Algebraic equivalent of AC

Question

I was reading this Wikipédia link https://en.wikipedia.org/wiki/Group_structure_and_the_axiom_of_choice and I have some questions I'd like someone to explain to me.

• For any $x \in X$ there exists an $\alpha \in \aleph(X)$ s.t. $x \bullet \alpha \in \aleph(X)$, where $\bullet$ is the operation on the group $X^{\prime} = X \cup \aleph(X)$. Is this because of the cancellative law of groups?
• Shouldn't I prove that $x < y$ iff $j(x) < j(y)$ is a wellordering on $X$?

Thanks

Answer 1

• The point is that if there was no $\alpha$ with $x \bullet \alpha \in \aleph(X)$ then all $x \bullet \alpha \in X$ and thus $x \bullet -$ is a function $\aleph(X) \rightarrow X$. This function is injective (as you say: by the cancellation law), contradicting the definition of $\aleph(X)$.
• The above section in the wikipedia article establishes that $j$ is an injection into the well-ordered set $\aleph(X)^2$. It therefore follows pretty quickly that $X$ is well-ordered: we can treat the ordering on $X$ as a subset of the ordering on $\aleph(X)^2$ and any subset of a well-ordered set is also well-ordered.

Answer 2

1. You are correct. The proof actually shows that a right-cancellative operation is enough. In other words, if every set $X$ has a function $f\colon X^2\to X$ such that for all $x$, $f(x,\cdot)$ is injective, the proof works as it is.

2. Sure. You should. But you can also note that $j$ is just a transport of structure, and therefore an isomorphism of orders. And being isomorphic to a well-order means that you're well-ordered too.