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Math Help - Topological Subgroup

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    Topological Subgroup

    If we let H be a subgroup of (X, x, T) where x is the binary operation. Show that
    (i)if H is normal, then H* is normal

    H* is the closure of H
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Turloughmack View Post
    If we let H be a subgroup of (X, x, T) where x is the binary operation. Show that
    (i)if H is normal, then H* is normal

    H* is the closure of H
    Of course this is supposed to be a topological group, right?

    So, suppose that G is a topological group with multiplication and inversion maps given by \mu,\iota respectively and H\trianglelefteq G. It is easy to prove that H\leqslant G\implies\overline{H}\leqslant G (to get an idea look here where I prove the inversion part.) Now, to prove it's normal notice that the conjugation map

    \alpha_{g_0}:G\to G:g\mapsto g_0gg_0^{-1} is continuous since since \alpha_{g_0}=\mu_{\mid \{g_0\}\times G}\circ\mu_{\mid G\times\{g_0^{-1}\}}

    So, using continuity we may conclude that

    g_0\overline{H}g_0^{-1}=\alpha_{g_0}\left(\overline{H}\right)\subseteq\  overline{\alpha_{g_0}\left(H\right)}=\overline{g_0  Hg_0^{-1}}=\overline{H}. Since g_0 was arbitrary it follows that \overline{H}\trianglelefteq G
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