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Thread: Group and subgroup

  1. #1
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    Group and subgroup

    (G;x) is a group ( e it's identity element). $\displaystyle H_1$ and $\displaystyle H_2$ are a subgroups from (G,x).
    $\displaystyle H_1H_2$={$\displaystyle x_1x_2$/$\displaystyle (x_1;x_2$}$\displaystyle \in$ $\displaystyle H_1$x$\displaystyle H_2$}

    1)-I must show that : $\displaystyle H_1H_2$ is a subgroup from(G;x) $\displaystyle \Longrightarrow H_1H_2=H_2H_1$
    2)- We assume that $\displaystyle H_1$ and $\displaystyle H_2$ are finished and $\displaystyle H_1\cap H_2$={e} and:
    $\displaystyle \phi$ : $\displaystyle H_1$x$\displaystyle H_2$ $\displaystyle \longrightarrow $$\displaystyle H_1$$\displaystyle H_2$
    ($\displaystyle x_1$;$\displaystyle x_2$)$\displaystyle \longrightarrow$ $\displaystyle x_1$$\displaystyle x_2$

    I must show that $\displaystyle \phi$ is surjective and Card($\displaystyle H_1H_2$)=Card($\displaystyle H_1$)Card($\displaystyle H_2$)
    I don't know anything!!!!!!! Can you give me some help please????
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bhitroofen01 View Post
    Let $\displaystyle H_1,H_2\leqslant G$ be such that $\displaystyle H_1\cap H_2=\{e\}$. Define $\displaystyle \phi:H_1\times H_2\to H_1H_2$ by $\displaystyle (h_1,h_2)\overset{\phi}{\longmapsto}h_1h_2$. Prove that $\displaystyle \phi$ is surjective.
    Is the above the second part?
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  3. #3
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    yes
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bhitroofen01 View Post
    yes
    Let's see some work.
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  5. #5
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    $\displaystyle H_1$$\displaystyle H_2$ is a subgroup from (G;x) $\displaystyle \Longleftrightarrow$ $\displaystyle H_1$x($\displaystyle H_2)^{-1}\in H_1H_2$
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  6. #6
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    Did you drop any assumptions? The statement doesn't seem true to me.
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