For simplicity, let $G$ be a finite group with discrete topology and $M$ be a smooth manifold.
Suppose that $G$ acts on $M$ smoothly. Then, it is kind of trivial to see that the action of $G$ is proper.
However, I further assume that $G$ does NOT act freely on $M$. That is, there exists some $g (\neq e_G) \in G$ and a point $p \in M$ such that $g \cdot p =p$.
Then, can I still define the quotient space $M/G$ as some kind of a manifold?
If I remember right, $M/G$ is NOT Hausdorff. Nevertheless, is $M/G$ still locally Euclidean and given a smooth structure in lieu with the projection map $\pi : M \to M/G$?
The link https://en.wikipedia.org/wiki/Non-Hausdorff_manifold somewhat gives hope to me in this aspect, but I cannot judge on my own..
Could anyone please help me?
The simplest example of a smooth finite group action such that the quotient space is not a topological manifold is $$ M={\mathbb R}^3, G\cong {\mathbb Z}_2, $$ and the generator $g$ of $G$ acts on $M$ by $$ g({\mathbf x})= -{\mathbf x}. $$ The quotient space $M/G$ is homeomorphic to the (open) cone over ${\mathbb R} P^2$, hence, is not a topological manifold.