I am trying to solve the problem:
Let \begin{align*} G \times X &\longrightarrow X \\ (g,x) &\longmapsto g \circ x \end{align*} be an action of $G$ on $X$. Suppose that $K \leq G$, and that the action of $K$ on $X$ \begin{align*} K \times X &\longrightarrow X \\ (k,x) &\longmapsto k \circ x \end{align*} obtained by restriction of the first variable is transitive.
Prove that if $g \in G$ and $x \in X$, then there exists $k \in K$ and $b \in \operatorname{Stab}G(x)$ such that $g = kb$.
My thoughts are the following: If the equality $g = kb$ holds, then $g \circ x = (kb)\circ x = k\circ (b \circ x) = k \circ x$. And how can I prove that those elements $k$ satisfying $k\circ x = g\circ x$ still belong to $K$?
Any help will be appreciated.
You can't prove that if $g \circ x = k \circ x$ then $k \in K$, because it doesn't have to be in $K$. However, you are given that $K$ acts transitively on $X$. Can you use that to find $k \in K$ such that $g \circ x = k \circ x$? Once you find such $k$, you can see that
\begin{align} k^{-1} \circ (g \circ x) &= k^{-1} \circ (k \circ x) \\ (k^{-1} g) \circ x &= (k^{-1} k) \circ x \\ (k^{-1} g) \circ x &= e \circ x \\ (k^{-1} g) \circ x &= x. \end{align}
So $b := k^{-1} g \in \text{Stab}_G (x)$ and $g = kb$.