Isn't it trivial (without any left or right multiplication) that is gh = hg then g^-1 h = h g^-1 ? Since g is an arbitrary element, and it's inverse is another arbitrary element within the group. Therefore either one can serve as "g" in the original condition, gh = hg?
If you have the statement Forall g in G, Forall h in G, gh = hg then yes, the group is just abelian there. But that isn't what we have here We are looking at only the g which satisfy Forall h in G, gh = hg, its not always true for any g that we input we don't necessarily know that g^-1 satisfies even if g does, thats why we have to use the multiplication
I teach this course from: abstract.ups.edu/ That being said, this is a pretty standard result that could be found in most Abstract Algebra books/courses.
