On connected components of a topological/Lie group

Question

Let $G$ be a topological group (or a Lie group). If $G^0$ is the identity component, we know that $G^0$ is a normal subgroup of $G$. But is it true that if $C$ is any connected component of $G$, there is $a\in G$ such that $C=aG^0$?

I tried to come up with some counter-example, to no avail (I'm low on imagination, I guess), and I couldn't prove it either.

Initially I thought about the problem for topological groups, but if the situation is somehow easier for Lie groups you can discuss this case. Help?

Answer 1

Yes it's true. Let $C$ be a connected component and take any $a\in C$. Consider the map $L_a:G\to G$ defined by $L_a(g)=ag$. Then, $L_a$ is a homeomorphism, so $L_a(G^0)$ is a connected component. Since $a\in L_a(G^0)\cap C$ we must have $L_a(G^0)=C$.