Consider the following situation: Let $G$ be a topological group with identity element $e \in G$ and let $G_0$ denote the identity component of $G$, i.e. the connected component of $e$ (actually, the topology of $G$ is induced by a metric $d$, i.e. $(G, d)$ is a metric space).
From standard results on topological groups, it follows that $G_0$ is a closed, characteristic subgroup of $G$, i.e. $G_0$ is a closed subset of the topological space $G$ and $\phi(G_0) \subseteq G_0$ for every group automorphism $\phi$ of $G$ (in particular, $G_0$ is a normal subgroup). Moreover, the factor group $G/G_0$ consists of all connected components of the topological space $G$ and is a totally disconnected topological group.
Furthermore, in my current situation, the following special properties hold:
- If the sequence $(g_n)_n \subseteq G$ converges to the identity element $e$ in $G$, then, for every $f \in G$, the sequence $(f \circ g_n)_n \subseteq G$ converges to $f$ in $G$.
- The metric $d$ (see above) is invariant under right multiplication, i.e. $d(f \circ g, h \circ g) = d(f, h)$ for all $f, g, h \in G$.
I'd like to show that the topological group $G$ is connected.
Therefore, I assume that $G$ is not connected and try to construct a contradiction. If $G$ is not connected, then $G_0$ is a proper normal subgroup of $G$, i.e. $G_0 \not = G$. Hence, there exists $f \in G \backslash G_0$ such that the connected component $C_f$ of $f$ and $G_0$ are disjont, i.e. $C_f \cap G_0 = \emptyset$ (note that $C_f = f \circ G_0 = G_0 \circ f$).
Can anybody give me a hint on how to proceed here, or does anybody see how to create the desired contradiction? Thanks in advance for any comment/hint etc.!
EDIT: As A.Rod pointed out, the first property is satisfied in every topological group, hence this is nothing special here. Furthermore, let me explain why I'd like $G$ to be connected: There are two proper topological subgroups $U_1, U_2 \leq G$ (i.e. $U_1, U_2$ are closed subgroups of $G$) that are both connected - hence, it may be likely that $G$ itself is connected as well.