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Math Help - Deduce Subgroup property

  1. #1
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    Deduce Subgroup property

    Hey I'm having some trouble with this question
    Deduce that any group of order 2p. where p is a prime p>2 Must contain a subgroup of order p
    Heres what I've come up with so far.

    Since G is closed this implies that for every element  g \in G

     g^q = I

    where q is a positive integer. Hence each element of g generates a cyclic subgroup of G of order q.

    By Lagranges therom q can only be 2 or p.

    The trouble I'm having now is proving G cannot just consist of elements of order 2 (or did I misunderstand the question?)
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  2. #2
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    Quote Originally Posted by thelostchild View Post
    Hey I'm having some trouble with this question


    Heres what I've come up with so far.

    Since G is closed this implies that for every element  g \in G

     g^q = I

    where q is a positive integer. Hence each element of g generates a cyclic subgroup of G of order q.

    By Lagranges therom q can only be 2 or p.

    The trouble I'm having now is proving G cannot just consist of elements of order 2 (or did I misunderstand the question?)
    If you apply the Cauchy's theorem, the group of order 2p ( p: a prime number) has an element of order 2 and an element of order p. Find the relations of the elements of this group. Then you will see that a group of order 2p is either a cyclic group or a dihedral group. In either case, this group contains a subgroup of order p.
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  3. #3
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    Quote Originally Posted by TheArtofSymmetry View Post
    If you apply the Cauchy's theorem, the group of order 2p ( p: a prime number) has an element of order 2 and an element of order p. Find the relations of the elements of this group. Then you will see that a group of order 2p is either a cyclic group or a dihedral group. In either case, this group contains a subgroup of order p.
    Yes there is a theorem that states a Group of order 2p is isomorphic to either
     Z_{2p} or  D_p . if its  Z_{2p} then it's cyclic and the answer is immediate. if it's  D_p then it contains a subgroup of order .... it's immediate also.
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  4. #4
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    Quote Originally Posted by thelostchild View Post
    Hey I'm having some trouble with this question


    Heres what I've come up with so far.

    Since G is closed this implies that for every element  g \in G

     g^q = I

    where q is a positive integer. Hence each element of g generates a cyclic subgroup of G of order q.

    By Lagranges therom q can only be 2 or p.

    The trouble I'm having now is proving G cannot just consist of elements of order 2 (or did I misunderstand the question?)

    I'm assuming you haven't yet heard of Cauchy's theorem of Sylow theorems, otherwise your question is trivial, so we can begin by checking cases ( and assuming p\neq 2 ):

    1) if G is abelian then:

    (i) if G=<x> is cyclic then <x^2> solves our problem (use here what you said: the order of any element in the group is exactly one of the following: 1,2, p, 2p)

    ii) if G is not cyclic: define G_2:=\{x\in G\;;\;x^2=1\,,\,x\neq 1\}. Check now that y:=\prod_{x\in G_2}x\notin G_2\cup\{1\} , so <y> solves our problem.

    2) If G is not abelian then it can't be that all its non-unit elements have order 2 (why? Prove that a group in which all its elements satisty the condition g^2=1 must be abelian), and thus there's an element of order p, and again we're done.

    Tonio
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