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Math Help - Subgroup of a fraction group

  1. #1
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    Subgroup of a fraction group

    Let p be a prime and let  \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \} .

    Prove that if there is no upper bound on the orders of elements in a subgroup J of  \mathbb {Z}(p^ \infty) , then J = \mathbb {Z}(p^ \infty) .

    Proof so far.

    Suppose that J is a subgroup, and suppose that for some elements j \in J, ord(j) = \infty . So I need to show that some how this element would generate the whole group, how do I do that? Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let p be a prime and let  \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \} .

    Prove that if there is no upper bound on the orders of elements in a subgroup J of  \mathbb {Z}(p^ \infty) , then J = \mathbb {Z}(p^ \infty) .
    if \frac{1}{p^n} + \mathbb{Z} \in J, for all integers n \geq 1, then J=\mathbb{Z}_{p^{\infty}}, and we're done. so let m=\min \{n \geq 1: \ \frac{1}{p^n} + \mathbb{Z} \notin J \}. now suppose \frac{a}{p^m} + \mathbb{Z} \in J, for some a \in \mathbb{Z} with \gcd(a,p)=1. choose r,s \in \mathbb{Z}

    such that ra + sp=1. then: \frac{1}{p^m}+\mathbb{Z}=\frac{ra}{p^m} + \frac{s}{p^{m-1}} + \mathbb{Z}\in J. contradiction! hence \frac{a}{p^m} + \mathbb{Z} \notin J, for all integers a coprime with p. so: \forall n \geq m, \forall a \in \mathbb{Z} \ \text{with} \ \gcd(a,p)=1: \ \frac{a}{p^n} + \mathbb{Z} \notin J,

    and therefore p^{m-1} is an upper bound for the orders of elements in J. contradiction! \Box
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