Can a Group Have Multiple Homomorphisms With the Same Identity?

[Ackbach]

Here is this week's POTW:

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Let $G$ be a group with identity $e$ and $\phi:G\rightarrow G$ 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 $x,y\in G$).

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[Ackbach]

Re: Problem Of The Week # 229 - Aug 17, 2016

This was Problem A-4 in the 1997 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows:

Spoiler

If $g\in G$ then $g^{-1}ge = e = g^{-1}eg$, and so $\phi(g^{-1})\phi(g)\phi(e) = \phi(g^{-1})\phi(e)\phi(g).$

Multiply both sides on the left by $(\phi(g^{-1}))^{-1}$ to get $\phi(g)\phi(e) = \phi(e)\phi(g).$ Thus $\phi(e)$ commutes with every element in the range of $\phi.$ Let $a = (\phi(e))^{-1}.$ Then $a$ also commutes with every element in the range of $\phi.$

Next, $gg^{-1}e = e = eee$, and so $\phi(g)\phi(g^{-1})\phi(e) = (\phi(e))^3,$ or $\phi(g)\phi(g^{-1})a^{-1} = a^{-3}.$ Since $a$ commutes with $\phi(g)$ and $\phi(g^{-1}),$ we can write that as $a\phi(g)\bigl(a\phi(g^{-1})\bigr) = e.$ If $\psi(g) = a\phi(g)$, that says that $\psi(g)\psi(g^{-1}) = e,$ or in other words $\psi(g^{-1}) = (\psi(g))^{-1}.$

Now let $g,h\in G$. Then $(gh)h^{-1}g^{-1} = e = eee.$ Therefore $\phi(gh)\phi(h^{-1})\phi(g^{-1}) = \phi(e)^3 = a^{-3}.$ Since $a$ commutes with each element on the left of that equation, we can write it as $a\phi(gh)a\phi(h^{-1})a\phi(g^{-1}) = e,$ or $\psi(gh) \psi(h^{-1}) \psi(g^{-1}) = e.$ But $\psi(g^{-1}) = (\psi(g))^{-1}$ and $\psi(h^{-1}) = (\psi(h))^{-1},$ so we have $\psi(gh)(\psi(h))^{-1}(\psi(g))^{-1} = e.$ Multiply both sides on the right by $\psi(g)$ and then by $\psi(h)$ to get $\psi(gh) = \psi(g)\psi(h)$. Thus $\psi$ is a homomorphism, as required.