If x is a cycle of length n, x^n is the identity.

[AxiomOfChoice]

Is it true that if \sigma \in S_n is a cycle of length k \leq n, then \sigma^k = \varepsilon, where \varepsilon is the identity permutation, and that k is the least nonzero integer having this property?

 

[matt grime]

That is surely obvious, isn't it?

 

[AxiomOfChoice]

That is surely obvious, isn't it?

Not to me. Maybe I'm missing something small...if you can get me started on why it's the case, I can probably finish it out.

 

[Hurkyl]

Special cases are always a good way to get started. Try k=1,2,3.

P.S. you meant "least positive integer"

 

[matt grime]

A k-cycle has order k - it really is trivial. You only need to consider the case of

(123..k)

which just rotates the elements 1,..,k cyclically.

 

[matt grime]

More geometrically, label the vertices of a k-gon with 1,..,k, then (1...k) rotates it by 2pi/k.

If you don't like that then just think what (1...k) does to the set 1,..,k it sends i to i+1 (wrapping k round to 1). So what happens if apply it r times?