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 $\displaystyle G$ is a topological group with multiplication and inversion maps given by $\displaystyle \mu,\iota$ respectively and $\displaystyle H\trianglelefteq G$. It is easy to prove that $\displaystyle H\leqslant G\implies\overline{H}\leqslant G$ (to get an idea look here where I prove the inversion part.) Now, to prove it's normal notice that the conjugation map
$\displaystyle \alpha_{g_0}:G\to G:g\mapsto g_0gg_0^{-1}$ is continuous since since $\displaystyle \alpha_{g_0}=\mu_{\mid \{g_0\}\times G}\circ\mu_{\mid G\times\{g_0^{-1}\}}$
So, using continuity we may conclude that
$\displaystyle g_0\overline{H}g_0^{-1}=\alpha_{g_0}\left(\overline{H}\right)\subseteq\ overline{\alpha_{g_0}\left(H\right)}=\overline{g_0 Hg_0^{-1}}=\overline{H}$. Since $\displaystyle g_0$ was arbitrary it follows that $\displaystyle \overline{H}\trianglelefteq G$