MHB Is G an Abelian Group Given Specific Conditions?

[alexmahone]

Let $G$ be a group such that for all $a$, $b$, $c$, $d$, and $y\in G$ if $ayb=cyd$ then $ab=cd$. Show that $G$ is an Abelian group.

HINTS ONLY as this is an assignment problem.

 

[Evgeny.Makarov]

Consider $ayb=cyd$ when $y$ is the inverse of something.

 

[alexmahone]

Consider $ayb=cyd$ when $y$ is the inverse of something.

Even if I suppose $y=a^{-1}$ or $b^{-1}$ or $a^{-1}b^{-1}$, how do I know that $ayb=cyd$ has to be true?

 

[Evgeny.Makarov]

Even if I suppose $y=a^{-1}$ or $b^{-1}$ or $a^{-1}b^{-1}$, how do I know that $ayb=cyd$ has to be true?

Given $a$ and $b$, you can make $ayb=cyd$ true by choosing $y$, $c$ and $d$ appropriately.

Another way to look at this is the following. You need to prove $ab=ba$ for all $a$ and $b$. Try to apply the implication that is given to you in post #1. For this you have to guess $y$ because it occurs only in the assumption and not the conclusion.

 

[alexmahone]

I think I got it!

Take $c=b$ and $d=a$
Take $y=a^{-1}$

$ayb=aa^{-1}b=b$
$cyd=ba^{-1}a=b$
So, $ayb=cyd$
$\implies ab=cd$
i.e. $ab=ba$
So, $G$ is an Abelian group.

 

[Evgeny.Makarov]

Yes, that's correct.