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^{-1}1 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^{-1}x^{-1}) = 1 = (x y)(y^{-1}x^{-1}) 1, so φ(x)φ(y)φ(y^{-1}x^{-1}) = φ(xy)φ(y^{-1}x^{-1})φ(1). But φ(1) commutes, so φ(xy)φ(y^{-1}x^{-1})φ(1) = φ(1)φ(xy)φ(y^{-1}x^{-1}) and hence φ(x)φ(y)φ(y^{-1}x^{-1}) = φ(1)φ(xy)φ(y^{-1}x^{-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