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Math Help - subgroups of a factor group

  1. #1
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    subgroups of a factor group

    Hi! I have a question concerning subgroups: Define H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}, where p is a prime. Then H_p is a subgroup of (\mathbb{Q},+). I want to show:

    (1) In the quotient group \mathbb{Q} / \mathbb{Z} the subgroup G_p=H_p / \mathbb{Z} is exactly the set of elements with order a power of p (EDIT).

    (2) What are the real subgroups of G_p?

    Does anyone have ideas for how to solve this? I am thankful for any hints.
    Banach
    Last edited by Banach; October 21st 2008 at 01:09 PM.
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  2. #2
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    Quote Originally Posted by Banach View Post
    Hi! I have a question concerning subgroups: Define H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}, where p is a prime. Then H_p is a subgroup of (\mathbb{Q},+). I want to show:

    (1) In the quotient group \mathbb{Q} / \mathbb{Z} the subgroup G_p=H_p / \mathbb{Z} is exactly the set of elements with order p.
    i haven't thought about yet but this part doesn't seem quite right to me, because \frac{1}{p^2} + \mathbb{Z} \in G_p is not of order p.
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  3. #3
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    Oh yes, thank you. It should be the set of elemtents with order a power of p. I will change it.
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  4. #4
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    Quote Originally Posted by Banach View Post
    Hi! I have a question concerning subgroups: Define H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}, where p is a prime. Then H_p is a subgroup of (\mathbb{Q},+). I want to show:

    (1) In the quotient group \mathbb{Q} / \mathbb{Z} the subgroup G_p=H_p / \mathbb{Z} is exactly the set of elements with order a power of p (EDIT).
    let r = \frac{a}{b} \in \mathbb{Q}, \ \gcd(a,b)=1. then the order of r + \mathbb{Z} is a power of p if and only if p^k r =\frac{p^ka}{b} \in \mathbb{Z}, for some integer k \geq 0, which is equivalent to say that b \mid p^k, since \gcd(a,b)=1.

    thus b=p^n, for some 0 \leq n \leq k. hence r=\frac{a}{p^n} \in H_p.

    (2) What are the real subgroups of G_p?
    the subgroups of G_p are exactly in the form N/\mathbb{Z}, where N is a subgroup of H_p, which (properly) contains \mathbb{Z}. since we're looking for proper subgroups, we have N \neq H_p. we need a lemma first:

    Lemma: if \frac{a}{p^k} \in N, for some k \geq 0 and a \in \mathbb{Z} with \gcd(a,p)=1, then \frac{n}{p^s} \in N, for all n \in \mathbb{Z}, \ 0 \leq s \leq k.

    Proof: since \gcd(a,p^k)=1, there exist integers u,v such that up^k+va=1. thus: \frac{1}{p^k}=u+v\frac{a}{p^k} \in N. hence, if n \in \mathbb{Z} and 0 \leq s \leq k, then \frac{n}{p^s}=np^{k-s}\frac{1}{p^k} \in N. \ \ \Box

    now let A=\{k \geq 0: \ \frac{a}{p^k} \in N, \ \text{for some} \ a \in \mathbb{Z} \ \text{with} \ \gcd(a,p)=1 \}. if A is unbounded, then by the Lemma, we will have N=H_p, which is not a proper subgroup. otherwise, let m=\max A.

    then again the Lemma gives us: N=\{\frac{n}{p^s}: \ n \in \mathbb{Z}, \ 0 \leq s \leq m \}=\frac{1}{p^m}\mathbb{Z}. since \mathbb{Z} \subsetneq N, we must have m \geq 1. so the set \left \{\frac{\frac{1}{p^m}\mathbb{Z}}{\mathbb{Z}}: \ m \geq 1 \right \} is exactly the set of all proper subgroups of G_p. \ \ \Box
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  5. #5
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    Thank you very, very much for your so well-structured answer. My problem was that i always thought of the wrong group operation, namely (\mathbb{Q},\cdot).

    Nice greetings
    Banach
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