Can we have some (new) examples of group extensions $G/N=Q$ with continuous (e.g. Lie groups) $G$ and $Q$, but a finite discrete $N$? Note that $1 \to N \to G \to Q \to 1$.
What I know already contains:
$$SU(2)/\mathbb{Z}_2=SO(3).$$
$$\frac{\mathbb{R}/{\mathbb{Z}}}{\mathbb{Z}_n}={\mathbb{R}}/{(n\mathbb{Z})}.$$
What else are the examples that you can provide?
A systematic answer to obtain new examples (a few or even a list) is of course most welcome. ; )
In general, there is a way to classify extensions of group $1 \to N \to G \to Q \to 1$ for given $Q$ and $N$. You need
a group morphism $\phi$ from $Q$ to $Out(N)$ (the outer automorphism group of $N$),
and the cohomological class of a $2$-cocycle in $H^2(Q,Z(N)_{\phi})$ ($Z(N)_{\phi}$ means that $Z(N)$ is considered as a $Q$-module with the action given by $\phi$).
You can read about the link between extensions of groups and second cohomology groups in Adem and Milgram’s $\textit{Cohomology of Finite Groups}$.
I assume that by $G$ and $Q$ continuous groups, you mean topological groups. As far as I can tell, once you are given such extension, where $N$ is finite and $Q$ is a topological group there is only one way to extend the topology, the same is true for a Lie group. This comes from the fact that whenever we have $1 \to N \to G \to Q \to 1$ where $N$ is finite and $G$ and $Q$ are topological groups, the projection map $G\to Q$ is necessarily a topological cover.
Edit : as written in the comment below, you also need to add the condition that $\phi$ is continuous to ensure that $G$ is a topological group (here $N$ has the discrete topology).
Let $G = Q = \{z; z \in \Bbb C, \left| z \right| = 1\}$ and $N = \{1,-1\}$.