Topological Subgroup

**Turloughmack** · November 8th 2010, 09:36 AM

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

**Drexel28** · November 8th 2010, 09:54 AM

Originally Posted by Turloughmack

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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