Relation between Galois representation and rational $p$-torsion Let $E$ be an elliptic curve over $\mathbb{Q}$. Does the image of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ under the mod $p$ Galois representation tell us whether or not $E$ has rational $p$-torsion or not?
 A: Yes, it does, and in a rather straightforward way. The $p$-torsion in $E(\mathbb{Q})$ is precisely the fixed vectors under the Galois action. In particular, $E$ has full rational $p$-torsion if and only if the mod $p$ representation is trivial.
A: Let $E$ be an elliptic curve defined over $\mathbb{Q}$, and let $p$ be a prime. The mod $p$ representation attached to $E$ is a homomorphism
$$\rho_{E,p} : \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{Aut}(E[p])$$
given by $\sigma \mapsto (\phi_\sigma : E[p] \to E[p])$, such that $\phi_\sigma(R)=\sigma(R)$ for every $R\in E[p]$. If we fix a $\mathbb{Z}/p\mathbb{Z}$-basis $\{P,Q\}$ of $E[p]$, then $\text{Aut}(E[p])\cong \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$, and so we can regard $\rho_{E,p}$ as
$$\rho_{E,p} : \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{GL}(2,\mathbb{Z}/p\mathbb{Z}).$$
Now, $R$ is in $E(\mathbb{Q})[p]$ if and only if $\sigma(R)=R$ for all $\sigma \in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, if and only if $R$ is fixed by every $\phi_\sigma \in \text{Aut}(E[p])$, if and only if $R=\lambda P+\mu Q$ and the vector $(\lambda, \mu)$ is a common eigenvector of every matrix in $\text{GL}(2,\mathbb{Z}/p\mathbb{Z})$ in the image of $\rho_{E,p}$, with eigenvalue $1$. In particular, $E(\mathbb{Q})[p]$ is non-trivial if and only if there is a $\mathbb{Z}/p\mathbb{Z}$-basis of $E[p]$ such that the image of $\rho_{E,p}$ in $ \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$ is a subgroup of
$$\left\{ \left(\begin{array}[cc] $1 & b \\ 0 & c \end{array}\right): b\in \mathbb{Z}/p\mathbb{Z},\ c\in (\mathbb{Z}/p\mathbb{Z})^\times\right\}.$$
