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Math Help - Correspondence Theorem

  1. #1
    Senior Member slevvio's Avatar
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    Correspondence Theorem

    Use the correspondence theorem to show that

    the only subgroups of \mathbb{Z} are  \mathbb{Z} and <m> \iff m is prime

    I understand how to show this in the <= direction. I have a proof here for the => direction which i don't understand.


    Conversely, suppose that the only subgroups of Z which contain <m> are <m> itself and Z.
    If m = rs for natural numbers r,s with 1 < r [I am guessing here we suppse m is composite] then  <m> \le <r> \not\le \mathbb{Z} [since <m> is the only proper subgroup of <r>, so <r> cannot be a subgroup of \mathbb{Z}] , and |<r> / <m> | = | <r> / <rs> | = s [I understand this bit] .

    f 1 < s then <r> \not= <m>, hence s = 1. I understand that s = 1 would prove that m must be prime, but where does this part come from? Where in the hypothesis does it say that <r>= <m> ? Isn't the whole point that they are not equal?

    Any help with this would be appreciated, thank you
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  2. #2
    Senior Member slevvio's Avatar
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    also surely <r> is a subgroup of the integers?
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    Quote Originally Posted by slevvio View Post
    also surely <r> is a subgroup of the integers?

    The original question is obviously false. What it should be, imo, is: G is a group whose only sbgps. are G,\,\{1\}\Longleftrightarrow |G|=p= a prime.

    Tonio
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Use the correspondence theorem to show that

    the only subgroups of \mathbb{Z} are  \mathbb{Z} and <m> \iff m is prime
    Quote Originally Posted by tonio View Post
    The original question is obviously false. What it should be, imo, is: G is a group whose only sbgps. are G,\,\{1\}\Longleftrightarrow |G|=p= a prime.

    Tonio
    Could this possibly be saying that the only non-trivial ideals I of \left(\mathbb{Z},\cdot,+\right) are I=\langle p\rangle with p prime?
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    Quote Originally Posted by Drexel28 View Post
    Could this possibly be saying that the only non-trivial ideals I of \left(\mathbb{Z},\cdot,+\right) are I=\langle p\rangle with p prime?
    No. nZ is always an ideal no matter n is prime or not. pZ just makes it maximal. But I believe that's unrelated to the question.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by FancyMouse View Post
    No. nZ is always an ideal no matter n is prime or not. pZ just makes it maximal. But I believe that's unrelated to the question.
    I meant to say maximal.
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    Quote Originally Posted by Drexel28 View Post
    I meant to say maximal.


    This, I think, is already assuming too much: the OP clearly wrote subgroups, nothing about rings, ideals or maximal.

    Let us give the OP a chance to clear out his own question, shall we?

    Tonio
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