Originally Posted by
mathguy25 Let G be a group and let e be the identity of G. Then for all a,b in G we have that e = aba^3b^2a. Let x be in G. Need to show that x = e. Use the condition where a = x and b = e. Then e = xex^3e^2x = x^5. Then |G| divides 5. Now use the condition where a = e and b = x. Then e = exe^3x^2e = x^3. Then |G| divides 3. Thus, |G| divides gcd(3, 5). Thus, |G| divides 1. Then |G| = 1. Thus, since e is in G, we see that G = {e} and so G is trivial.
Remember, if |G| = n and x^m = e, then n | m. If m = qn + r for some integers q and r where 0 <= r < n, then e = x^m = x^(nq + r) = x^nq * x^r = (x^n)^q * x^r = e^q * x^r = e * x^r = x^r. Then x^r = e. If r > 0, then this contradicts the minimality of |G| = n. Thus, r = 0. Then m = qn. Then n | m.