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

1. May 27, 2009

### 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?

2. May 27, 2009

### matt grime

That is surely obvious, isn't it?

3. May 27, 2009

### AxiomOfChoice

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.

4. May 27, 2009

### Hurkyl

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

P.S. you meant "least positive integer"

5. May 27, 2009

### 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.

6. May 28, 2009

### 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?

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