Let (X, *) be a semigroup. Furthermore, for every a and b in X, if a ≠ b, then…

Given: for all a,b in X, if a ≠ b then a*b ≠ b*a. Equivalently, a*b = b*a implies a = b. Use associativity freely.

• Proof of A (a*a = a):

  Assume a*a ≠ a and set x = a*a. Then x ≠ a, so by the given property x*a ≠ a*x. But associativity gives x*a = (a*a)*a = a*(a*a) = a*x, a contradiction. Hence a*a = a for every a.

• Proof of B (a*b*a = a):

  Let z = a*b*a. If z ≠ a then z*a = (a*b*a)*a = a*b*(a*a) = a*b*a = z and a*z = a*(a*b*a) = (a*a)*b*a = a*b*a = z, so z*a = a*z, contradicting the hypothesis that distinct elements do not commute. Therefore z = a, so a*b*a = a for every a,b.

• A brief symmetry remark:

  Applying the same argument with the roles of a and b swapped gives b*a*b = b. Thus both x*y*x = x and y*x*y = y hold for all x,y.

• Proof of C (a*b*c = a*c):

  Let u = a*b*c and v = a*c. Compute u*v = (a*b*c)*(a*c) = a*b*(c*a)*c. Since c*a*c = c (from the identity x*y*x = x), we get u*v = a*b*c = u. Similarly, v*u = (a*c)*(a*b*c) = a*c*a*b*c = a*b*c = u, using a*c*a = a. Thus u*v = v*u = u. If u and v were distinct this would contradict the hypothesis that distinct elements do not commute. Therefore u = v, i.e. a*b*c = a*c for all a,b,c.

• Conclusion:

  All three equalities A, B and C hold for every choice of elements in X. Hence the correct answer is that A, B and C are all true.