Midterm Exam. There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! Best 5

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1 Department of Mathematical Sciences Instructor: Daiva Pucinskaite Modern Algebra June 22, 2017 Midterm Exam There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! Best 5

2 1. Prove that the set R with the operation given by is an abelian group. Recall: A set G with an operation is called a group if: x y = x + y (a b) c = a (b c) for all a, b, c G, i.e. is associative, 2 in G exists (unique element) e having the property a e = e a = a for all a G; this element is called the identity element, 3 for each element a G there exists an element a 1 G with the property a a 1 = a 1 a = e; the element a 1 is called the inverse element of a. A group G, is called abelian if a b = b a for all a, b G. We have to prove that R with satisfies 1, 2, 3 and a b = b a for all a, b R. 1 Let a, b, c R, then (a b) c = (a + b + 37) c = a + b c + 37 = a + b + c + 74 We have and a + b + c + 74 (a b) c for all a, b, c R. Therefore is associative. a (b c) = a (b + c + 37) = a + b + c = a + b + c + 74 = a + b + c + 74 a (b c) 2 If there is an element e R, such that e a = a for all a R, then e a = e + a + 37 = a e = 37 Since for any a R, we have e a = e + a + 37 = a if e = 37, the real number e = 37 is the identity element in R, w.r.t.. 3 Let a R. If there is an element b R, such that b a = e = 37, then Obviously 74 a R for each a R, and ( 74 a) + a + 37 b a b a = b + a + 37 = 37 b + a = 74 b = 74 a = a + ( 74 + a) + 37 = 37 = e a b for each a R. Thus a 1 = 74 a is the inverse element of a for each a R w.r.t. ( ) Let a, b R, then a b = a + b + 37 = b + a + 37 = b a. Thus R with is an abelian group.

3 2. Prove that the set of integers that are multiples of 9 is a subgroup of the set of integers under addition, i.e. H = {9 z : z Z = {..., 27, 18, 9, 0, 9, 18, 27,... is a subgroup of Z, +. A non-empty subset H of a group G, is called a subgroup if I. H is closed under, i.e. for any two elements a, b from H, the element a b is in H. II. H is closed with respect to inverses, i.e. if a is in H, then a 1 have to be in H. Since 9 1 = 9 H implies that H is not a non-empty set, we have to prove that H satisfies I. and II, w.r.t. +. I. Let a and b be some elements from H, i.e. a = 9 x and b = 9 y for some x, y Z. (In order to show that a + b H, we have to show, a + b = 9 z for some z Z) a + b = 9 x + 9 y = 9 (x + y) (since x, y Z, we have z = x + y Z) = 9 z for z = x + y Therefore a + b H for all a, b H. Thus H is closed under +. II. Let a be some element from H, i.e. a = 9 x for some x Z. (In order to show that a 1 H, we have to show, a H, because the identity element in Z w.r.t. + is e = 0, and the inverse element of a w.r.t. + is a) a = 9 x = 9 ( x) (since x Z, we have x Z) = 9 z for z = x Therefore a 1 = a H for all a H. Therefore H is closed with respect to inverses. [That is what we needed to show]

4 3. Consider the cycles α = (1, 5, 4, 6) and β = (2, 5, 1) in S 6. a. Write α 2 β 1 as a product of disjoint cycles. α 2 β 1 = (1, 5, 4, 6) (1, 5, 4, 6) (1, 5, 2) = = (1, 6, 5, 2, 4) ( b. Express α 2 β 1 as a product of transpositions of the form (3, 1), (3, 2), (3, 4), (3, 5), (3, 6). ) Recall. Consider S n: (1) Any cycle in S n is a product of transpositions: (a 1, a 2,..., a m 1, a m) = (a 1, a m) (a 1, a m 1) (a 1, a 2). (2) If a 3 and b 3, then the transposition (a, b) in S n is a product of transpositions of the form (3, i) where i {1,..., n: (a, b) = (3, a) (3, b) (3, a). (3) For all transpositions we have (x, y) (x, y) = (identity of S n). α 2 β 1 b. = (1, 6, 5, 2, 4) (1) = (1, 4) (1, 2) (1, 5) (1, 6) (2) = (3, 1) (3, 4) (3, 1) (3, 1) (3, 2) (3, 1) (3, 1) (3, 5) (3, 1) (3, 1) (3, 6) (3, 1) (1,4) (1,2) (1,5) (1,6) (3) = (3, 1) (3, 4) (3, 2) (3, 5) (3, 6) (3, 1) c. Give an explanation for the statement The group S 6, is generated by transpositions (3, 1), (3, 2), (3, 4), (3, 5), (3, 6). (In order to explain that The group S 6, is generated by (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), we have to bring forward the argument that (3, 1), (3, 2), (3, 4), (3, 5), (3, 6) are elements in S 6 (that s obvious), and that each permutation in S 6 can be written as a product of some transpositions of the form (3, 1), (3, 2), (3, 4), (3, 5), (3, 6).) Let f be some permutation in S 6. If f is the identity in S 6, then f = (3, 1) (3, 1), thus f is product of transpositions of the form we wish. If f is not the identity, then f can be written as a product of disjoint cycles. Each cycle can be written as a product of transpositions (for example in sense of (1)). ( ) If for a transpositions (a, b) in each of those cycles we have a 3 and b 3, then (a, b) = (3, a) (3, b) (3, a), i.e. (a, b) can be written as a product of transpositions having form we wish. ( ) If for a transpositions (a, b) in each of those cycles we have a = 3 or b = 3, then the transposition (a, b) is already of the form (3, i) (if a = 3, then (a, b) = (3, b), and if b = 3, then (a, b) = (a, 3) = (3, a)).

