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Math Help - cyclic subgroup

  1. #1
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    cyclic subgroup

    I should be able to do this one but I'm stuck...

    I have a cyclic subgroup G of order q-1 where q is prime.

    Let p be a prime dividing q-1. Show G contains a unique cyclic subgroup of order p.

    Now I know Cauchy's Theorem tells me G indeed has at least one subgroup of order p.

    Also I know G\simeq\mathbb{Z}_{pm} for some positive integer m.

    I may be making this too complicated...Any hints?
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  2. #2
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    Quote Originally Posted by ziggychick View Post
    I should be able to do this one but I'm stuck...

    I have a cyclic subgroup G of order q-1 where q is prime. i guess you meant "group"!

    Let p be a prime dividing q-1. Show G contains a unique cyclic subgroup of order p.

    Now I know Cauchy's Theorem tells me G indeed has at least one subgroup of order p.

    Also I know G\simeq\mathbb{Z}_{pm} for some positive integer m.

    I may be making this too complicated...Any hints?
    something is not right here! any finite cyclic group of order n has a unique (cyclic) subgroup of any order which divides n. so, in your problem, we don't need to assume that p, q are prime .

    are you sure G is cyclic?
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    any finite cyclic group of order n has a unique (cyclic) subgroup of any order which divides n.
    Ok, i need to prove the above statement.

    I was assuming p and q where prime because this is actually part of a larger proof. ( that there are p distinct homomorphisms from \mathbb{Z}_p to Aut(\mathbb{Z}_q) where p divides q-1.)

    I just got stuck on proving Aut(\mathbb{Z}_q) has a unique subgroup of order p.
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  4. #4
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    Quote Originally Posted by ziggychick View Post

    Ok, i need to prove the above statement.
    that's not hard to prove:

    first of all recall that every subgroup of a cyclic group is cyclic. now let G=<a> be a group of order n and d \mid n. then clearly <a^{\frac{n}{d}}> is a subgroup of G of order d.

    now suppose H=<a^k> is any subgroup of G of order d. then a^{kd}=1 and thus n \mid kd, i.e. \frac{n}{d} \mid k. let k=m\frac{n}{d}. then a^k=a^{m \frac{n}{d}} \in <a^{\frac{n}{d}}> and hence H \subseteq <a^{\frac{n}{d}}>.

    therefore, since |H|=|<a^{\frac{n}{d}}>|, we must have H=<a^{\frac{n}{d}}>.
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  5. #5
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    ok got it. thanks
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