Results 1 to 5 of 5

Thread: proof for multiplicative subgroups

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    96

    proof for multiplicative subgroups

    I don't have any idea where to start.
    Please give me hint(s) to prove the following:


    Let $\displaystyle A$ be a group with $\displaystyle h \in A$ be of infinite order.
    Let $\displaystyle x \in A \backslash \langle h^{s} \rangle$ for every positive integer $\displaystyle s$.
    If there exists a normal subgroup of finite index $\displaystyle N$ in $\displaystyle A$ such that $\displaystyle N \cap \langle h \rangle=\langle h^{rs} \rangle$ for some positive integer $\displaystyle r$, show that $\displaystyle x \notin \langle h^{s} \rangle N$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by deniselim17 View Post
    I don't have any idea where to start.
    Please give me hint(s) to prove the following:


    Let $\displaystyle A$ be a group with $\displaystyle h \in A$ be of infinite order.
    Let $\displaystyle x \in A \backslash \langle h^{s} \rangle$ for every positive integer $\displaystyle s$.
    If there exists a normal subgroup of finite index $\displaystyle N$ in $\displaystyle A$ such that $\displaystyle N \cap \langle h \rangle=\langle h^{rs} \rangle$ for some positive integer $\displaystyle r$, show that $\displaystyle x \notin \langle h^{s} \rangle N$.

    Surely you meant $\displaystyle x\notin h^sN$ in your last line above...

    Sime ideas: $\displaystyle x\in h^sN\Longleftrightarrow x=h^sn\,,\,\,n\in N$ . Now, suppose $\displaystyle [G:N]=k\Longrightarrow g^k\in N\,\,\forall g\in G$ , and

    thus $\displaystyle x^k\in N\,,\,say\,\,x^k=n_1\,,\,n_1\in N\Longrightarrow x^k\in \left(h^s N\right)^k=h^{ks} N\Longrightarrow n_1=h^{ks}m\,,\,m\in N$ ...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    96
    I'm very sure that it's $\displaystyle x \notin \langle h^{s} \rangle N$.
    Well, I'm not really understand where to use the 2nd line.
    But I did try. Can you please check for me?

    Suppose $\displaystyle x \in \langle h^{s} \rangle N$.
    Then $\displaystyle x=h^{st}n$ for some positive integer $\displaystyle t$ and $\displaystyle n \in N$.
    Since $\displaystyle x \in \langle h \rangle$, we let $\displaystyle x=h^{p}, p\in \mathbb{Z}$.
    Now we have $\displaystyle x=h^{p}=h^{st}n$.
    This implies $\displaystyle n \in N \cap \langle h \rangle = \langle h^{rs} \rangle$.
    So, $\displaystyle x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle$.
    Thus contradiction occurs.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by deniselim17 View Post
    I'm very sure that it's $\displaystyle x \notin \langle h^{s} \rangle N$.
    Well, I'm not really understand where to use the 2nd line.
    But I did try. Can you please check for me?

    Suppose $\displaystyle x \in \langle h^{s} \rangle N$.
    Then $\displaystyle x=h^{st}n$ for some positive integer $\displaystyle t$ and $\displaystyle n \in N$.
    Since $\displaystyle x \in \langle h \rangle$


    Why? From where did you conclude this?

    Tonio



    , we let $\displaystyle x=h^{p}, p\in \mathbb{Z}$.
    Now we have $\displaystyle x=h^{p}=h^{st}n$.
    This implies $\displaystyle n \in N \cap \langle h \rangle = \langle h^{rs} \rangle$.
    So, $\displaystyle x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle$.
    Thus contradiction occurs.
    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    Posts
    96
    Quote Originally Posted by tonio View Post
    .
    Sorry. There's a typo.
    $\displaystyle x \in \langle h \rangle \backslash \langle h^{rs} \rangle$.

    I think my proof works.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multiplicative Identity Proof
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Feb 9th 2011, 02:30 PM
  2. Replies: 4
    Last Post: Jan 26th 2011, 08:21 AM
  3. Subgroups and Intersection of Normal Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 1st 2010, 08:12 PM
  4. Abstract Algebra: subgroups and order proof
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: Nov 10th 2010, 01:38 AM
  5. Multiplicative Proof
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Oct 14th 2010, 02:29 PM

Search Tags


/mathhelpforum @mathhelpforum