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Thread: 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 $\displaystyle G $ of order $\displaystyle q-1$ where $\displaystyle q$ is prime.

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

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

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

    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 $\displaystyle G $ of order $\displaystyle q-1$ where $\displaystyle q$ is prime. i guess you meant "group"!

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

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

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

    I may be making this too complicated...Any hints?
    something is not right here! any finite cyclic group of order $\displaystyle n$ has a unique (cyclic) subgroup of any order which divides $\displaystyle n.$ 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 $\displaystyle n$ has a unique (cyclic) subgroup of any order which divides $\displaystyle n.$
    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 $\displaystyle \mathbb{Z}_p$ to $\displaystyle Aut(\mathbb{Z}_q)$ where p divides q-1.)

    I just got stuck on proving $\displaystyle Aut(\mathbb{Z}_q)$ has a unique subgroup of order $\displaystyle p$.
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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 $\displaystyle G=<a>$ be a group of order $\displaystyle n$ and $\displaystyle d \mid n.$ then clearly $\displaystyle <a^{\frac{n}{d}}>$ is a subgroup of $\displaystyle G$ of order $\displaystyle d.$

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

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