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Thread: another group theory question

  1. #1
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    another group theory question

    Let G be a group and H be a subgroup of G. For any $\displaystyle g_1 , g_2 \in G$ prove that:
    $\displaystyle g_2 \in g_1H \Rightarrow g_1H = g_2H$
    Use this to prove that any two left cosets of H are equal or disjoint.

    help would be much appreciated! thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by situation View Post
    Let G be a group and H be a subgroup of G. For any $\displaystyle g_1 , g_2 \in G$ prove that:
    $\displaystyle g_2 \in g_1H \Rightarrow g_1H = g_2H$
    Use this to prove that any two left cosets of H are equal or disjoint.

    help would be much appreciated! thanks
    I'm sure you're aware that $\displaystyle (ab)H=a(bH)$ and $\displaystyle hH=H$ for all $\displaystyle h\in H$. So, note that $\displaystyle g_2\in g_1 h\implies g_2=g_1h$ for some $\displaystyle h\in H$ and so $\displaystyle g_1h^{-1}=g_2$. Thus, $\displaystyle g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H$
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    Quote Originally Posted by Drexel28 View Post
    I'm sure you're aware that $\displaystyle (ab)H=a(bH)$ and $\displaystyle hH=H$ for all $\displaystyle h\in H$. So, note that $\displaystyle g_2\in g_1 h\implies g_2=g_1h$ for some $\displaystyle h\in H$ and so $\displaystyle g_1h^{-1}=g_2$. Thus, $\displaystyle g_2H=(g_1h^{-1})H=g_1(h^{-1}H)=g_1H$

    thanks man, just a quick one about the result $\displaystyle hH=H$ how is this the case? if you wouldnt mind explaining that further it would be much appreciated!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by situation View Post
    thanks man, just a quick one about the result $\displaystyle hH=H$ how is this the case? if you wouldnt mind explaining that further it would be much appreciated!
    No problem! By definition $\displaystyle hH=\left\{hh':h'\in H\right\}$ clearly then since each element is the multiplication of elements in $\displaystyle H$ we have that $\displaystyle hH\subseteq H$. But, if $\displaystyle h'\in H$ then $\displaystyle h^{-1}h'\in H$ and so $\displaystyle h(h^{-1}h')=h'\in hH$ and so $\displaystyle H\subseteq hH$. Therefore, $\displaystyle H=hH$ (this is a result usually proven along with Cayley's theorem, that the map $\displaystyle \sigma_h:H\to H:h'\mapsto hh'$ is a permutation)
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