Let be a group such that for all and for three consecutive integers . Prove is abelian.
We are given three equations. For all
From (1), we have
From (2), by multiplying by the inverse found from (1), we have
From (3), again multiplying by the inverse found from (1), we have
Take (4) and mutliply by on the left and (5) and multiply by on the left and we get
Now cancelling the on the right, we get
Therefore, for every , it is true that commutes with every element of . Using that information and (4), we get
Therefore, for every element of commutes with every other element of .