a. Not a group. Only 0 has an inverse so G4 fails.
c. Group. It is clearly closed and
proves associativity. The unity is −1, and the inverse of a is −a − 2. Note that G is also abelian.
e. Not a group. It is not closed: (1 2)(1 3) = (1 3 2) is not in G. Note that ε is a unity and each element is self inverse, so only G1 fails.
g. Group. The unity is 16; associativity from . For inverses and closure —see the Cayley table:
i. Not a group. It is closed (by Theorem 3 §0.3), and associative, and ε is the unity. However G4 fails. If has σn = 2n for all then σ has no inverse because it is not onto.
3. a.First ad = c, a2 = d by the Corollary to Theorem 6. Next ba ≠ b, a, d; and ba = c ⇒ b = ac = a(ba) = (ab)a = 1a = a, a contradiction. So ba = 1. Then bd = a, bc = d, b2 = c. Next, ca = b, cd = 1, c2 = a, cb = d. Finally, da = c, db = a, dc = 1, d2 = b.
5. A monoid is a ...