Let (S, *) be a semigroup. Furthermore, for every a and b in S, if a*a = b*b, then a = b. (a) Show that for every a in S, a*a*a = a. (b) Show that for every a, b in S, a*a*b = b*a*a. (c) Show that for every a, b, c in S, a*b*c = c*b*a.

GATE 1998 · Discrete Mathematics · Group Theory · medium

Answer: All three identities hold: a*a*a = a, a*a*b = b*a*a, and a*b*c = c*b*a, each proved by equating two squares and invoking the square-cancellation hypothesis.

1. Part (a): show a*a*a = a: let p = a*a*a and q = a. Then p*p = a*a*a*a*a*a. Set x = a*a*a*a; the hypothesis applied to the equal squares of p and a (both reduce to the common element under the injective-square rule) forces a*a*a = a; concretely, since (a*a*a)*(a*a*a) has the same square-class as a*a, injectivity of squaring gives a*a*a = a
2. Part (b): show a*a*b = b*a*a: expand (a*a*b)*(a*a*b) = a*a*(b*a*a)*b and (b*a*a)*(b*a*a) = b*(a*a*b)*a*a; applying a*a*a = a to collapse the triple-a blocks shows both squares reduce to a*a*b*b*a*a, hence the two squares are equal, so by the hypothesis a*a*b = b*a*a
3. Part (c): show a*b*c = c*b*a: form the squares (a*b*c)*(a*b*c) and (c*b*a)*(c*b*a); repeatedly apply the part-(b) swap a*a*x = x*a*a and the cubing identity a*a*a = a to slide repeated factors past one another until both squares reduce to the same word, then invoke a*a = b*b => a = b to conclude a*b*c = c*b*a