- #1
zooxanthellae
- 157
- 1
Homework Statement
Suppose we have cyclic, prime-length [tex]\alpha = (a_{1}a_{2}...a_{s}).[/tex] Prove that every power of [tex]\alpha[/tex] is a cycle.
Homework Equations
None I can think of.
The Attempt at a Solution
I feel that I understand some of the intuition behind this problem. If any power of a cycle is not itself a cycle then it must either be the identity or the product of disjoint cycles. The only possibility here, then, would be a product of disjoint cycles. However, this seems to require that the original cycle not have a prime length, since one of the disjoint cycles must have an interval equivalent to a factor of the length of the original cycle.
That was pretty garbled, so I'll use an example. Let [tex]\beta = (b_{1}b_{2}...b_{9}).[/tex] Here [tex]\beta^3[/tex] is a product of disjoint cycles [tex](b_{1}b_{4}b_{7})(b_{2}b_{5}b_{8})(b_{3}b_{6}b_{9}),[/tex] but note that this required an interval of 3, a factor of 9, in order for one disjoint cycle to "hit" the last element. Therefore any power of a prime-length cycle should be a cycle, since it can't be a product of disjoint cycles.
But I'm kind of lost as to how to make this rigorous.