Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Abstract Algebra: Dummit and Foote Exercise

  1. Jun 25, 2015 #1
    This isn't homework, I'm just trying to refresh my memory on cyclic groups.

    My question is, in this problem solution, how does ##{\sigma_i}^m=1## follow from ##\sigma_i## being disjoint?
     
  2. jcsd
  3. Jun 25, 2015 #2
    It rather follows from ##\sigma_i## being a cycle.

    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.
     
  4. Jun 26, 2015 #3
    Got it thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Abstract Algebra: Dummit and Foote Exercise
  1. Abstract Algebra (Replies: 0)

  2. Abstract Algebra (Replies: 9)

Loading...