I have come across an interesting problem from the Putnam 1997 test, question A4:
Problem: Let $G$ be a group with identity $e$ and $\phi: G \to G$ a mapping such that $\phi(g_1)\phi(g_2)\phi(g_3) = \phi(h_1)\phi(h_2)\phi(h_3)$ when $g_1 g_2 g_3 = e = h_1 h_2 h_3$. Prove there exists $a \in G$ such that $\psi = a\phi$ is a homomorphism.
I worked it out and found an 'official' solution that uses the identity $x^{-1}xe = e^3$ to derive the desired homomorphism, but I solved it in an entirely different manner and I would like the sharp minds of StackExchange to help verify that it is correct (translation: Make sure I didn't make a stupid algebra mistake...). It also differs from the solution presented here. I'm a little curious about my initial assumption (If 'a' exists, then showing the hypothesis is true...)
My solution: We start with the definition of $\psi = a\phi$. If $a$ exists, then by the identity in the hypothesis, we have $\psi(e) = a\phi(e) = e$, which tells us that $a = \phi^{-1}(e)$. Thus we can reformulate the problem to show that $\psi(x) = \phi^{-1}(e)\phi(x)$ is a homomorphism. We know that, for all $g \in G$, that $geg^{-1} = egg^{-1}$. So, by hypothesis, we also have $\phi(g)\phi(e)\phi(g^{-1}) = \phi(e)\phi(g)\phi(g^{-1})$, implying that $\phi(g)\phi(e) = \phi(e)\phi(g)$, so $\phi(e)$ commutes. We will use the fact that $\phi(e) \in \mathcal{Z}(\phi(G))$ later in the proof.
In $G$, we have the equivalence $xyy^{-1}x^{-1} = e = exyy^{-1}x^{-1}$, from which we get $\phi(x)\phi(y)\phi(y^{-1}x^{-1}) = \phi(e)\phi(xy)\phi(y^{-1}x^{-1})$, which tells us that $\phi(x)\phi(y) = \phi(e)\phi(xy)$. Now, we return to the fact that $\phi(e) \in \mathcal{Z}(\phi(G))$. We have:
$\phi(e)\phi(xy) = \phi(x)\phi(y) = \phi(x)\phi(e)\phi^{-1}(e)\phi(y) = \phi(e)\phi(x)\phi^{-1}(e)\phi(y)$, which tells us:
$\phi(e)\phi(xy) = \phi(e)\phi(x)\phi^{-1}(e)\phi(y)$, which we can generate a further equivalence by 'conveniently' multiplying out both sides repeatedly by $\phi^{-1}(e)$ to get: $\phi^{-1}(e)\phi(xy) = \phi^{-1}(e)\phi(x)\phi^{-1}(e)\phi(y)$.
But since we defined $a = \phi^{-1}(e)$, the last equation can be rewritten as $\psi(xy) = \psi(x)\psi(y)$, i.e. it is a homomorphism. QED