Let S = {0, 1, 2, 3, 4, 5, 6, 7} and ⊗ denote multiplication modulo 8, that is, x ⊗ y = (x * y) mod 8. (a) Prove that ({0, 1}, ⊗) is not a group. (b) Write three distinct groups (G, ⊗) where G is a proper subset of S and G has 2 elements.
GATE 2000 · Discrete Mathematics · Group Theory · medium
Answer: (a) ({0,1}, ⊗) is not a group because 0 has no multiplicative inverse. (b) Three distinct 2-element groups are ({1,3}, ⊗), ({1,5}, ⊗), and ({1,7}, ⊗).
- Part (a): 0 has no inverse, so ({0,1}, ⊗) is not a group: in ({0,1}, ⊗) the identity must be 1 (since 1 ⊗ 0 = 0 and 1 ⊗ 1 = 1). For 0 to be invertible we would need some x in {0,1} with 0 ⊗ x = 1, but 0 ⊗ 0 = 0 and 0 ⊗ 1 = 0; no inverse of 0 exists, so the inverse axiom fails and ({0,1}, ⊗) is not a group
- Part (b): find 2-element groups {1, x} with x ⊗ x = 1: test odd residues: 3 ⊗ 3 = 9 mod 8 = 1, 5 ⊗ 5 = 25 mod 8 = 1, 7 ⊗ 7 = 49 mod 8 = 1. Each pair {1,3}, {1,5}, {1,7} is closed (1 ⊗ x = x, x ⊗ x = 1), associative (inherited from integer multiplication), has identity 1, and gives every element an inverse - so all three are valid groups isomorphic to Z_2