G is a group, not necessarily abelian. We write the operation as juxtapostion and the identity as 1. There is a function φ : G → G such that if abc = def = 1, then φ(a)φ(b)φ(c) = φ(d)φ(e)φ(f). Prove that there exists an element k ∈ G such that kφ(x) is a homomorphism.
Solution
Easy.
x-11 x = 1 = x-1 x 1, so φ(x-1)φ(1)φ(x) = φ(x-1)φ(x)φ(1). Hence φ(1)φ(x) = φ(x)φ(1), so φ(1) commutes with all φ(x), and hence k = φ(1)-1 also commutes with all φ(x).
x y (y-1x-1) = 1 = (x y)(y-1x-1) 1, so φ(x)φ(y)φ(y-1x-1) = φ(xy)φ(y-1x-1)φ(1). But φ(1) commutes, so φ(xy)φ(y-1x-1)φ(1) = φ(1)φ(xy)φ(y-1x-1) and hence φ(x)φ(y)φ(y-1x-1) = φ(1)φ(xy)φ(y-1x-1). So φ(x)φ(y) = φ(1)φ(xy), and hence k φ(xy) = kφ(x)kφ(y), as required.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998