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Math Help - Finite abelian p-group

  1. #1
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    Finite abelian p-group

    Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.

    \forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}

    (\Rightarrow)
    Suppose G is cyclic. Then for a g\in G, G=<g>.

    What next? I am at a loss.
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    MHF Contributor Drexel28's Avatar
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    Re: Finite abelian p-group

    Quote Originally Posted by dwsmith View Post
    Suppose G is a finite abelian p-group. Prove that if G is cyclic iff G has exactly p-1 elements of order p.

    \forall g\in G, \ \ g^{p^{\alpha}}=1, \ \ |G|=p^{\alpha}

    (\Rightarrow)
    Suppose G is cyclic. Then for a g\in G, G=<g>.

    What next? I am at a loss.
    If G is cyclic, you know it's isomorphic to \mathbb{Z}_{p^\alpha} how many elements o order p does this have (hint:recall cyclic groups have only one subgroup of a given order)? Assume now that G has only p-1 elements of order p. Choose g\in G with |g|=p, now assume that x\in G-\langle g\rangle, you know that \langle x\rangle intersects trivially with \langle g\rangle (why?) and so \langle x\rangle is a cyclic subgroup of G whose order is divisible by p--so?
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    Re: Finite abelian p-group

    if G is cyclic, it has only one subgroup of order p. since any element of order p generates a subgroup of order p, all elements of order p must generate the same subgroup of G (since G has only ONE subgroup of order p, because its cyclic). so all elements of order p must all lie in that one subgroup of order p, which has (of course) p-1 elements of order p.

    now suppose that G has exactly p-1 elements of order p. pick one of them, say x, we have a cyclic subgroup of G of order p <x>. this is the ONLY subgroup of order p of G.

    now suppose H is a subgroup of order p^2, which has to be cyclic, because the direct product of 2 subgroups of order p is not possible. let y be a generator of H. then y^p has order p, so y^p is in <x>. hence <x> is a subgroup of H.

    now suppose that K is a subgroup of order p^3....can you see where this is going? show that any subgroup of order p^k has to be cyclic, or else it contains more than one subgroup of order p.

    conclude that G must be cyclic.
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