So, abstract algebra beginner here. The proof I have uploaded supposedly only works for abelian groups, although the actual theorem is true of all groups. Call me stupid, but I can't figure out why it wouldn't work even for nonabelian groups. I've tried hard and long, but I just can't figure it out. Not that I'm very good at abstract algebra anyway. So, could you please help me and tell me where the proof goes wrong? I'll be grateful.
The crucial part of the proof is the rewriting at the bottom of the first page $$ \prod_{s\in S_x} s = \prod_{g\in G} xg = x^n\prod_{g\in G} g = x^n \prod_{s\in S_x}s $$ The products here do not explain which order to multiply the elements of $S_x$ or $G$ in, so they are only even meaningful if we know the order doesn't matter -- that is, if the group is abelian.
Even if we fix a particular order of multiplication for the initial $\prod_{s\in S_x} s$, the second equals sign depends on moving all of the $x$s in the product out in front of all of the $g$s, which is only valid if the group is abelian, and the very point of the third equals sign is to rearrange the order of the elements to be multiplied, because we know that the $g$s and the $xg$s are the same elements but generally in a different order.