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Thread: proving in cosets

  1. #1
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    proving in cosets

    I have 2 questions on cosets.

    Let $\displaystyle G$ be a group with subgroup $\displaystyle \langle h \rangle$.
    Let $\displaystyle g \in G \backslash \langle h \rangle$.
    If there exists $\displaystyle N \lhd G$ such that $\displaystyle [a,h] \notin N$, show that $\displaystyle g \notin \langle h \rangle N$.

    I have tried this problem for a long time, but I still can't get a convinced proof.
    Please give me some hint.


    Another question is if $\displaystyle aN=bN$, is it true that $\displaystyle a=b$?
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  2. #2
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    Quote Originally Posted by deniselim17 View Post
    I have 2 questions on cosets.

    Let $\displaystyle G$ be a group with subgroup $\displaystyle \langle h \rangle$.
    Let $\displaystyle g \in G \backslash \langle h \rangle$.
    If there exists $\displaystyle N \lhd G$ such that $\displaystyle [a,h] \notin N$, show that $\displaystyle g \notin \langle h \rangle N$.

    I have tried this problem for a long time, but I still can't get a convinced proof.
    Please give me some hint.


    I assume your "a" must in fact be "g", so $\displaystyle [g,h]\notin N\Longrightarrow g^{-1}h^{-1}gh\notin N$ , so if $\displaystyle g\in \langle h\rangle N$ then

    $\displaystyle g=h^rn\,,\,r\in\mathbb{Z}\,,\,n\in N\Longrightarrow [g,h]=n^{-1}h^{-r}h^{-1}h^rnh=n^{-1}h^{-1}nh=[n,h]$ , but

    this last rightmost expression is in N (why? Hint: normality), so we get a contradiction.



    Another question is if $\displaystyle aN=bN$, is it true that $\displaystyle a=b$?


    Not at all: $\displaystyle aN=bN\Longleftrightarrow b^{-1}a\in N$ and that's all.

    For example, take $\displaystyle N=5\mathbb{Z}\,,\,a=2\,,\,b=7\Longrightarrow 2+N=7+N$

    Tonio


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