Suppose $G$ is a finite group and G acts transitively on some set $X$. Let $a$ and $b$ be two distinct elements of $X$ and $G_{a}$ and $G_{b}$ be stabilizers of $a$ and $b$ respectively.Show that $G_{a}$.$G_{b}\subsetneq $$G$.
By orbit-stabilizer theorem I know that $\bigm|G_{a}\bigm|$ $\bigm|X\bigm|= \bigm|G\bigm|$ and $\bigm|G_{b}\bigm|$ $\bigm|X\bigm|= \bigm|G\bigm|$ combining both of these equalities gives:$\bigm|G_{a}\bigcup G_{b}\bigm|$ $\le$2($\frac{\bigm|G\bigm|}{\bigm|X\bigm|}-1)$ $+$$1$ But i don't have any idea how to proceed further?
Suppose that $g\in G$ sends $a$ to $b$. If $G=G_aG_b$, then $g=uv$, where $u\in G_a$ and $v\in G_b$. Then $b = a^g = a^{uv} = a^v,$ so $a = b^{v^{-1}}$. But, $G_b$ is a subgroup of $G$, so $v\in G_b$ implies that $v^{-1}\in G_b$, so $a = b$, a contradiction.