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Math Help - Give an example..

  1. #1
    ynj
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    Give an example..

    An infinite group has no proper infinite subgroup. Give an example.
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    Quote Originally Posted by ynj View Post
    An infinite group has no proper infinite subgroup. Give an example.
    our very own Prufer group: fix a prime number p and define G=\left \{\frac{m}{p^n} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset \mathbb{Q}/\mathbb{Z}. every proper subgroup of G is finite and cyclic. in fact H_n=<1/p^n + \mathbb{Z}>, where the

    integer n \geq 0, is fixed, are exactly all proper subgroups of G. this is easy to prove and i'll leave it for you. a standard notation for G is \mathbb{Z}(p^{\infty}).
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    ynj
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    Quote Originally Posted by NonCommAlg View Post
    our very own Prufer group: fix a prime number p and define G=\left \{\frac{m}{p^n} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset \mathbb{Q}/\mathbb{Z}. every proper subgroup of G is finite and cyclic. in fact H_n=<1/p^n + \mathbb{Z}>, where the

    integer n \geq 0, is fixed, are exactly all proper subgroups of G. this is easy to prove and i'll leave it for you. a standard notation for G is \mathbb{Z}(p^{\infty}).
    How about H=\left \{\frac{m}{p^{2n}} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset G?
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    Quote Originally Posted by ynj View Post
    How about H=\left \{\frac{m}{p^{2n}} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset G?
    this is not a "proper" subgroup because then \frac{m}{p^{2n-1}} + \mathbb{Z} =\frac{mp}{p^{2n}} + \mathbb{Z} \in H, for all integers m,n with n \geq 1, and so H=G.
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    ynj
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    Quote Originally Posted by NonCommAlg View Post
    this is not a "proper" subgroup because then \frac{m}{p^{2n-1}} + \mathbb{Z} =\frac{mp}{p^{2n}} + \mathbb{Z} \in H, for all integers m,n with n \geq 1, and so H=G.
    hmm..I have got it... If \frac{p^km}{p^n}appears, where \gcd (m,p)=1, then \frac{1}{p^{k}},k\leq n-mappears.. If there is infinite elements in a subgroup, then n-m\leq Mfor every element. And this will be a contradiction...
    Great! I have thought of your example before, but I was stuck on my example..Now I get it !
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