In a lecture my professor quickly went over the problem:
Let $ a $ and $ b $ be elements of a group G. If $ a^3 b = ba^3 $ and if a has order 7, show that $ ab = ba $ .
The sketch of a proof he wrote out went like this (this is what I have in my notes)
Does $ (a^3)^2 = a^6 $ commute with $ b $?
Well, $ a^6 = a^{-1} $ because $ a^7 = e $
So $ a^{-1}b = b a^{-1} $ so $ b = aba^{-1} $ so $ ba = ab $
I don't completely follow. It seems like this should be basic, but it isn't for me. By writing out $ (a^3)^2 $ it seem to imply that he could substitute $ (a^3)^2 $ for $ a^3 $ and put it into the original equation. That doesn't make sense to me because I figured $ a^3 $ is a distinct element, not a variable.
I am not sure how he is synthesizing the premises with $ a^6 = a^{-1} $ to find $ a^{-1}b = b a^{-1} $. I am hoping for some help with that.