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Math Help - proof for multiplicative subgroups

  1. #1
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    proof for multiplicative subgroups

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


    Let A be a group with h \in A be of infinite order.
    Let x \in A \backslash \langle h^{s} \rangle for every positive integer s.
    If there exists a normal subgroup of finite index N in A such that N \cap \langle h \rangle=\langle h^{rs} \rangle for some positive integer r, show that x \notin \langle h^{s} \rangle N.
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  2. #2
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    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 A be a group with h \in A be of infinite order.
    Let x \in A \backslash \langle h^{s} \rangle for every positive integer s.
    If there exists a normal subgroup of finite index N in A such that N \cap \langle h \rangle=\langle h^{rs} \rangle for some positive integer r, show that x \notin \langle h^{s} \rangle N.

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

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

    thus 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
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  3. #3
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    I'm very sure that it's 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 x \in \langle h^{s} \rangle N.
    Then x=h^{st}n for some positive integer t and n \in N.
    Since x \in \langle h \rangle, we let x=h^{p}, p\in \mathbb{Z}.
    Now we have x=h^{p}=h^{st}n.
    This implies n \in N \cap \langle h \rangle = \langle h^{rs} \rangle.
    So, x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle.
    Thus contradiction occurs.
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  4. #4
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    Quote Originally Posted by deniselim17 View Post
    I'm very sure that it's 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 x \in \langle h^{s} \rangle N.
    Then x=h^{st}n for some positive integer t and n \in N.
    Since x \in \langle h \rangle


    Why? From where did you conclude this?

    Tonio



    , we let x=h^{p}, p\in \mathbb{Z}.
    Now we have x=h^{p}=h^{st}n.
    This implies n \in N \cap \langle h \rangle = \langle h^{rs} \rangle.
    So, x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle.
    Thus contradiction occurs.
    .
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  5. #5
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    Quote Originally Posted by tonio View Post
    .
    Sorry. There's a typo.
    x \in \langle h \rangle \backslash \langle h^{rs} \rangle.

    I think my proof works.
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