ1) −1(a2, σ2)−1 = (a1, σ1)(a2, σ2)(−a1, σ−11 )(−a2, σ −1 2 ) = (a1 + a2 − a1 − a2, σ1σ2σ−11 σ −1 2 ). = (0, σ1σ2σ−11 σ −1 2 ). All permutations σ1σ2σ−11 σ −1 2 for σ1, σ2 ∈ S3 generate the commutator subgroup [S3, S3] = A3 of S3. So the commutator subgroup of G = Z3 × S3 is [G,G] = [Z3 × S3, Z3 × S3] = [Z3,Z3]× [S3, S3] = {0} ×A3. 15.19 Solution: a. T. b. F. e.g. S3 is noncyclic, but S3/A3 ' Z2 is cyclic. c. F. 0.5 + Z is an element of R/Z that has order 2. d. T. e. F. R/Z only has two elements of order 4, i.e. 0.25+Z and 0.75+Z. f. T. g. F. If G/H is abelian, then H contains the commutator subgroup C of G. h. F. G must be nonabelian. i. T. j. F. An for every n ≥ 5 is a nontrivial finite simple group, but it has the order |An| = n!/2, which is not a prime. 15.20 Let F be the additive group of all functions mapping R into R. Let K be the subgroup of F consisting of the constant functions. Find a subgroup of F to which F/K is isomorphic. 32 Solution: Let F0 = {f ∈ F | f(0) = 0}. Then F0 is a group. Define a mapping φ : F → F0 by φ(g)(x) := g(x) − g(0). Then φ is a surjective group homomorphism with kerφ = K. Therefore, F/K = F/ kerφ ' Imφ = F0. 15.22 Let F be the additive group of all functions mapping R into R. Let K be the subgroup of continuous functions in F . Can you find an element of F/K having order 2? Why or why not? Solution: You cannot find an element f + K of F/K having order 2. Because if f + K has order 2, then f /∈ K but 2f ∈ K. So 2f is a continuous function. So f is a continuous function. So f ∈ K. A contradiction! 15.28 Give an example of a group G having no elements of finite order > 1 but having a factor group G/H, all of whose elements are of finite order. Solution: For example, take G := Z and H := 2Z. 15.30 Describe the center of every simple a. abelian group Solution: Let G be a simple abelian group. Then Z(G) = G since G is abelian. Notice that the simple abelian group G must be of the form Zp for some prime p. b. nonabelian group. Solution: Let G be a simple nonabelian group. Then Z(G) = {e} since Z(G) can only be {e} or G by the simplicity of G, but Z(G) 6= G since G is nonabelian. 15.31 Describe the center of every simple a. abelian group Solution: Let G be a simple abelian group. Then [G,G] = {0} since the generators of [G,G] are x+ y+ (−x) + (−y) = 0 for all x, y ∈ G. b. nonabelian group. Solution: Let G be a simple nonabelian group. Then [G,G] = G since the subgroup [G,G] can only be {e} or G by the simplicity of G, but [G,G] 6= {e} since G is nonabelian. 33
