On the Hausdorff property, and if it passes to quotient space. I wrote in my notes that the Hausdorff property in topological spaces (given two distinct point belonging to the topology one can always find two disjoint neighbourhoods that contain them) does not pass to quotient spaces.
Then I have an example that says: Consider $X = R^2$,  $G = GL(2, R)$  then 
$$Y= X/G =\{ (0,0), (1,0) \}$$
But I am unsure of what $X/G$ means since I have always been dealing with equivalence relations when I take the quotient of a space and here $G$ is a group. 
Could somebody be so kind to explain how $X/G$ is defined and give me my first example?
 A: Suppose that we have the action of a group on a topological space, $X$.
For a given $g\in G$ we can assign $f_g: X \rightarrow X$ to it such that $f_e=1_X$ and this choice is associative and so on.
Then we can consider for each point $x \in X$ the orbit of $x$ under $f_g$. $Orb(x)=\{f_g (x)$ with  $g \in G\}$
It's easy to see that this induces an equivalence relation $x \sim y $ iff $y \in Orb(x)$.
Now $X/ \sim$ it's what you're looking for.
As a first example consider $\mathbb{Z}$ action on $\mathbb{R}$ by $f_n (x)=x+n$. Try to show for example that $\mathbb{R}/\sim \cong S^1$
Extra: There's an useful lemma to see if the quotient is also Hausdorf.
Consider the product space $X \times X$ and the diagonal given by the pairs that are in the same equivalence class in the quotient. If that diagonal is closed in $X \times X$ then $X/\sim$ is Hausdorf.
That happens for example in this case.
A: Abellan's answer defines what $X / G$ is, $X$ modded out by the equivalence relation generated by $x \sim y$ iff there is some $g \in G$ with $gx = y$, when $G$ works on $X$. 
Yes, $X$ has two orbits: $(0,0)$ stays fixed for all $g \in G$. And for every other vectors $x,y \neq 0$, we can find some $g \in G$ (some linear invertable map) such that $gx = y$. So all other vectors are in one orbit. And by the quotient topology $\{(0,0)\}$ is closed, as it equals its class, which is closed and not open in the plane. And so $\{(0,1)\}$ (in the quotient topology) is open and not closed (as it corresponds to the class of the plane minus the origin).
So $X/G$ is not Hausdorff, but homeomorphic to the Sierpinski space $X = \{0,1\}$ with topology $\{\emptyset, X, \{0\}\}$ 
