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jmjlt88
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Proposition: Suppose σ is a cycle of length s. Then σ2 is a cycle if and only if s is odd.
Quick remark... The following "proof" seems more like an explanation. This is for a self-study and I am looking back through my notes to refine/correct/improve. If you were correcting this, say out of 10pts, what would you give it?
Proof:
Let σ be a cycle of length s. That is, σ=(a1a2a3...as-1as). Suppose s is odd. Now, we calculate σ2 and we obtain that σ2=(a1a3a5a7...asa2a4a6...as-1). We notice that all odd numbered terms appear first in our cycle. Then, as, our last odd numbered term, gets sent to a2. The term a2 starts a sequence of moving even numbered terms to the next even numbered term, which concludes with as-1, which is also even. The term as-1 gets sent to a1 which completes the cycle. Now if s was even, then the term as-1 would be odd. But, as-1 would still get sent to a1. Hence, σ2 would be written as the product of disjoint cycles. That is,
[Equation 1] σ2=(a1a3a5a7...as-1)(a2a4a6...as). Thus, σ2 would not be a cycle.
Similarly, if we have that σ2 is cycle, then we deduce that s must be odd or else a situation similar to [Equation 1] would arise.
QED
Thanks for the help! :)
Quick remark... The following "proof" seems more like an explanation. This is for a self-study and I am looking back through my notes to refine/correct/improve. If you were correcting this, say out of 10pts, what would you give it?
Proof:
Let σ be a cycle of length s. That is, σ=(a1a2a3...as-1as). Suppose s is odd. Now, we calculate σ2 and we obtain that σ2=(a1a3a5a7...asa2a4a6...as-1). We notice that all odd numbered terms appear first in our cycle. Then, as, our last odd numbered term, gets sent to a2. The term a2 starts a sequence of moving even numbered terms to the next even numbered term, which concludes with as-1, which is also even. The term as-1 gets sent to a1 which completes the cycle. Now if s was even, then the term as-1 would be odd. But, as-1 would still get sent to a1. Hence, σ2 would be written as the product of disjoint cycles. That is,
[Equation 1] σ2=(a1a3a5a7...as-1)(a2a4a6...as). Thus, σ2 would not be a cycle.
Similarly, if we have that σ2 is cycle, then we deduce that s must be odd or else a situation similar to [Equation 1] would arise.
QED
Thanks for the help! :)