# Abstract Nonsense

## Putnam Problem

1997 Putnam-A4

Problem: Let $G$ be a group with identity $e$ and $\phi:G\to G$ a mapping such that

$\phi(g_1)\phi(g_2)\phi(g_3)=\phi(h_1)\phi(h_2)\phi(h_3)$

whenever

$g_1g_2g_3=e=h_1h_2h_3$

Prove there exists $\alpha\in G$ such that $\psi=\alpha\phi$ is a homomorphism.

Proof: It’s clear that if $\alpha$ does exist that

$e=\psi(e)=\alpha\phi(e)\implies \alpha=\phi(e)^{-1}$

So, we claim that $\psi(x)=\phi(e)^{-1}\phi(x)$ is a homomorphism. To do this we first note that

$geg^{-1}=e=egg^{-1},\text{ }\forall g\in G$

and so by assumption

$\phi(g)\phi(e)\phi\left(g^{-1}\right)=\phi(e)\phi(g)\phi\left(g^{-1}\right)\implies \phi(g)\phi(e)=\phi(e)\phi(g)$

Thus,

$\phi(e)\in\mathcal{Z}\left(\phi\left(G\right)\right)$

(the center of the image). Thus, we then note that

$(x)(y)(y^{-1}x^{-1})=e=(e)(xy)(y^{-1}x^{-1})$

where the parentheses are merely for indicative grouping, so then

$\phi(x)\phi(y)\phi\left(y^{-1}x^{-1}\right)=\phi(e)\phi(xy)\phi\left(y^{-1}x^{-1}\right)\implies \phi(x)\phi(y)=\phi(e)\phi(xy)$

but, using the fact that $\phi(e)\in\mathcal{Z}\left(\phi\left(G\right)\right)$ we see that

$\phi(e)\phi(xy)=\phi(x)\phi(y)=\phi(x)\phi(e)\phi(e)^{-1}\phi(y)=\phi(e)\phi(x)\phi(e)^{-1}\phi(y)$

Namely,

$\phi(e)\phi(xy)=\phi(e)\phi(x)\phi(e)^{-1}\phi(y)\implies \overbrace{\phi(e)^{-1}\phi(xy)}^{(1)}=\overbrace{\phi(e)^{-1}\phi(x)\phi(e)^{-1}\phi(y)}^{(2)}$

But, by definition $(1)=\psi(xy)$ and $(2)=\psi(x)\psi(y)$ from where the conclusion follows.