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Thread: Cyclic Groups

  1. #1
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    Cyclic Groups

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    Last edited by dabien; Dec 8th 2009 at 02:18 AM.
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  2. #2
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    Nevermind, I misread the question.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by dabien View Post
    Prove that if p is a prime then $\displaystyle (Z/p^2)^x \cong C_p$ $\displaystyle x$ $\displaystyle C_{p-1} \cong C_{p(p-1)}$ where $\displaystyle C_n$ denotes the cyclic group of order n.

    My idea was to consider a homomorphism $\displaystyle Z/p^2 \rightarrow Z/p $ and observe that $\displaystyle (Z/p)^x$ is a cyclic group of order p-1. Then maybe use proof by induction that if p is an odd prime then $\displaystyle (Z/p^n)^x$ is cyclic for all n.

    Next part is to describe the Galois structure of the field of $\displaystyle p^2$ roots of unity, p is prime for the cases of a. p=5, b. p=7. How many intermediate fields are there in these cases?
    I'm a bit confused as to what the question is asking here - is it saying "$\displaystyle (Z/p^2)^x \cong C_p \times C_{p-1} \cong C_{p(p-1)}$"? That is, $\displaystyle \underbrace{(Z/p^2) \times (Z/p^2) \times \ldots \times (Z/p^2)}_\textrm{x times} \cong C_{p(p-1)}$?

    If so, I don't think I believe this. $\displaystyle C_{p(p-1)}$ is finite, but if this is isomorphic to the direct product of $\displaystyle Z/p^2$ with itself an arbitrary number of times we get that $\displaystyle C_{p(p-1)} \cong C_{p(p-1)} \times C_{p(p-1)}$, a contradiction...

    I am willing to be proved wrong though.
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    Quote Originally Posted by Swlabr View Post
    I'm a bit confused as to what the question is asking here - is it saying "$\displaystyle (Z/p^2)^x \cong C_p \times C_{p-1} \cong C_{p(p-1)}$"? That is, $\displaystyle \underbrace{(Z/p^2) \times (Z/p^2) \times \ldots \times (Z/p^2)}_\textrm{x times} \cong C_{p(p-1)}$?

    If so, I don't think I believe this. $\displaystyle C_{p(p-1)}$ is finite, but if this is isomorphic to the direct product of $\displaystyle Z/p^2$ with itself an arbitrary number of times we get that $\displaystyle C_{p(p-1)} \cong C_{p(p-1)} \times C_{p(p-1)}$, a contradiction...

    I am willing to be proved wrong though.

    $\displaystyle \left({\mathbb{Z}}\slash{p^2\mathbb{Z}}\right)^{*} =$ the group of units of the ring of residues modulo $\displaystyle p^2$

    Tonio
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  5. #5
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    Quote Originally Posted by dabien View Post
    Prove that if p is a prime then $\displaystyle (Z/p^2)^x \cong C_p$ $\displaystyle x$ $\displaystyle C_{p-1} \cong C_{p(p-1)}$ where $\displaystyle C_n$ denotes the cyclic group of order n.

    My idea was to consider a homomorphism $\displaystyle Z/p^2 \rightarrow Z/p $ and observe that $\displaystyle (Z/p)^x$ is a cyclic group of order p-1. Then maybe use proof by induction that if p is an odd prime then $\displaystyle (Z/p^n)^x$ is cyclic for all n.

    Next part is to describe the Galois structure of the field of $\displaystyle p^2$ roots of unity, p is prime for the cases of a. p=5, b. p=7. How many intermediate fields are there in these cases?

    We know $\displaystyle \left|\mathbb{Z}/p^2\mathbb{Z}\right|=\phi(p^2)=p(p-1)$ ==> as we're dealing with abelian groups, there're elements a,b of order p, p-1, resp. As $\displaystyle (ord(a), ord(b))=1$ , we get that $\displaystyle ord(ab)=p(p-1)$ and we're done.

    Tonio
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  6. #6
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    Thanks I will take a look and see if I have more questions.
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  7. #7
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    Quote Originally Posted by tonio View Post

    ... there're elements a,b of order p, p-1, resp.
    why does the element $\displaystyle b$ exist? if your arguemnt was true, then every abelian group of order $\displaystyle p(p-1)$ would be cyclic, which is a false result.

    [to dabien] here's a hint: let $\displaystyle c$ be a generator of $\displaystyle (\mathbb{Z}/p)^{\times}.$ show that either $\displaystyle c$ or $\displaystyle c+p$ will generate $\displaystyle (\mathbb{Z}/p^2)^{\times}.$
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  8. #8
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    ...
    Last edited by dabien; Dec 8th 2009 at 02:18 AM.
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