Math 417 Exam 1 (Solutions) Prof. I.Kapovich February 27, 2009 Problem 1.[20 points] For each of the following statements indicate whether it is true or false. You DO NOT need to provide explanations for your answers in this problem. (1) The set {a : a ∈ Z, a ≥ 0} is a subgroup of (Z, +). (2) Every group is abelian. (3) If (M, ·) is a nonabelian monoid, then M is not a group. (4) If m, n, x, y ∈ Z are such that mx + ny = 3 then gcd(m, n) = 3. (5) The set M2(R) of all 2×2-matrices with real coefficients, with respect to the operation of matrix addition, is a group. Answers: (1) False. (2) False. (3) False. (4) False. E.g. 2 · 3 + 1 · (−3) = 3 but gcd(2, 1) = 1 6= 3. (5) True. Problem 2.[20 points] (1) Give an example of a commutative monoid which is not a group. Justify that your example has the required properties. (2) Let (M, ·) be a finite monoid and let a ∈ M be a unit. Prove that there exists n ≥ 1 such that an = e, where e ∈ M is the identity element in M. Solution. (1) For example M = (R, ·) is a commutative monoid which is not a group. Multiplication of real numbers is associative and commutative and there is an identity element, namely 1 ∈ R. Thus M is a commutative monoid. However 0 ∈ R has no multiplicative inverse, and therefore M is not a group. (2) Look at list of all the positive powers of a in M:

a, a2, a3, . . . , am,... Since this is an infinite list of elements from a finite set M, some two of these elements must be equal. That is, there exist 1 ≤ i < j such that ai = aj. Since a is a unit, it has an inverse a−1 ∈ M. Multiply both sides of the equation ai = aj on the left by a−i = (a−1)i and get a−iai = a−iaj =⇒ e = aj−i. By assumption j > i ≥ 1 so that n := j − i ≥ 1 and an = e, as required. Problem 3.[20 points] Find all the integer solutions of the congruence equation 13x ≡ 9 mod 35. 1 2

Show the details of your work. Solution. We have gcd(13, 35) = 1 and therefore 13 is invertible in Z35. We first need to find the multiplicative inverse of 13 in Z35. First, we apply the Euclidean algorithm to express 1 = gcd(13, 35) as an integer linear combination of 13 and 35:

35 = 2 · 13 + 9 13 = 1 · 9 + 4 9 = 2 · 4 + 1 and hence 1 = 9 − 2 · 4 = 9 − 2(13 − 1 · 9) = −2 · 13 + 3 · 9 = −2 · 13 + 3(35 − 2 · 13) = 3 · 35 − 8 · 13.

Thus −8 · 13 + 3 · 35 = 1. Hence −8 · 13 ≡ 1 mod 35 and so −813 = 1 in Z35. Thus −8 is the multiplicative inverse of 13 in Z35. Now take the equation 13x ≡ 9 mod 35, that is 13x = 9 and multiply both sides by −8. We get:

−813x = −89 1x = −72 = −2 x = −2.

Therefore the solution of the equation 13x = 9 is x = −2, that is, the set of all integers x satisfying the original equation is:

{−2 + 35k : k ∈ Z}.

Problem 4.[20 points] Let 1 a H = { : a ∈ }. 0 1 R Prove that H is a subgroup of the group GL(2, R) (where GL(2, R) is the group of all 2 × 2 matrices with entries from R and nonzero determinant, considered with the operation of matrix multiplication; you do not need to prove that GL(2, R) is a group).

Solution. 1 0 First, note that the identity matrix I = ∈ H (by taking a = 0). 2 0 1 3

We now have to check that H is closed under matrix multiplication. Let 1 a A, B ∈ H be arbitrary elements, so that they have the form A = 0 1 1 b and A = for some a, b ∈ . 0 1 R Then 1 a + b AB = ∈ H 0 1 as required since a + b ∈ R. Now check that H is closed under taking matrix inverses. Let A = 1 a ∈ H, where a ∈ , be arbitrary. Then det(A) = 1 and we have 0 1 R 1 −a A−1 = ∈ H 0 1 as required. Thus H is indeed a subgroup of GL(2, R). Problem 5.[20 points] Let α = (2 3 4 5)(1 2)(2 4 5) ∈ S5. (a) Compute α in the two-row notation, find a factorization of α into disjoint cycles and determine the parity of α. (b) Prove that α cannot be represented as a product α = γ1γ2γ3γ4γ5γ6γ7 where γi are 4-cycles.

Solution. (a) We have 1 2 3 4 5 α = = (1 3 4 2 5). 3 5 4 2 1 Thus α is a 5-cycle and hence α is even (since the length 5 of the cycle is odd). (b) A 4-cycle is odd and the product of an odd number of odd permu- tations is odd. Therefore any product of seven 4-cycles is odd. Since α is even, it follows that α cannot be represented as a product of seven 4-cycles.