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Homework 7 October 21, 2005 Math 521 Direction: This homework is due on October 28, 2005. In order to receive full credit, answer each problem completely and must show all work. 1. Consider the group G = {1, i, −1, −i} under multiplication. For an arbitrary but fixed element a in G, define a map ha : G → G by ha (x) = xa for all x ∈ G. Find the maps hi , h1 , h−1 and h−i . Show that they are permutations on the set of elements in G. Answer: Let G be {1, i, −1, −i}. The Cayley table of G is the following: · 1 i −1 −i 1 i Hence h1 = 1 i 1 i −1 hi = −i 1 i tations on G. 1 1 i −1 −i i i −1 −i 1 −1 −1 −i 1 i −i −i 1 i −1 −1 −i 1 i −1 −i 1 i −1 −i , hi = , h−1 = , −1 −i i −1 −i 1 −1 −i 1 i −i . Since these functions are one-to-one and onto, they are permu−1 2. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa for all x in G. Prove that ha is a permutation on the set of elements in G. Answer: For each a ∈ G, define the map ha : G → G by ha (x) = xa for all x ∈ G. To show ha is a permutation we have to show ha is well-defined, one-to-one and onto. First we show ha is well-defined. Suppose x = y. Then xa = ya. Hence ha (x) = ha (y). Therefore ha is well-defined. Next, we show ha is one-to-one. Suppose x 6= y. Then xa 6= ya. This implies that ha (x) 6= ha (y). So ha is one-to-one. Finally, we show that ha is onto. Let b ∈ G and find an element x ∈ G such that ha (x) = b. This implies that xa = b which yields x = ba−1 . This shows that ha is onto. 3. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa for all x. Show that the set of permutations H = {ha | a ∈ G} is a group under function composition. Answer: Let H = { ha | a ∈ G }. We want to show that H is a group under function composition. If e is the identity of G, then he is the identity of H. This can be seen from the followings: he ha (x) = he (ha (x)) = he (xa) = (xa)e = xa = ha (x). So he ha = ha for any a ∈ G. Similarly, we have ha he = ha for every a ∈ G. Next, we show that if a ∈ G, then ha−1 is the inverse ha . To see this consider ha−1 ha (x) = ha−1 (xa) = (xa)a−1 = x = xe = he (x). Hence ha−1 ha = he and similarly ha ha−1 = he . Therefore ha−1 is the inverse ha in H. Since the function composition is associative, the associativity law holds in H. Thus H is a group consisting of a set of all permutations on G. 4. For each a in the arbitrary group G, define a mapping ha : G → G by ha (x) = xa for all x in G. Show that the permutation group H = {ha | a ∈ G} is anti-isomorphic to the group G. [Note: A function φ : H → G is an anti-isomorphism if and only if φ is one-to-one, onto, and satisfies φ(xy) = φ(y)φ(x) for all x, y ∈ H.] Answer: We want to show G is anti-isomorphic to H. That is, we want to produce an anti-isomorphism from G to H. Consider the mapping φ : G → H defined by φ(a) = ha for all a ∈ G. First, we show φ is well-defined. Suppose a = b. This implies xa = xb for x ∈ G. Hence from the definition of ha , we get ha (x) = hb (x) for all x ∈ G. This yields ha = hb and therefore φ(a) = φ(b). Hence we see that φ is well-defined. Second, we show that φ is oneto-one. Suppose a 6= b. Then xa 6= xb. Hence ha (x) 6= hb (x) and thus ha 6= hb . Therefore φ(a) 6= φ(b). That is φ is one-to-one. Third, we show that φ is onto. Since φ(a) = ha for each a ∈ G, φ is clearly onto. Finally, we show that φ(ab) = φ(b)φ(a) for all a, b ∈ G. For this first we show hab = hb ha . Since hab (x) = xab = (xa)b = hb (xa) = hb (ha (x)) = hb ha (x), we see that hab = hb ha Since φ(ab) = hab = hb ha = φ(b)φ(a), the mapping φ is an antiisomorphism and G and H are anti-isomorphic to each other. 