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Solutions to assignment 4 p. 53: #9 If g ∈ G and H is a subgroup of G, we claim that gHg −1 is a subgroup of G. First, we show closure: if ghg −1 , gh0 g −1 ∈ gHg −1 , then (ghg −1 )(gh0 g −1 ) = ghg −1 gh0 g −1 = ghh0 g −1 ∈ gHg −1 . Next, we show that it contains inverses: if ghg −1 ∈ gHg −1 , then (ghg −1 )−1 = gh−1 g −1 ∈ gHg −1 . Therefore, gHg −1 is a subgroup. Furthermore, by the cancellation lemma, ghg −1 = gh0 g −1 if and only if h = h0 , so there is a one-to-one correspondence between H and gHg −1 , and o(H) = o(gHg −1 ). Therefore, if H is the only subgroup of G with order o(H), then gHg −1 = H. #3 If m ∈ R and m 6= 0, we define Hm = Hfm,0 = {f1,b fm,0 | b ∈ R} = {fm,mb | b ∈ R} = {fm,b | b ∈ R}. Then Hm Hn = Hfm,0 fn,0 = Hfmn,0 = Hmn . Let ψ : A1 /H → R∗ be the map ψ(Hm ) = m. This is one-to-one and onto, and by the calculation above, this is a homomorphism, so it is an isomorphism. (Alternatively, let φ : A1 → R∗ be the map φ(fm,b ) = m. Then φ is surjective and φ(fm,b fn,a ) = φ(fmn,nb+a ) = mn = φ(fm,b )φ(fn,a ), so φ is a homomorphism. Therefore, by the First Isomorphism Theorem, A1 /H ∼ = R∗ .) p. 64, #1 1. If x, y ∈ R∗ , then φ(xy) = (xy)2 = x2 y 2 = φ(x)φ(y), so this is a homomorphism. The kernel is φ−1 (1) = {±1}. 2. If x, y ∈ R∗ , then φ(xy) = 2xy 6= 2x 2y , so this is not a homeomorphism. 3. If x, y ∈ R, then φ(x + y) = x + y + 1 6= (x + 1) + (y + 1), so this is not a homeomorphism. 4. If x, y ∈ R, then φ(x + y) = 13(x + y) = φ(x)φ(y), so it is a homomorphism. The kernel is φ−1 (0) = {0}. 5. If x, y ∈ G, then φ(xy) = (xy)5 = x5 y 5 = φ(x)φ(y), so it is a homomorphism. The kernel is the set of elements of G with order 1 or 5. p. 64, #2 For all x, y ∈ G, we have φ(x)φ(y) = gxg −1 gyg −1 = gxyg −1 = φ(xy), so φ is a homomorphism. Furthermore, φ is invertible; if ψ : G → G is the map ψ(x) = g −1 xg, then φ ◦ psi = ψ ◦ φ = idG . Therefore, φ is one-to-one and onto, so it is a homomorphism. #5 1. Let g ∈ G. In the previous problem, we showed that the map φ : G → G given by φ(x) = gxg −1 is a homomorphism. Therefore, if a, b ∈ G, then g(aba−1 b−1 )g −1 = φ(aba−1 b−1 ) = φ(a)φ(b)φ(a)−1 φ(b)−1 ∈ G0 Every element of G0 is a product of elements of the form aba−1 b−1 and their inverses, so every element of φ(G0 ) = gG0 g −1 is a product of elements of the form φ(aba−1 b−1 ) and their inverses. 1 But all of these are elements of G0 , so every element of gG0 g −1 is an element of G0 , and G0 is normal. If G is abelian, then aba−1 b−1 = e for every a, b ∈ G, so G0 = h{e}i = {e}. On the other hand, if G0 = {e}, then aba−1 b−1 = e for every a, b ∈ G. Consequently, aba−1 b−1 = e ab = ba, so G is abelian. 2. If N contains G0 and a, b ∈ G, then ab(ba)−1 = aba−1 b−1 ∈ N , so (N a)(N b) = N ab = N ba (#5 of the conditions that imply Hg1 = Hg2 ). On the other hand, if (N a)(N b) = (N b)(N a), then N ab = N ba, and using condition #5 again, ab(ba)−1 = aba−1 b−1 ∈ N . #6 1. We claim that f (H1 ) is closed under the group operation and under inverses. If x, y ∈ f (H1 ), then there are u, v ∈ H1 such that f (u) = x and f (v) = y. Since uv ∈ H1 , we have f (uv) = f (u)f (v) = xy ∈ H1 , so f (H1 ) is closed under the operation. Similarly, we have f (u−1 ) = f (u)−1 = x−1 ∈ H1 , so f (H1 ) is closed under taking inverses. It doesn’t have to be normal; for instance, if f : Z → Sym(∆) is given by f (n) = F1n and H1 = Z, then H1 is normal, but f (H1 ) = {R0 , F1 } is not normal. 2. If x, y ∈ f −1 (H2 ), then f (x), f (y) ∈ H2 . The set f −1 (H2 ) is then closed under multiplication because f (xy) = f (x)f (y) ∈ H2 , and it is closed under inverses because f (x−1 ) = f (x)−1 ∈ H2 . Furthermore, if g ∈ G1 and x ∈ f −1 (H2 ), then f (gxg −1 ) = f (g)f (x)f (g)−1 . If H2 is normal, then f (g)f (x)f (g)−1 ∈ H2 , so gxg −1 ∈ f −1 (H2 ) and f −1 (H2 ) is normal. 2