If H is a totally disconneted normal subgroup of a connected Hausdorff group G, show that H is contained in the center of G.
My attempt. I have to prove that $H \subseteq Z(G)$. Let's suppose $H \nsubseteq Z(G)$ so exists $a \in H$ such that $a \notin Z(G)$ then exists $b \in G$ such that $ab \neq ba$ and since $G$ is Hausdorff exist $U$ and $V$ open such that $ab \in U$ and $ba \in V$ and also $U \cap V=\emptyset $. Now let $f: G \to H$ defined as $f(x)=xax^{-1}$ since $H$ is normal and $a \in H$ we have that $xax^{-1} \in H$ but I'm not sure if $f$ is continous. If $f$ were continuous we would have that $f(G)$ is connected in $H$ and since H is a totally disconneted we have that $f(G)=\{h\}$ where $h \in H$. And this is what I have for the moment.
But $f(e) = eae^{-1} = a$ so $f(G) = \{a\}$.
($f$ is continuous because it is a composition of multiplications and inverses which are continuous functions.)