(1) Let G = R \ {1}. the set of all real numbers except 1. Show that G, together with the operation * given by x * y = x + y = xy for all x, y Є G, is a group. [5: 1 for each of the four axioms, and 1 for the conclusion] Hint: See Question 3.2 in the Course Notes for an example of how to write this down formally. See also Exercise 7 in Section 4.7 of [Groups, C. R. Jordan and D. A. Jordan], available online via the Library Question 3.2. Let G = Z. Show that G, together with the operation x * y = x + y – 1, forms a group. Answer: We check the axioms: (G1) (Closure) For all x, y = Z, x * y Є Z. We can't work with the elements one-by-one, so we start with 'Let x, y = Z' and show the result directly. Proof of (G1): Let x, y Є Z. Then x * y = x + y − 1 € Z since x, y and 1 are all integers. (G2) Associativity: For all x, y, z € Z, (x * y) * z = x * (y * z). Similarly, since (G2) is a 'For-all' statement, we start by taking x, y, z in Z and then showing what is needed. COURSE NOTES 8 Proof of (G2): Let x, y, z Є Z. Then (x * y) * z = (x + y − 1) +≈ − 1 = x + y + z −2, while x * (y * z) = x + (y + z − 1) − 1 = x + y + z − 2, so (x * y) * z = x*(y * z). (G3) Identity: There exists an element e = Z such that, for all x Є Z, xe = x and e* x = x. Informally, we do some experimentation: we want x+y−1 = x, so we need to take y = 1. But we must write out a formal proof: Proof of (G3): Let e = 1, and let x E G. Then xe = x*1 = x+1−1 = x and ex = 1 x = 1 + x − 1 = x. Notice that in order to prove the 'For-all x = G' part of the statement we start with 'Let x = G'. (G4) Inverses For all x € Z, there is an element y ≤ Z such that x × y = e and y* x = e. - Informally: given x, we need to solve x + y − 1 y = 2-x. = 1, so we should take Proof of (G4): Let x Є Z. Then set y = 2− x. We have x*y = x+y−1= x + (2 − x) − 1 = 1 = e and y × x = y + x − 1 = (2 − x) + x − 1 = 1 = e. Since (G,*) satisfies all four group axioms, (G, *) forms a group as required. ☐