This will be part $1$ of a $4$ part question regarding proving certain properties about the group action.
Part $2$ - The kernel of the action is the intersection of the stabilizers
Part $3$ - Set of orbits of $X$ form a partition of $X$
Part $4$ - Group action transitive iff $G \cdot x = X$
For this problem the statement is:
Suppose $G \times X \to X$ is a group action. Prove that $\text{stab}_{G}(x)$ is a subgroup of $G$
For this we note that since $e \cdot x = x$ by the 'function definition' of an action then $\text{stab}_{G}(x)$ is nonempty since the identity of $G$, $e \in \text{stab}_{G}(x)$
Next, let $g,h \in \text{stab}_{G}(x)$ then we have, by 'function definition' of the action that: $$(gh) \cdot x = g \cdot (h \cdot x) = g \cdot (x) = g \cdot x = x$$ Thus $gh \in \text{stab}_{G}(x)$
Finally, let $g \in \text{stab}_{G}(x)$. Since $g \in G$ we know, by the definition of a group, there exists an inverse element $g^{-1} \in G$ such that $gg^{-1} = g^{-1}g = e$. Now since $e = gg^{-1} = g^{-1}g \in \text{stab}_{G}(x)$ and by the definition of an action we have: $$x = e \cdot x = (g^{-1}g) \cdot x = g^{-1} \cdot (g \cdot x) = g^{-1} \cdot x = x$$ Thus $g^{-1} \in \text{stab}_{G}(x)$. Consequently $\text{stab}_{G}(x) < G$. $\square$
How does this proof look? Any improvements/suggestions if it is indeed correct?