Abstract Nonsense

Crushing one theorem at a time

Putnam Problem

1997 Putnam-A4

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




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)



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


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(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.


October 4, 2010 - Posted by | Algebra, Fun Problems, Group Theory, Uncategorized | , ,

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