I got caught up on the question before that was about cyclic groups!

As for the function being a bijection:

We have

lots of

. We also have

lots of

(proof as follows):

I really don't have any idea what can you possibly mean by "having lots of..."...
Since we have

lots of

, we have at most

lots of

. We will only have less if, for some

,

(since inverses are unique).

This is incorrect: it can perfectly well be that without being true that . You've to use the definition of the equivalence relation ...what?!
Therefore we have

lots of

.

Also, since inverses are unique,

is unique to a

. Therefore the function is injective.

We have a mapping from a set of

elements to a set of

elements which is injective. By the pigeonhole principle, the function is a bijection.