Group structure and the axiom of choice

I'm trying to understand the proof of the following theorem: "Every nonempty set is a group for some operation iff the Axiom of Choice holds."

I've read the post on Stack Exchange, the post on Overflow and the Wikipedia entry, but I'd like some other references on the subject, perhaps a book or article.

• At the very least, links to the other things that you have read would be helpful. Even better, an executive summary of each of those sources would be ideal. – Xander Henderson Sep 22 '17 at 1:34
• I'm trying to link it with my cellphone, but it's kinda hard hahah – user362073 Sep 22 '17 at 1:45
• Then maybe wait until you're home and can write a proper post, instead of writing a half-assed post on a phone? – Asaf Karagila Sep 22 '17 at 2:52
• The proof is not very difficult if you know Hartogs' theorem, the fact that the axiom of choice is equivalent to the well-ordering theorem, and what is a group. So please elaborate what exactly is unclear in the many versions of the proof that you will find everywhere. This will also make recommending a book easier. – Asaf Karagila Sep 22 '17 at 2:57
• You understand that groups are cancellative, so $a\cdot b=a\cdot c$ implies that $b=c$, right? That's the elementary part. It allows us to define the injection from any set into a sufficiently large ordinal, which then implies the set is well-ordered. – Asaf Karagila Sep 22 '17 at 6:15