Assignment 8: Selected Solutions

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1 Section 4.1 Assignment 8: Selected Solutions 1. and 2. Express each permutation as a product of disjoint cycles, and identify their parity. (1) (1,9,2,3)(1,9,6,5)(1,4,8,7)=(1,4,8,7,2,3)(5,9,6), odd; (2) (1,2,9)(3,4)(5,6,7,8,9)(4,9)=(1,2,9,3,4,5,6,7,8), even; (3) (1,4,8,7)(1,9,6,5)(1,5,3,2,9)=(1,4,8,7)(2,6,5,3), even; (4) (1,4,2,3,5)(1,3,4,5)=(1,5,4)(2,3), odd; (5) (1,3,5,4,2)(1,4,3,5)=(1,2)(3,4,5), odd; (6) (1,9,2,4)(1,7,6,5,9)(1,2,3,8)=(1,4)(2,3,8,7,6,5), even; (7) (2,3,7)(1,2)(3,5,7,6,4)(1,4)=(1,7,6,4,3,5,2), even; (8) (4,9,6,7,8)(2,6,4)(1,8,7)(3,5)=(1,4,2,7)(3,5)(6,9), odd. 14. List the elements of A 3 in cycle notation. A 3 = (1, 2, 3) = {(1), (123), (321)} S Let G be a group. For each element a G, let k a : G G be the map k a (g) = ga 1. (1) Prove that each k a is a permutation of G; (2) Prove that K = {k a a G} is a group under composition of maps; (3) Define ϕ : G K by ϕ(a) = k a. Determine if ϕ is an isomorphism. (1) Observe that k a is surjective, since given g G, k a (ga) = (ga)a 1 = g. Also, k a is injective since, if ga 1 = k a (g) = k a (h) = ha 1, then g = (ga 1 )a = (ha 1 )a = h. (2) We show that K Sym(G) is a subgroup (I am using the notation Sym(G) for the group of permutations of G). Indeed, k 1 : G G is the identity map, so K contains the identity. Next, one calculates k 1 a = k a 1: k a k a 1(g) = k a (ga) = (ga)a 1 = g and k a 1k a (g) = k a 1(ga 1 ) = (ga 1 )a = g. Finally, we check that k a k b = k ab : k a k b (g) = k a (gb 1 ) = gb 1 a 1 = g(ab) 1 = k ab (g). (3) First note that ϕ is obviously bijective. The computation in the previous line shows that ϕ(ab) = k ab = k a k b = ϕ(a)ϕ(b) 1

2 2 so ϕ is an isomorphism Let G be the group of the rigid motions of the tetrahetron: Determine G Actually, we can show that G = A 4. Indeed, G is generated by the 3- cycles (123), (124), (134), (234). Therefore, we just need to show that A n, n 3 is generated by 3-cycles. To this end we prove Claim: Any even permutation can be written as the product of 3-cycles. Indeed, any even permutation can be written as product of an even number of transpositions, so we just need to show that any pair of transpositions can be written as a product of 3-cycles. To this end, we consider the two possible cases: (1) Product of disjoint transpositions: (a, b)(c, d) = (a, b)(b, c)(b, c)(c, d) = (a, b, c)(b, c, d); (2) Product of non-disjoint transpositions: (a, b)(b, c) = (a, b, c). This proves the claim. Therefore, G = A 4, so G = A 4, since S 4 / A 4 = [S 4 : A 4 ] = 2, it follows that A 4 = 4!/2 = Let H be the subgroup (1, 2) S 3. (1) Find the distinct left cosets of H in S 3. (2) Find the distinct right cosets of H in S 3. (1) The left cosets are: (123)H = {(123), (13)}. H = {1, (12)} (23)H = {(23), (321)}

