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Math Help - Transitive subgroup

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    Transitive subgroup

    Let G be a transitive subgroup of S_n where n>1. Show that G contains a permutation without fixpoints, i.e., where all cycles has length >1.
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    Quote Originally Posted by peteryellow View Post

    Let G be a transitive subgroup of S_n where n>1. Show that G contains a permutation without fixpoints, i.e., where all cycles has length >1.
    that's a quick result of Cauchy-Frobenius lemma (wrongly known as Burnside's lemma). let X = \{1,2, ... , n \}. since G is transitive, the number of orbits is 1. so we must have |G|= \sum_{g \in G} |X^g|.

    let \iota be the identity permutation. we have X^{\iota}=|X|=n > 1. if for every g \in G: \ |X^g| \geq 1, then |G| > \sum_{g \in G} 1 = |G|. contradiction! so there exists g \in G such that |X^g|=0. Q.E.D.
    Last edited by NonCommAlg; April 1st 2009 at 10:13 AM.
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    Quote Originally Posted by NonCommAlg View Post
    since G is transitive, the number of orbits is 1.
    How do you have thet number of orbit is 1.

    Quote Originally Posted by NonCommAlg View Post
    |G| > \sum_{g \in G} 1 = |G|. contradiction!
    How does this follow?
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    Quote Originally Posted by peteryellow View Post
    How do you have thet number of orbit is 1.
    Because it is transtive. Two elements are in the same orbit if and only if y=gx, since the group is transitive it means for any x,y there is g so that y=gx. Hence all elements are in the same obrit.

    How does this follow?
    Because |G| = \sum_{g\in G}|X^g| , \sum_{g\in G} |X^g| > \sum_{g\in G}1 (just see what happens for g=e). But \sum_{g\in G}1 = |G| and so |G| > |G|.
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    Thanks NoncommutativeAlgebra and PerfectHacker.
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