In my studies I have seen these results that do not make sense to me. Will be grateful for any advice. I will also give my ideas of why they could be true.
1) For connected group $G$ of finite Morley rank, and $x\in G$, $x^{G}$ is indecomposable.
2) If $x$ is not in the centre of $G$, then $x^{G}\cup \{1_{G}\}$ is indecomposable.
Here if a set $W$ is (left)-indecomposable it means that for definable subgroup $H$ of $G$ and subset $W\subseteq G$, $\{gH \in G/H| gH\cap W \neq \emptyset \}$ can't have finite cardinal $>1$. So indecomposable means both left and right-indecomposable with this definition.
So my try for the first part:
$x^{G}$ cannot have finite index in $G$ since $G$ connected. Then assume (trying to contradict) that $\{gH \in G/H| gH\cap W \neq \emptyset \}$ has finite cardinal $>1$. This means that a finite collection $g_{1}H,...,g_{m}H$ has nonempty intersection with $x^{G}$ so $x^{g}\subseteq \cup_{i=1}^{m}g_{i}H$. I am not sure how to proceed from there but I want to show that connectedness is contradicted by how the finite cardinal $>1$.
For the second part am a bit stuck. Was thinking maybe to use the same method as the first part, but I have not made use of the fact that $x$ is not in the centre of $G$ which mean I cannot just reuse this method?
Thank you for any advice.