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15.14 Find Both the Center and the Commutator Subgroup of Z3 × S3

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15.14 Find Both the Center and the Commutator Subgroup of Z3 × S3 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}. −1 −1 The commutator subgroup [G, G] is generated by all g1g2g1 g2 for all g1, g2 ∈ G. Suppose g1 = (a1, σ1) and g2 = (a2, σ2). Then −1 −1 −1 −1 g1g2g1 g2 = (a1, σ1)(a2, σ2)(a1, σ1) (a2, σ2) −1 −1 = (a1, σ1)(a2, σ2)(−a1, σ1 )(−a2, σ2 ) −1 −1 = (a1 + a2 − a1 − a2, σ1σ2σ1 σ2 ). −1 −1 = (0, σ1σ2σ1 σ2 ). −1 −1 All permutations σ1σ2σ1 σ2 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 ker φ = K. Therefore, F/K = F/ ker φ ' 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 15.35 Let φ : G → G0 be a group homomorphism, and let N be a normal subgroup of G. Show that φ[N] is normal subgroup of φ[G]. Solution: Given arbitrary φ(g) ∈ φ[G], by homomorphism property and the normality of N in G, φ(g)φ[N]φ(g)−1 = φ[gNg−1] = φ[N]. Thus φ[N] is normal in φ[G]. 15.36 Let φ : G → G0 be a group homomorphism, and let N 0 be a normal subgroup of G0. Show that φ−1[N 0] is a normal subgroup of G. Solution: For every x ∈ G, the set xφ−1[N 0]x−1 satisfies that φ(xφ−1[N 0]x−1) = φ(x)φ(φ−1[N 0])φ(x−1) ⊆ φ(x)N 0φ(x)−1 ⊆ N 0 since N 0 ¢ G0. So xφ−1[N 0]x−1 ⊆ φ−1[N 0] for every x ∈ G. This shows that φ−1[N 0] is a normal subgroup of G. 15.40 Let N be a normal subgroup of G and let H be any subgroup of G. Let HN = {hn | h ∈ H, n ∈ N}. Show that HN is a subgroup of G, and is the smallest subgroup containing both H and N. Solution: Let us check that HN is a subgroup of G: 1. Closed: Let h1n1, h2n2 ∈ HN. Then n1h2 ∈ Nh2 = h2N. There is n3 ∈ N such that n1h2 = h2n3. So h1n1h2n2 = h1(n1h2)n2 = h1(h2n3)n2 = (h1h2)(n3n2) ∈ HN. Thus HN is closed under multiplication. 2. Identity: Since e ∈ H and e ∈ N, it is true that e = ee ∈ HN. 3. Inverse: If hn ∈ HN, then (hn)−1 = n−1h−1 ∈ Nh−1 = h−1N ⊆ HN. So the inverse of hn is also in HN. 34 Therefore, HN is a subgroup of G. Every subgroup K containing H and N will contain HN. That is, HN is the smallest subgroup that contains both H and N. Finally, every subgroup of G containing H and N must contains the elements in HN. So HN is the smallest subgroup containing both H and N. 35.
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