Math 280 Modern Algebra Assignment 3 Solutions

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1 Math 280 Modern Algebra Assignment 3 s 1. Below is a list of binary operations on a given set. Decide if each operation is closed, associative, or commutative. Justify your answers in each case; if an operation does not have a particular property, provide a counterexample. (a) on Z is defined by a b = a + b. This is a closed operation on Z. If we add any two integers, we get another integer. It s also associative and commutative, both of which are properties inherited from the addition of real numbers. (b) on Z is defined by a b = a b. This is a closed operation on Z. The difference of any two integers is another integer. It isn t associative, however: 2 (3 4) = 2 (3 4) = 3 nor is it a commutative operation: (2 3) 4 = (2 3) 4 = = 2 3 = 1 1 = 3 2 = 3 2 (c) on N is defined by a b = a b. This is a closed operation on N: raising one natural number to the power of another will always result in a natural number (remember, exponents are really just a short way of writing products, and multiplying positive integers yields positive integers). As the following counterexample shows, the operation is not associative: 2 (3 4) = 2 (34) = (2 3) 4 = (2 3 ) 4 = 8 4 = 2 12 nor is it commutative: 2 3 = = 3 2 (d) on Z (the set of nonzero integers) is defined by a b = a/b. This is not closed: 2 3 = 2/3 is not an integer. It is associative: a (b c) = a/(b/c) = ac/b = (a/b)/c = (a b) c but not commutative: 2 3 = 2/3 3/2 = 3 2

2 2. Give two reasons why the set of odd integers under addition is not a group. The operation is not closed: 3 and 5 are both odd integers, but their sum is not. In addition, there is no identity; the additive identity in a group of real numbers is 0, but that s not an odd integer. The operation is associative, and there are inverses (if a is odd, so is a), but we do not have a group. 3. Translate each of the following multiplicative expressions into its additive counterpart. Assume that the operation is commutative. (a) a 2 b 3 Recall that a 2 just means apply the operation on a with itself. If we re writing that additively, then a 2 represents a + a. Similarly, b 5 would be b + b + b + b + b, and so on. Replace every product with a + sign: a + a + b + b + b = 2a + 3b (b) (ab 2 ) 3 c 2 = e 3(a + 2b) + 2c = e

3 4. Prove that a group G is Abelian if and only if (ab) 1 = a 1 b 1 for all a, b G. Remember, to prove a biconditional statement, we need to prove both directions: ( ) Suppose G is an Abelian group with elements a and b, and let a 1, b 1 denote the inverses of those elements, respectively. We need to show that (ab) 1 = a 1 b 1 ; that is, we want to show that a 1 b 1 is the inverse of ab. Since and similarly then (ab) 1 = a 1 b 1. (ab)(a 1 b 1 ) = (ba)(a 1 b 1 ) by definition of Abelian = b(aa 1 )b 1 by associativity = b(e)b 1 by definition of inverse = bb 1 by definition of identity = e by definition of inverse (a 1 b 1 )(ab) = (a 1 b 1 )(ba) = a 1 (b 1 b)a = a 1 (e)a = a 1 a = e ( ) Suppose G is a group such that for all a, b G, we have (ab) 1 = a 1 b 1. We know that in any group, (ab) 1 = b 1 a 1 (the shoes-and-socks principle). If we also have (ab) 1 = a 1 b 1, then b 1 a 1 = a 1 b 1 Multiplying both sides of this equation by ab on the left, we have Now multiply both sides on the right by ba: abb 1 a 1 = aba 1 b 1 e = aba 1 b 1 eba = aba 1 b 1 ba ba = ab Since a, b were generic group elements, then G must be an Abelian group.

4 5. Prove that if a and b are elements of a group G, (ab) 2 = a 2 b 2 if and only if ab = ba. (Note: Be careful with assumptions. We cannot assume G is Abelian in either direction ). ( ) Suppose G is a group such that for all a, b G, we have (ab) 2 = a 2 b 2. We can rewrite this expression (carefully - remember we can t assume elements commute) as abab = aabb Now, multiplying both sides on the right by b 1 and on the left by a 1 : a 1 ababb 1 = a 1 aabbb 1 we immediately have as required. ba = ab ( ) Suppose G is a group such that for some a, b G, we have ab = ba. (Note: we cannot assume G is Abelian, since we can t assume ab = ba for all a, b G). Multiplying both sides on the right by ab yields (ab)(ab) = (ba)(ab) (ab) 2 = baab Since we re assuming ab = ba, we can rearrange the terms on the right: as required. (ab) 2 = abab = aabb = a 2 b 2

5 6. A modern algebra professor meant to type a list of six integers that formed a group under multiplication mod 35. However, one of the integers was inadvertently left out, so the list appeared as {1, 6, 11, 16, 31} Which integer was missing from the list? Show that your completed set forms a group under multiplication mod 35 (you may assume associativity holds). We ll do this the hard way first: by figuring out what number is missing by checking closure. This means we need to check that for every a, b in the set, a b = ab mod 35 is also in the set. However, multiplication of real numbers mod n is always commutative, so that at least eliminates the need to check both ab and ba. Letting a b denote the product of a and b mod 35, we have: 1 1 = = = = = = = = = = = = = = = 16 It looks like we have a problem with 6 16 and In order to be a group, the set must contain 26. We have to check, however, that simply including 26 in the set doesn t cause further problems. If a 26 is not in the set for some a, we still won t have closure. However, 1 26 = 26, 6 26 = 16, = 6, = 31, = 1 so we really do have a closed operation on the set {1, 6, 11, 16, 26, 31}. Note that by checking the work we ve already done, we can see that there is an identity (1) and that each element has an inverse: 1 1 = 1, 6 1 = 6, 11 1 = 16, 16 1 = 11, 26 1 = 31, 31 1 = 26 Associativity is inherited, so we do have a group. There is a (slightly) easier method here. Instead of figuring out what is needed to fulfill the closure requirement, we could have checked what we needed to fulfill the inverse requirement. If we first tried to find the inverse of each element, we d have seen that every element had an inverse except 31. Then, by solving the equation 31x = 1 mod 35 (in other words, finding an x so that 31x is one more than a multiple of 35) we d have found the missing element right away. Note that this group is not U(35), since that would contain all numbers that are relatively prime to 35, such as 2, 4, and so on. The group in this problem is a subgroup of U(35).

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