Results 1 to 5 of 5

Thread: subgroups of a factor group

  1. #1
    Member
    Joined
    May 2008
    Posts
    75

    subgroups of a factor group

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

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

    (2) What are the real subgroups of $\displaystyle G_p$?

    Does anyone have ideas for how to solve this? I am thankful for any hints.
    Banach
    Last edited by Banach; Oct 21st 2008 at 12:09 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Banach View Post
    Hi! I have a question concerning subgroups: Define $\displaystyle H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $\displaystyle H_p$ is a subgroup of $\displaystyle (\mathbb{Q},+)$. I want to show:

    (1) In the quotient group $\displaystyle \mathbb{Q} / \mathbb{Z}$ the subgroup $\displaystyle G_p=H_p / \mathbb{Z}$ 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 $\displaystyle \frac{1}{p^2} + \mathbb{Z} \in G_p$ is not of order $\displaystyle p.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    75
    Oh yes, thank you. It should be the set of elemtents with order a power of p. I will change it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Banach View Post
    Hi! I have a question concerning subgroups: Define $\displaystyle H_p=\{\frac{n}{p^\alpha}: n, \alpha \in \mathbb{Z}, \alpha \geq 0\}$, where p is a prime. Then $\displaystyle H_p$ is a subgroup of $\displaystyle (\mathbb{Q},+)$. I want to show:

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

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

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

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

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

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

    then again the Lemma gives us: $\displaystyle N=\{\frac{n}{p^s}: \ n \in \mathbb{Z}, \ 0 \leq s \leq m \}=\frac{1}{p^m}\mathbb{Z}.$ since $\displaystyle \mathbb{Z} \subsetneq N,$ we must have $\displaystyle m \geq 1.$ so the set $\displaystyle \left \{\frac{\frac{1}{p^m}\mathbb{Z}}{\mathbb{Z}}: \ m \geq 1 \right \}$ is exactly the set of all proper subgroups of $\displaystyle G_p. \ \ \Box$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    75
    Thank you very, very much for your so well-structured answer. My problem was that i always thought of the wrong group operation, namely $\displaystyle (\mathbb{Q},\cdot)$.

    Nice greetings
    Banach
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Symmetric Group Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 18
    Last Post: Oct 8th 2011, 12:22 PM
  2. subgroups of the group <x,y:x=y=(xy)>
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 13th 2010, 02:23 AM
  3. Normal Subgroups and Factor Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 22nd 2009, 04:00 PM
  4. A group with no subgroups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Sep 30th 2007, 08:22 AM
  5. Normal Subgroups of A Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 29th 2006, 06:43 AM

Search Tags


/mathhelpforum @mathhelpforum