Is G an Abelian Group Given Specific Conditions?

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SUMMARY

The group \( G \) is proven to be an Abelian group under the condition that for all elements \( a, b, c, d, y \in G \), if \( ayb = cyd \) then \( ab = cd \). By selecting \( c = b \), \( d = a \), and \( y = a^{-1} \), it is demonstrated that \( ab = ba \) holds true for all \( a \) and \( b \). This confirms that \( G \) satisfies the properties of an Abelian group, specifically the commutative property of group multiplication.

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alexmahone
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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.
 
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Consider $ayb=cyd$ when $y$ is the inverse of something.
 
Evgeny.Makarov said:
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?
 
Alexmahone said:
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.
 
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.
 
Yes, that's correct.
 

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