5 Therefore each cycle can be written as a product of transpositions having form (3, i). Here symbolic representation: (permutation) = ( cycle ) 1 product of transpositions (3, ) ( cycle ) 2 product of transpositions (3, ) Consequently each permutation is a product of some transpositions of the form (3, 1), (3, 2), (3, 4), (3, 5), (3, 6). ( cycle ) m product of transpositions (3, )

6 4. Let, and be the operations on G defined by the operation tables e b c d f g e e b c d f g b b c d f g e c c d f g e b d d f g e b c f f g e b c d g g e b c d f and G, is an abelian group, and G, is non-abelian group a. Show that G, is generated by b, i.e. b = G w.r.t.. The subgroup of G w.r.t.. generated by b is b = b, b2, b 3 c d, b 4 e b c d f g e e b c d f g b b e g f d c c c d e b g f d d c f g b e f f g d c e b g g f b e c d, b 5, b 6 f g e Since each element of G, is a power of b, the group G, is generated by b. b. Show that b is not the only element which generates G,. The subgroup of G w.r.t.. generated by g is g = g, g2, g 3 f d, g 4 c, g 6 b e, g 5 Since each element of G, is a power of g, the group G, is generated by g. Thus g is the second element in G, that generates G,. c. Show that b does not generate G,. Since b b = e, we have b n = e if n is even, and b n = b if n is odd w.r.t.. Thus there is no n such that b n = c, therefore b G w.r.t.. d. Find two elements which generate G,. The elements b and c generate G, because b, c = e, b, c, d b b b c c b, f b c b, g = G i.e. each element in G can be written in some way as a product of some powers of b and c w.r.t.. b c

7 5. Let G, be a group, and a G. Show that the function is bijective. f : G G given by f(x) = a x Recall: A function F : G G is bijective, if F is injective and F is surjective. (i) F is injective, if for all g, g G the equation F (g) = F ( g) is true only if g = g. (s) F is surjective if for each g G there exists c G such that F (c) = g. Proof for f is bijective: (i) f is injective: Let g and g be some elements in G, such that f(g) = f( g). f(g) = f( g) a g f(g) = a g f( g) a 1 (a g) = a 1 (a g) where a 1 is the inverse element of a (a 1 a) g = (a 1 a) g (the associativity of ) (a 1 a) g = (a 1 a) g (the property of the identity element e) e e g = g. Since f(g) = f( g) implies g = g the function f is injective. (s) f is surjective: Let g G, then for some x G we have f(x) = g if f(x) = g a x = g f(x) a 1 (a x) = a 1 g (a 1 a) x = a 1 g e x = a 1 g x = a 1 g Since x = a 1 g G, and f(a 1 g) = a (a 1 g) = (a a 1 ) g = e g = g for every g G, we have that f is surjective. The function f : G G given by f(g) = a g is bijective, because it is injective and surjective.

