Show that there are 2 fixed points and count the number of the orbits Let $H = (R − {0}, ·)$ be the group of nonzero real numbers with the usual product. Consider the direct product $G := H × H$. Let L be the set of all lines in the xy-plane.
(a) Show that G acts on L by * where for any $(r, s) ∈ G$ and any line $l ∈ L$ the action * is defined as follows:
if the equation of the line l is ax + by = c then the equation of the line $(r, s) * l$ is $rax + sby = rsc$.
(b) Find the orbits and stabilizers of the following lines:
y = x + 1, x = 1, x = 0, y = x.
(c) Show that there are two fixed points, that is $|L^G| = 2$.
(d) Show that there are 6 orbits.
So I did the a and b parts, but I just can't find the second fixed point. I have found the first point as 0=0, but I am not really sure if that is actually true. And I have absolutely no idea how to find those 6 orbits. I mean I found like 5: $rax=0$, $sby=0$, $x=\frac{sc}{a}$, $y=\frac{rc}{b}$ and $y=\frac{rax}{sb}+\frac{rc}{b}$. Can you help me understand, please?
Edit: I think i know those 2 lines, they are the x-axis and y-axis, right?
 A: The only fixed points are indeed the $x$-axis and the $y$-axis this is because if $ax+by=c$ is a fixed point then for every $r,s\in G$ you have $rax+sby=rsc$.
If $b\not = 0$ then take $r=2$ and $s=1$ then $(r,s)*(ax+by=c) = (2ax+by=2c)$ and this can't be equal to $(ax+by=c)$ unless $b=0$.
If $a\not = 0$ take $r=1$ and $s=2$.
Hence either $a$ or $b$ are zero. If $b=0$ then you have $ax=c$, this is invariant only if it is invariant under $r=1$ and $s=2$, then $ax=c$ is equivalent to $ax=2c$ this is only possible if $c=0$ hence this line is the $x$-axis, and the same with $y$.
Now you are right that we can take $a=b=0$ but then you won't get a line.

about the orbits: so it's a good approach to start guessing and then make sure we guessed everything.
I claim there are 6 orbits and they are


*

*The orbit of $x=0$

*The orbit of $y=0$

*The orbit of $x=1$

*The orbit of $y=1$

*The orbit of $x+y=0$

*The orbit of $x+y=1$


It is a direct computation that these are 6 different orbits, so I will only show that there are no more, that is give $ax+by=c$ we have to show that it is an element in one of these orbits. There are $6$ options


*

*If $a=0$ (then $b\not = 0$) and $c=0$ then $ax+by=c$ is an element in the orbit of $y=0$

*If $b=0$ (then $a\not = 0$) and $c=0$ then $ax+by=c$ is an element in the orbit of $x=0$

*If $a=0$ but $c\not = 0$ then $ax+by=c$ lies in the orbit of $y=1$

*if $b=0$ but $c\not = 0$ then $ax+by=c$ lies in the orbit of $x=1$


This 4 cases answers all the options in which either $a$ or $b$ are zero, the only options left is that $a$ and $b$ are different then zero


*if $a,b\not = 0$ and $c=0$ then $ax+by=c$ lies in the orbit of $x+y=0$

*if $a,b\not = 0$ and $c\not = 0$ then $ax+by=c$ lies in the orbit of $x+y=1$.


and there are no more options.
