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Quiz 1 solutions 1. (a) What is a group? A group is a set G with an operation · that satisfies • • • • Closure: a, b ∈ G ⇒ a · b ∈ G Associativity: a, b, c ∈ G ⇒ (a · b) · c = a · (b · c) ∈ G Identity: there is an e ∈ G such that a · e = e · a = a for all a ∈ G Inverses: for every a ∈ G, there is an a−1 ∈ G such that a · a−1 = a−1 · a = e. (b) Suppose N is a subgroup of a group G. When is N normal? (Give just one of the equivalent conditions we discussed in class.) One of: • • • • gN g −1 ⊂ N for all g ∈ G gN g −1 = N for all g ∈ G every left coset is a right coset the product of any two right cosets is a right coset 2. Let G = (Q, +) be the group of rational numbers, with addition as its operation. Let H = (Z, +) be the subgroup of integers. Show that H has infinite index in G. For any rational number q, Z + q is a coset of H in G, and Z + q = Z + r if and only if q − r is an integer. Therefore, Z + 1/2, Z + 1/3, Z + 1/4, . . . are all distinct cosets of H. There are infinitely many of these, so [G : H] = ∞. 3. Suppose that G is a group of order 60, H is a subgroup of order 5, and K is a subgroup of order 12. Show that G = HK. We want to show that o(HK) = 60. We have o(HK) = 60 o(H)o(K) = o(H ∩ K) o(H ∩ K) so we need to show that H ∩ K = {e}. It is a subgroup of H, so o(H ∩ K) is a divisor of 5, and it is a subgroup of K, so o(H ∩ K) is a divisor of 12. The only common divisor of 5 and 12 is 1, so o(H ∩ K) = 1.