Let $G$ be a group with the following property: If $a, b,$ and $c$ belong to $G$ and $ab=ca$,then $b=c$. Prove that $G$ is abelian.

I think I have the answer for the finite $G$ case. Then, for a given $a \in G$, we can list all the elements $aG$ and $Ga$ and using cancellation law and the above property show that $ag=ga$, for arbitrary $g$ and $a$.

But is there a better proof that does not need the special consideration of finite and infinite.

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    $\begingroup$ Hint: Multiply by $a^{-1}$ from the right on both sides and see what the condition then says. $\endgroup$ – Tobias Kildetoft Aug 5 '14 at 13:59

Since it is associative you have $a(ba) = (ab)a$. Thus, $ba=ab$ by your property, and you are done.

Thus, it is in fact true that any associative structure (in particular, you do not need that elements are invertible) with that property is commutative.

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  • $\begingroup$ Nice and simple :) $\endgroup$ – rschwieb Aug 5 '14 at 14:03
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    $\begingroup$ This is awesome. I thought inverses would have been needed. $\endgroup$ – Tobias Kildetoft Aug 5 '14 at 14:03
  • $\begingroup$ @TobiasKildetoft yes that is surprising. I edited it in. $\endgroup$ – quid Aug 5 '14 at 14:06
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    $\begingroup$ What an elegant solution! $\endgroup$ – rockstar123 Aug 5 '14 at 14:10
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    $\begingroup$ I wonder if there is any study of magmas satisfying this property, or if it too weak to be of interest. $\endgroup$ – Tobias Kildetoft Aug 5 '14 at 14:15

Here is another way to do this, which shows that we also do not need the operation to be associative, as long as we instead require the following property (which is also satisfied by groups):

For all $x,y\in G$ there is a $z\in G$ such that $x = zy$ (or such that $x = yz$, either one will suffice).

Now, if the operation is not commutative there will be two elements $a$ and $b$ such that $ab\neq ba$. But there is some $c$ such that $ab = ca$ (by the above property with $x = ab$ and $y = a$). Since we cannot have $c = b$ this breaks the assumption that whenever we had $ab = ca$ we would have $b = c$.

A bit more about the property I assume here, to "demystify" it:
To make the property a bit more natural, it might help to see a slightly different way to define a group: We start in the usual way with a binary operation that is associative and has a unit. But as an alternative (though it is equivalent) to the inverses, we require that for each element $g\in G$ both left- and right- multiplication by $g$ is a bijective map (I leave it as a nice exercise to show this is indeed equivalent to the usual definition).

Now, the property I assume here can be restated as: For each $g\in G$, right multiplication by $g$ is surjective (or very shortly just $Gg = G$ for all $g\in G$).

As noted in the comments, the condition is essentially "half" (or maybe closer to a quarter of) the property required for having a quasigroup (if we just keep the property that both left- and right- multiplication by any element is bijective and remove associativity and unit, we get to what a quasigroup is).

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  • $\begingroup$ That's an interesting additional aspect. $\endgroup$ – quid Aug 5 '14 at 18:21
  • $\begingroup$ Might be worth stating explicitly: this is "half" of the quasigroup axiom. $\endgroup$ – Eric Stucky Aug 6 '14 at 4:04
  • $\begingroup$ I think this is valid only for the finite $G$ case, since the property that $x=zy$ can be proven only for the finite case. Am I right? $\endgroup$ – rockstar123 Aug 6 '14 at 13:19
  • $\begingroup$ @ajinkya No, this property is satisfied by any group (in fact, any quasigroup as mentioned by Eric Stucky). If I get the time, I will write a bit more about the property later. $\endgroup$ – Tobias Kildetoft Aug 6 '14 at 13:32
  • $\begingroup$ @EricStucky I added a bit more about the property (which is probably closer to a quarter than half). $\endgroup$ – Tobias Kildetoft Aug 6 '14 at 18:33

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