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.
asked Aug 5, 2014 at 13:58 rockstar123 rockstar123 1,301 1 1 gold badge 9 9 silver badges 23 23 bronze badges$\begingroup$ Hint: Multiply by $a^<-1>$ from the right on both sides and see what the condition then says. $\endgroup$
Commented Aug 5, 2014 at 13:59Since 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.
answered Aug 5, 2014 at 14:02 42.4k 9 9 gold badges 63 63 silver badges 104 104 bronze badges $\begingroup$ Nice and simple :) $\endgroup$ Commented Aug 5, 2014 at 14:03 $\begingroup$ This is awesome. I thought inverses would have been needed. $\endgroup$ Commented Aug 5, 2014 at 14:03 $\begingroup$ @TobiasKildetoft yes that is surprising. I edited it in. $\endgroup$ Commented Aug 5, 2014 at 14:06 $\begingroup$ What an elegant solution! $\endgroup$ Commented Aug 5, 2014 at 14:10$\begingroup$ I wonder if there is any study of magmas satisfying this property, or if it too weak to be of interest. $\endgroup$
Commented Aug 5, 2014 at 14:15 $\begingroup$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).