Transitive action A group $G$ acts on a nonempty set $X$. Suppose $H$ is a subgroup of $G$ and $H$ acts transitively on $X$. In a proof, an author claims: If $H$ acts transitively on $X$, then obviously $G$ acts transitively on $X$. But why? I don't see this. Another thing which is said is: Fix $x \in X$. For $g \in G$, $gx=hx$ for some $h \in H$, so $h^{-1}g \in \text{Stab}_x$. My question concerning this: Why is $gx=hx$ and why can we imply $h^{-1}g \in \text{Stab}_x$? I thought, multiplying with $h^{-1}$ is only allowed in groups and $gx,\ hx$ are no group elements.

For the first question, this comes directly from the definition of a transitive group action.

Second, $gx \in X$ and as $H$ acts transitively on $X$, there is _some_ $h \in H$ such that $gx = hx$. Remember the definition of a group action- we have for every $g_1, g_2 \in G$ and $x \in X$, $g_1 * (g_2 * x) = (g_1g_2) * x $ and $e * x = x$. So if $g*x = h*x$, $$h^{-1}*(g*x) = h^{-1}*(h*x) \Rightarrow (h^{-1}g)*x = (h^{-1}h)*x = x$$ Thus, $h^{-1}g \in \text{Stab}_x$

Edit: Here, $*$ denotes the group action of $G$ on $X$. I've used $*$ so that we can explicitly see what's going on.