Putnam 1997, Problem A-4

Let $G$ be a group with identity $e$ and $\phi: G \to G$ a function such that
$$\phi(g_1)\phi(g_2)\phi(g_3) = \phi(h_1)\phi(h_2)\phi(h_3)$$
whenever $g_1g_2g_3 = e = h_1h_2h_3$. Prove that there exists an element $a \in G$ such that $\psi(x) = a\phi(x)$ is a homomorphism (i.e. $\psi(xy) = \psi(x)\psi(y)$ for all $x,y \in G$).

Solution:
In order to have \( \psi(x) = a \phi(x) \) for all \( x \), we must in particular have this for \( x = e \), and so we take \( a = \phi(e)^{-1} \). We first note that  
\[
\phi(g) \phi(e) \phi(g^{-1}) = \phi(e) \phi(g) \phi(g^{-1})
\]  
and so \( \phi(g) \) commutes with \( \phi(e) \) for all \( g \). Next, we note that  
\[
\phi(x) \phi(y) \phi(y^{-1} x^{-1}) = \phi(e) \phi(xy) \phi(y^{-1} x^{-1})
\]  
and using the commutativity of \( \phi(e) \), we deduce  
\[
\phi(e)^{-1} \phi(x) \phi(e)^{-1} \phi(y) = \phi(e)^{-1} \phi(xy)
\]  
or \( \psi(xy) = \psi(x) \psi(y) \), as desired.