5. For each a in an arbitrary group G, define a mapping φa (x) = axa−1 for all x ∈ G. Prove that φa is an isomorphism from G onto G. Answer: Define φa : G → G by φa (x) = axa−1 for all x ∈ G. First, we show φa is welldefined. Suppose x = y. This implies ax = ay and axa−1 = aya−1 . Hence φa (x) = φa (y). Therefore φa is well-defined. Second, we show φa is one-to-one. Suppose x 6= y. Then axa−1 6= aya−1 and hence φa (x) 6= φa (y). So φa is one-to-one. Third, we show φa is onto. Pick an arbitrary element b in G and find an element x in G such that φa (x) = b. If an element x exists, then axa−1 = b. Hence x = a−1 ba. Since a−1 ba ∈ G, such an element x always exists. Thus φa is onto. Finally, we show φa preserves group operation. Consider φa (xy) for all x, y ∈ G. Since φa (xy) = axya−1 = axa−1 aya−1 = φa (x)φa (y). Therefor φa is an isomorphism from G onto G. [ Recall that φa is known as the inner automorphism of the group G.] 6. Let G be a cyclic group of order 105. Find all subgroups of G. Answer: Since G is a cyclic group of order 105, therefore G =< g > for some g ∈ G, and G ' Z105 . The divisors of 105 are 1, 3, 5, 7, 15, 21, 35, 105. Hence the subgroups of G are: < g 105 >, < g 35 >, < g 21 >, < g 15 >, < g 7 >, < g 5 >, < g 3 > and < g 1 >. Note that < g 105 > is same as < e >. 7. Find an isomorphism from the group of integers (Z, +) to the group of even integers (2Z, +). Answer: We want to find an isomorphism from φ : Z → 2Z. Define φ(x) = 2x for each x ∈ Z. Then φ is one-to-one and also onto. Next show that φ(x + y) = φ(x) + φ(y) which is really easy. Hence Z ' 2Z. 8. Show that U (8) is isomprphic to U (12). Answer: We want to show U (8) ' U (12). Since U (8) = {1, 3, 5, 7} and U (12) = {1, 5, 7, 11} define a mapping φ : U (8) → U (12) by φ= 1 3 1 5 5 7 7 11 . Then clearly π is one-to-one and onto. Next we show φ(xy) = φ(x)φ(y) for all x, y ∈ U (8). U (8) 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 U (12) 1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 7 11 11 7 5 1 We see that if we replace each entry x in the multiplication table for U (8) by φ(x), then we get the multiplication table for U (12). Hence φ satisfies φ(xy) = φ(x)φ(y) for all x, y ∈ U (8). 9. Let C be the set of complex numbers and M= a −b b a | a, b ∈ R . Show that C and M are isomorphic under addition. Answer: We want to show that C ' M . Define a function φ : C → M as φ(a + ib) = a −b b a for all a + ib ∈ C. First, we show that φ is well-defined. Suppose a1 + ib1 = a2 + ib2 . Then we have a1 = a2 a1 −b1 a2 −b2 and b1 = b2 . Therefore = and φ(a1 + ib1 ) = φ(a2 + ib2 ). Thus b1 a1 b2 a2 φ is well-defined. Second, we show φ is one-to-one. Suppose φ(a1 + ib1 ) = φ(a2 + ib2 ). a1 −b1 a2 −b2 Then = . Therefore a1 = a2 and b1 = b2 . This implies that b1 a1 b2 a2 a1 + ib1 = a2 + ib2 . Hence φ is one-to-one. It is easy to see φ is onto. Finally, we show φ satisfies φ(z1 + z2 ) = φ(z1 ) + φ(z2 ) for all z1 , z2 ∈ C. For z1 = a1 + ib1 and z2 = a2 + ib2 consider φ(z1 + z2 ) = φ(a1 + ib1 + a2 + ib2 ) = φ(a1 + a2 + i(b1 + b2 )) a1 + a2 −b1 − b2 = b1 + b2 a1 + a2 a1 −b1 a2 −b2 = + b1 a1 b2 a2 = φ(a1 + ib1 ) + φ(a2 + ib2 ) = φ(z1 ) + φ(z2 ). Therefore φ is an isomorphism from C onto M , and C ' M .