Imagine a group generated by this 3 elements
1, x, y
Say there is no relationship whatsoever between x and y and say both are cyclic of order 2. We know that the group must have at most 4 elements right?
I mean, 1, x, y, xy.
Am I correct here? So there is only 1 element "farthest" from identity, namely, xy.
I must be a bit wrong here because yx may not be an inverse of xy, but I think I am on to something.
So it seems that the fartheest element on a finite group should only be 1 at most.
Am I wrong?
Actually, if a group has generators with limited cycles, is there a way to measure the number of all distinct elements in the group? Why is it so difficult to proof that the God's number for rubic is 20?
I mean all rubic position, assuming that the center never moves, can be generated with 6 elements with 4 cycle each right. So is there a formula that can compute the number of positions that can be solved in 1, 2, 3, 4, 5, 6, ... steps?