Killing vectors in flat FLRW metric I have the flat FLRW metric,
$$ ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2) $$
and a geodesic $\gamma(s)=(t(s),x(s),y(s),z(s))$ with parameter $s$. Two of the Killing vectors of the metric are $ \partial_x$ and $\partial_y$ which give rise to the constants of motion $ a^2\dot x$ and $ a^2 \dot y  $ respectively. This is how far I am. Now I need to show that from this follows that I can assume that the geodesic has $y(s)=0=x(s)$. This is not obvious for me from the above constants of motions, because they also permit $\dot x=\frac{const.}{a^2}$
If I can rule this out, I get $\dot x =0$ and therefore $x=const.$ and because I can set $x$ to zero locally, it is zero on the whole spacetime.
So my question is: Why is $\dot x =0$
Edit: Or am I completely on the wrong track?
 A: The correct statement is that we can always construct a geodesic such that  $x(s)=y(s)=0$ for every value of the affine parameter $s$. All that independently from our initial choice of the origin and orientation of orthogonal Cartesian coordinates $x,y,z$ in the $3$-manifolds normal to $\partial_t$ (the natural rest space of the considered spacetime).
The geodesics are solutions of the Euler-Lagrange equations of the Lagrangian
$${\cal L} = \sqrt{|-\dot{t}^2 + a(t(\xi))^2(\dot{x}^2+\dot{y}^2+\dot{z}^2)|}\:,\tag{1}$$
where the used parameter is a generic one $\xi$ ad the dot denotes the $\xi$-derivative.
As ${\cal L}$ does not explicitly depend on $x,y,z$, from E-L equations, we have the three constants of motion:
$$\frac{\partial {\cal L}}{\partial \dot{x}}\:, \quad \frac{\partial {\cal L}}{\partial \dot{y}}\:, \quad \frac{\partial {\cal L}}{\partial \dot{z}}\:.$$
Passing to describe the curves with the geodesical length $s$, with
$$ds = \sqrt{|-\dot{t}^2 + a(t(\xi))^2(\dot{x}^2+\dot{y}^2+\dot{z}^2)|} d\xi$$
these constants read, in fact,
$$a(t(s))^2 \dot{x}(s)\:,\quad a(t(s))^2 \dot{y}(s)\:, \quad a(t(s))^2 \dot{z}(s)\:,$$
where now the dot denotes the $s$-derivative.
In other words, there is a constant vector $\vec{c}\in\mathbb R^3$, such that, for every $s$:
$$a(t(s))^2 \frac{d\vec{x}}{ds} = \vec{c}\tag{2}$$
where $\vec{x}(s) = (x(s),y(s),z(s))$. The geodesics are described here by curves
$$\mathbb R \ni s \mapsto (t(s), \vec{x}(s)) \tag{3}\:.$$
Looking at the Lagrangian (1), one sees that it is invariant under spatial rotations. That symmetry extends to solutions of E-L equations. In other words we have that, if (3) is a geodesics, for $R\in SO(3)$,
$$\mathbb R \ni s \mapsto (t(s), \vec{x}'(s)) :=  (t(s), R\vec{x}(s)) \tag{4}$$
is a geodesic as well. 
Correspondingly, due to (2) we have the new constant of motion
$$a(t(s))^2 \frac{d\vec{x}'}{ds}=  a(t(s))^2 \frac{dR\vec{x}}{ds} = a(t(s))^2 R\frac{d\vec{x}}{ds} = R\vec{c}\tag{2'}$$
Unless $\vec{c}=0$(*), we can rotate this constant vector in order to obtain, for instance, $R\vec{c} = c \vec{e}_z$. This means that the new geodesic verifies 
$$a(t(s))^2 \frac{d\vec{x}'}{ds}\:\:  ||\:\: \vec{e}_z$$
the spatial part is parallel to $\vec{e}_z$. I will omit the prime $'$ in the following and I assume to deal with a geodesic with spatial  part parallel to $\vec{e}_z$ and thus, as $a\neq 0$, it holds $x(s)= x_0$, $y(s)=y_0$ constantly.
Let us finally suppose that the initial point of the geodesic is  $\vec{x}(0) = \vec{x}_0$. As the Lagrangian is also invariant under spatial translations, we also have that if (3) is a geodesic, for $R\in SO(3)$,
$$\mathbb R \ni s \mapsto (t(s), \vec{x}'(s)) :=  (t(s), \vec{x}(s)+ \vec{r}_0) \tag{5}$$
is a geodesic as well. Choosing $\vec{r}_0 := - \vec{x}_0$, we have a geodesic with $x(s)=y(s)=0$ as requested.
(*) We can always choose $\vec{c}\neq 0$ assuming that the initial tangent vector of the geodesic verifies this requirement (notice that $a^2 \neq 0$). And we know that there is a geodesics for every choice of the initial conditions.
A: My only issue is that, while the work above is obviously correct, there are actually 6 Killing vectors for the FLRW metrics, for spatial homogeneity (the 3 spatial ones above), and the 3 for rotations, which is not immediately obvious from the form of the metric above. Indeed, it may be more helpful to perform a coordinate transformation, and write the FLRW metric in the more standard form:
$ds^2 = -dt^2 + a^2(t) \left[dr^2 + r^2 \left(d\theta^2 + \sin^2 \theta d \phi^2\right)\right]$, from this form, the $G_{3}$ isotropy group around every point is quite clear which implies by definition, the existence of a $G_{3}$ simply transitive subgroup, but any comments on this would be helpful.
