TITLE: $1997$ Putnam - Let $G$ be a group and let $\Phi : G \rightarrow G$ be a function
QUESTION [7 upvotes]: Let $G$ be a group and let $\Phi : G \rightarrow G$ be a function such that
$\Phi(g_1)~ \Phi (g_2)~ \Phi(g_3) = \Phi(h_1)~ \Phi (h_2)~ \Phi(h_3)$ whenever
$ g_1~g_2~g_3 = h_1~h_2~h_3=e$ . 
Show that there exists an element $a$ in $G$ such that $\Psi(x) = a~ \Phi(x)$ is a homomorphism.
Attempt: We need to show that $\forall ~~x,y \in G, a ~\Phi(xy) = a~ \Phi(x)~a~\Phi(y)$
let $x,y,z \in G ~~|~~ (xy)^{-1}=z$
Then, $e \cdot xy \cdot z = e\cdot e\cdot e$
$\implies \Phi(e) ~\Phi(xy)~ \Phi(z) = [\phi(e)]^3 ..............(1)$
Similarly, $x \cdot y \cdot z =e \cdot e \cdot e$
$\implies \Phi(x) ~\Phi(y)~ \Phi(z) = [\phi(e)]^3 ..............(2)$
from $(1),(2): \Phi(e) ~\Phi(xy) = \Phi(x) ~\Phi(y)$ 
Now, I need to prove that '$\Phi(x) ~\Phi(y)=\Phi(e)~\Phi(x)~\Phi(e)~\Phi(y)$'
How do I move forward? Help will be appreciated. Thank you

REPLY [7 votes]: If $\Psi$ is a homomorphism then you must have $\Psi(e) = e$. This shows that the only $a$ that can possibly work is $a = \Phi(e)^{-1}$.
$\Phi(x^{-1})\Phi(x)\Phi(e) = \Phi(x^{-1})\Phi(e)\Phi(x)$, so $\Phi(e) = a^{-1}$ commutes with $\Phi(x)$ for all $x$. Then so does $a$ because $a\Phi(x) = a\Phi(x)a^{-1}a = aa^{-1}\Phi(x)a = \Phi(x)a$.
Then you have to show that forall $x,y$, $a\Phi(xy) = a\Phi(x)a\Phi(y)$.
This can be rewritten using cancellation and commutativity as $\Phi(e)\Phi(xy) = \Phi(x)\Phi(y)$, which is true as you have shown.