- #1

nateHI

- 146

- 4

My question is, in this problem solution, how does ##{\sigma_i}^m=1## follow from ##\sigma_i## being disjoint?

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- Thread starter nateHI
- Start date

In summary, in this problem, the question is asking for the smallest number m that satisfies ##\sigma^m = \text{id}## where ##\sigma## is a permutation written as a product of disjoint cycles. This is because the order of a cycle is its length, and the smallest m must be a multiple of the length of each cycle.

- #1

nateHI

- 146

- 4

My question is, in this problem solution, how does ##{\sigma_i}^m=1## follow from ##\sigma_i## being disjoint?

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- #2

geoffrey159

- 535

- 72

It says that a permutation can be written as a product of disjoint cycle ##\sigma = \sigma_1 \circ ... \circ \sigma_p ##.

The question asks you to find the smallest ##m \ge 1## such that ##\sigma ^ m = \text{id}##.

Since the cycles are disjoint, you can commute all the cycles, and rewriting things nicely, ##m## has to be such that ##\sigma_i ^ m = \text{id} ##, for all ##i = 1 ... p ##.

Since the order of a cycle is its length, ##m## must be a multiple of the length of each cycle ##\sigma_i##. The smallest such ##m## is the least common multiple of all the lengths.

- #3

nateHI

- 146

- 4

Got it thanks!

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