 Solution

For a short Weierstrass equation of an elliptic curve with coefficients in the p-adic integers, a point with nonintegral $x$ has valuations $v_p(x)=-2m$ and $v_p(y)=-3m$ for some integer $m\ge 1$: the term $x^3$ strictly dominates $Ax+B$ in the equation, so $2v_p(y)=3v_p(x)$. Its formal parameter $T=-x/y$ therefore has valuation $m\ge 1$. These points, together with the identity, form the formal neighborhood $E_1({\mathbb{Q}}_p)$ of the identity, with parameter in $p{\mathbb{Z}}_p$ and the elliptic formal group law over the p-adic integers. For a point with $x\in {\mathbb{Z}}_p$, the equation gives $y\in {\mathbb{Z}}_p$ too.

The formal neighborhood has no nontrivial torsion. Multiplication by an integer coprime to $p$ has series $[m](T)=mT+O(T^2)$ and preserves $|T{|}_p$, so any torsion there must have $p$-power order. If $p\ge 3$, part (i) gives $|T{|}_p\ge p^{-1/(p-1)}>p^{-1}$, contradicting $T\in p{\mathbb{Z}}_p$.

The printed hypotheses also allow $p=2$. In that case, a nontrivial torsion point in the formal neighborhood would have order a power of two and hence yield a point of order two in that neighborhood. But every nonidentity point of order two on a short Weierstrass equation of an elliptic curve has $y=0$ and $x$ a root of the monic cubic over the p-adic integers $X^3+AX+B$. Such a root lies in the p-adic integers, so it cannot belong to the formal neighborhood, whose nonidentity points have nonintegral coordinates. This excludes the remaining case. This is the torsion-free formal subgroup for a short Weierstrass equation. Thus every affine torsion point satisfiesWith good reduction, the reduction homomorphism $E({\mathbb{Q}}_p)\to \tilde{E}({\mathbb{F}}_p)$ has kernel exactly this torsion-free formal neighborhood. Its restriction to the torsion subgroup is therefore injective, proving the required subgroup isomorphism. The short model over the p-adic integers is essential to the prime-two argument; no blanket torsion-injectivity assertion for arbitrary models is needed.