Let $E/\mathbb{Q}$ be an elliptic curve. We know that by Serre in the non-CM case, for $p\geq5$,
$$\rho_p:Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow Aut(T_p(E))$$
is surjective iff
$$ \bar{\rho}_p:Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow Aut(E[p])$$
is surjective.
How does surjectivity imply that $E(\mathbb{Q})(p)$ is trivial?
I know that the maps are injective, so surjectivity then gives bijectivity. I'm trying to understand the how the maps actually operate. Any hints on this or links to articles where this is explained will be helpful.
If $E(\Bbb Q)(p)$ is not trivial, then $E(\Bbb Q)$ has a nontrivial point $P$ of order $p$. Because $P$ is rational, it is invariant under the action of $\bar{\rho_p}$.
On the other hand for any nonzero point $Q \in E[p]$ there are maps in $Aut(E[p])$ that don't fix $Q$, so $\bar{\rho_p}$ can't be surjective.