Image of homomorphism that acts transitively and contains the stabilizer is surjective.

Question

Let $\phi:G_1\rightarrow G_2$ be a homomorphism of groups. Let $G_2$ act on a set $X$, and let $\phi(G_1)$ act transitively on the same set $X$. Finally, let $p\in X$ be arbitrary, and suppose that the stabilizer $G_{2}(p)=\{g\in G_2: gp=p\}\subseteq \phi(G_1)$. How do I show that $\phi(G_1)=G_2$?

I have one idea: let $\alpha\in G_2$ be arbitrary. Then $\alpha\cdot p,p\in X$. Since $\phi(G_1)$ acts transitively on $X$, $\exists \beta\in G_1$ such that $\phi(\beta)\cdot p=\alpha\cdot p$, so then $p=\phi(\beta)^{-1}\alpha\cdot p=\phi(\beta^{-1})\alpha\cdot p$, so then $\phi(\beta^{-1})\alpha\in G_2(p)$. I'd like to conclude that $\alpha=\phi(\beta)$, but I'm stuck. Can anyone help?

Answer 1

Since $\phi(\beta^{-1})\alpha \in G_2(p)\subset \phi(G_1)$, there exists $\gamma \in G_1$ such that $\phi(\beta^{-1})\alpha = \phi(\gamma)$. So $\alpha = \phi(\beta\gamma) \in \phi(G_1)$.