3 3 (2) The right cosets are: H = {1, (12)} H(23) = {(23), (123)} H(321) = {(321), (13)}. 7. Let H be a subgroup of the group G. Prove that if two right cosets Ha and Hb are not disjoint, then Ha = Hb. Assume x Ha Hb. Then x = h 1 a = h 2 b for some h 1, h 2 H. Therefore a = h 1 1 x = h 1 1 h 2 b, showing that a Hb and therefore Ha Hb. By a symmetric argument H b H a, so equality holds. 22. Let G be a group of order pq where p and q are primes. Show that any proper nontrivial subgroup of G is cyclic. Let H G be a proper nontrivial subgroup. Then, by Lagrange s theorem, H is either p or q. In particular, H has prime order, so H is cyclic Show that every subgroup of an abelian group is normal. Let G be an abelian group and H a subgroup of G. Then, for h H and g G, ghg 1 = gg 1 h = 1h = h H. Hence, H is normal. 15. If {H λ } λ L is a collection of normal subgroups, for some index set L, then λ L H λ is a normal subgroup. Let h H λ and g G. Then, h H λ for all λ. Since each H λ is normal, ghg 1 H λ for each λ. Thus, ghg 1 H λ as required. 21. Prove that if H and K are normal subgroups of a group G such that H K = {1}, then hk = kh for all h H and k K. Note that hk = kh if, and only if hkh 1 k 1 = 1. But, (since K is normal) (hkh 1 k 1 K, and (since H is normal) h(kh 1 k 1 ) H. Therefore, hkh 1 k 1 H K = {1} proving the result. 32. Let H be a subgroup of G of index 2. (1) Prove that H is normal in G. (2) Prove that g 2 H for all g G. (1) To prove that H is normal, we first observe that G/H = {H, xh} where x is any element of G that is NOT in H. Now, assume g G and h H. If

4 4 g H, then of course ghg 1 H as H is a subgroup. We may therefore consider the case where g / H. To show that ghg 1 H, we show that ghg 1 H = H. Indeed, suppose not. Then ghg 1 H = gh (as g / H is one such x as above). But this means that g 1 (ghg 1 ) = hg 1 = h H. It now follows that g = (h ) 1 h H, a contradiction. We must therefore have ghg 1 H. (2) Since H is normal, G/H is a group of order 2. This means that for every g G, g 2 H = (gh) 2 = H. But this simply says that g 2 H Let H be a normal subgroup of a finite group G. If the order of the quotient group G/H is m, prove that g m H for every g G. This is a generalization of the last problem. Indeed, as G/H = m, gh divides m for every g G. This just means that g m H = (gh) m = H, so g m H. 22. Let H be a normal subgroup of the group G. Prove that G/H is abelian if, and only if a 1 b 1 ab H for every a, b G. The group G/H is abelian if, and only if, abh = ahbh = bhah = bah. This is equivalent to saying ab H ba, or a 1 b 1 ab = (ba) 1 ab H. 32. Let Inn(G) be the group of inner automorphisms of G. It consists of the automorphisms t a : G G (a G) defined by t a (g) = aga 1. Prove that Inn(G) is a normal subgroup of Aut(G). Let a G and β Aut(G). We show that βt a β 1 = t β(a). Indeed, suppose that g G. Then (βt a β 1 )(g) = (βt a )(β 1 (g)) = β(aβ 1 (g)a 1 ) = β(a)β(β 1 (g))β(a 1 ) = β(a)gβ(a) 1 = t β(a) (g). 34. If H and K are normal subgroups of a group G such that G = HK and H K = {1}, then G is said to be the internal direct product of H and K, written G = H K. If G = H K, prove that ϕ : H G/K defined by ϕ(h) = hk is an isomorphism.

5 First, note that ϕ(h 1 h 2 ) = h 1 h 2 K = h 1 Kh 2 K = ϕ(h 1 )ϕ(h 2 ), so ϕ is a homomorphism. Note that if h ker ϕ, then ϕ(h) = hk = K, so h H K = {1} showing that ϕ is injective. To see that ϕ is surjective, note that any g G can be written as g = hk for some h H and k K. Therefore, gk = hkk = hk = ϕ(h). This proves that ϕ is an isomorphism. 5

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