8 5. Let G, be a group, and g G. (a) What it means when we write ord(g) = 1 234? This means that g i = g g g i times where e is the identity element in G. e for any i {1, 2, 3,..., 1 233, and g = g g g = e, times (b) Let ord(g) = 1 234, and G = g. Prove: If m divides 1 234, then there exists an element in G having order m. Since ord(g) = 1 234, and G = g we have G = g = {e, g, g 2, g 3,..., g i,..., g with g i e for all i {1, 2, 3,..., If m divides 1 234, then m = n is a natural number, and 1 n Thus G = g = {e, g, g 2, g 3,..., g n 1, g n, g n+1,..., g We will prove ord(g n ) = m. Thus we have to show: For each j {1, 2,..., m 1 we have (g n ) j = g n g n g n j times e, and (g n ) m = g n g n g {{ n = e m times If 1 j m 1, then 1 n j < 1 234, because n j = m Therefore j = j m and 0 < j m < 1. (g n ) j = g n g n g {{ n = g n j e, because g i e for any 1 i < j times Furthermore (g n ) m = g n g n g {{ n = g n m = g m m = g = e, since ord(g) = m times This is what we needed to show. (c) For each divisor d of 8 find an element in Z 8 of order d. The group Z 8, + is generated by 1, i.e. Z 8 = 0, 1, 2, 3, 4, 5, 6, The divisors of 8 are m 1 = 1, m 2 = 2, m 3 = 4, m 4 = 8. Thus ( ) ord(1 8 1 ) = ord(8) = ord(0) = 1 = m 1, because 0 is the identity element in Z 8, + ( ) ord(1 8 2 ) = ord(1 4 ) = ord(4) = 2 = m 2, because 4 0, and 4 2 = 8 = 0 ( ) ord(1 8 4 ) = ord(1 2 ) = ord(2) = 4 = m 3, because 2 0, 2 2 = 4 0, 2 3 = 6 0, and 2 4 = 8 = 0 ( ) ord(1 8 8 ) = ord(1) = 8 = m 4, because 1 i = i 0, for all i {1,..., 7 and 1 8 = 8 = 0.

9 6. Consider the group Z 49, +, and m Z 49 (a) Show: If gcd(49, m) = 1, then ord(m) = We will show: If m x = 49 y for some x, y N, with x 49, then x = 49, and y = m. Fundamental theorem of arithmetic (FTA). Every positive integer n > 1 can be represented in exactly one way as a product of prime numbers (up to order of the prime numbers). (Thus if p 1p 2 p r n = q 1q 2 q r n then for any j there is some i such that q j = p i.) and p 1, p 2,..., p r, q 1, q 2,..., q r are prime numbers, Since gcd(49, m) = 1, and 49 = 7 7, we have that in the prime decomposition m = p 1 p 2 p l (here p 1, p 2,... p l are prime numbers), p i 7 for all i (otherwise gcd(49, m) 1). If m x = 49 y, i.e. p 1 p 2 p l x m x = 7 7 y 49 y we should have x = 7 7 (natural number), because in the prime decomposition of the number m x we have p 1 p 2 p l x = 7 7 (prime decomposition of y). m x 49 y Since we are looking for x 49, we have that x = 7 7 (natural number) 49 if (natural number) = 1, thus x = 7 7 = 49 (and therefore m 49 = 49 y when y = m). 2. We will show: ord(m) 49. Since ord(m) = r is the smallest positive integer with the property m r = m + m + {{ + m = 0 r times and m 49 = m + m + {{ + m = 49 m = {{ = {{ + 0 = 0, 49 times m times m times we have that r should be less or equal to 49, i.e. r = ord(m) We will show: ord(m) = r = 49. If i = 0 in Z 49, then i = 49 y for any y Z. Since m r = m + {{ + m = 0, we have r times m r = 49 y Since r 49 (see 2.), and by the assumption gcd(49, m) = 1, the part 1. implies that the equation m r = 49 y is true if r = 49.

10 (b) Determine all subgroups of Z 49. (You can use the fact that every subgroup of a cyclic group is cyclic) Since Z 49 is cyclic, every subgroup of Z 49 is generated by some element in Z 49. Thus the set of subgroups Sub(Z 49 ) is Sub(Z 49 ) = { 0, 1, 2,..., 48 Since gcd(49, n) 1 only if n = 7 (natural number), we have gcd(49, m) = 1 for any m {1, 2,..., 48 such that m 7, 14, 21, 28, 35, 42. Thus ord(m) = 49 for all m {1, 2,..., 48 with m 7, 14, 21, 28, 35, 42. Therefore the subgroup of Z 49 generated by m has 49 elements. Since m is a subset of Z 49, and Z 49 has 49 elements we have m = Z 49 for all m {1, 2,..., 48 with m 7, 14, 21, 28, 35, 42. { Since 7 = 7, 7 2, has 7 elements, and 7 is a prime number, =0 we have that each non-identity element in 7 generates 7. Therefore 7 = 14 = 21 = 28 = 35 = 42 Thus Z 49 has three subgroups Sub(Z 49 ) = { 0, 1, 7. Moreover, (c) Draw the subgroup diagram.