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Math Help - abstract algebra: subgroups question

  1. #1
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    abstract algebra: subgroups question

    If G is a finite Abelian group, H, K ≤ G, and if x ∈ H K with o(x) = p a prime, show that H or K must contain an element of order p.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: abstract algebra: subgroups question

    can you tell me what o(x) is? is it an element of inner automorphisms of (HK) ?
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    Super Member ILikeSerena's Avatar
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    Re: abstract algebra: subgroups question

    Hi henderson7878!

    There are 3 cases:
    1. x in H
    2. x in K
    3. x neither in H nor in K, and x = hk

    What conclusions can you draw in each of these cases for xp?


    (o(x) is a notation for the order of x.)
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  4. #4
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    Re: abstract algebra: subgroups question

    o(x) is the order
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  5. #5
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    Re: abstract algebra: subgroups question

    Here is a stronger statement:

    Let G be a finite group, H and K subgroups of G. Assume the product set HK = { hk : h is in H and k is in K } is also a subgroup of G. (If G is abelian, this is true for any two subgroups H, K.) Let p be a prime and hk in HK of order p. Then either H contains an element of order p or K contains an element of order p.

    Proof. Since hk is in HK, p divides the order of HK. |HK|=|H|\times{|K|\over |H\cap K|}. (Actually, this is true even without the supposition that HK is a subgroup. I leave the proof to you.) So p divides |H| or p divides |K|. By Cauchy's Theorem, one of H and K contain an element of order p.

    Extra Credit: Show by counterexample, this statement is false if HK is not a subgroup of G. Hint - consider the alternating group A4.
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  6. #6
    Junior Member Nehushtan's Avatar
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    Re: Abstract algebra: Subgroups question

    Let x=hk, h\in H,k\in K, and suppose \mathrm o(h)=a,\mathrm o(k)=b.

    Then x^{ab} = (hk)^{ab} = h^{ab}k^{ab} (as G is Abelian) = e. Thus p=\mathrm o(x) divides ab and so either p\mid a or p\mid b.

    Suppose p divides a, say a=cp. Then e=h^a=(h^c)^p. Moreover h^c\ne e as c<a=\mathrm o(h). Hence h^c\in H has order p.

    Similarly if p divides b then k^{b/p}\in K has order p.
    Last edited by Nehushtan; March 9th 2013 at 01:53 PM.
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