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Thread: Fun with normal subgroups.

  1. #1
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    Fun with normal subgroups.

    This stuff drives me bonkers. Anyway, I have been given or have proved the following:

    $\displaystyle \phi : X \rightarrow Y$ is onto.

    $\displaystyle x, \bar{x} \in X$

    $\displaystyle U \subset X$ is a subgroup

    $\displaystyle \phi (U)$ is a subgroup of $\displaystyle Y$

    ---------------------

    My last question (for this section) is:

    Verify that if $\displaystyle U$ is normal in $\displaystyle X$, then $\displaystyle \phi (U) $ is normal in $\displaystyle Y$

    -------------

    Can I say this:

    If $\displaystyle U$ is normal in $\displaystyle X$ then:

    $\displaystyle uXu^{-1}=X$ for all $\displaystyle u \in U$ , so:

    $\displaystyle \phi(uXu^{-1})= \phi (X)$

    $\displaystyle \phi(u)\phi(X) \phi(u^{-1})= \phi (X)$

    $\displaystyle \phi(u) Y \phi(u)^{-1}= Y$ for all $\displaystyle \phi(u) \in \phi(U)$

    Hence, $\displaystyle \phi(U)$ is normal in $\displaystyle Y$
    Last edited by MichaelMath; Oct 25th 2010 at 10:59 AM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    That's perfect. Of course, it all comes down to the fact that $\displaystyle \phi(u)$ can be any element of $\displaystyle Y$. You should perhaps write that down explicitly.
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