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Group Theory

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that any group of order ##15## is cyclic.

    2. The attempt at a solution

    I am looking at a link here: (http://www.math.rice.edu/~hassett/teaching/356spring04/solution.pdf) and I am confused why "there must be one orbit with five elements and three orbits with three elements" and "fixed points of this action are just elements of the center of ##G.## In our situation, it suffices to show that ##G## has just one fixed point ##\ne e##."

    Why are those the cases?
     
  2. jcsd
  3. Nov 8, 2013 #2

    pasmith

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    Homework Helper

    Read the previous two sentences: "Suppose there is no such fixed point: Each orbit of G under conjugation has order divisible by G, so the only possibilities are 3 and 5."

    The assumption (which will be shown to be false) is that the only orbit of one element is that of the identity, so you must partition the remaining 14 elements into disjoint orbits of 3 or 5 elements.

    What does it mean for an element to be a fixed point of the conjugation action?
    What does it mean for an element to be in the center?

    What is the relationship between the two conditions?

    The center is a subgroup, so if it isn't trivial or of order 15 (ie G is abelian) then it must be of order 3 or order 5, and therefore cyclic. In fact it follows from this that G is abelian, which I presume was what the exercise cited in the proof asked you to show.
     
  4. Nov 9, 2013 #3
    Thank you very much, pasmith!
     
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