= Solution
Assume <good reduction>, and write $N_p=\#\widetilde E(\mathbb F_p)$. The <surjectivity of good reduction over a local field> and the <formal group law> determine the torsion as follows. If $p\nmid n$, multiplication by $n$ is an automorphism of $E_1$, because its multiplication series has an integral compositional inverse with unit linear coefficient. Therefore reduction gives an isomorphism
$$
E(\mathbb Q_p)[n]\cong\widetilde E(\mathbb F_p)[n].
$$
To construct the inverse, lift a reduced point to $P$, solve $[n]Q=[n]P$ uniquely in $E_1$, and use $P-Q$. This proves the <prime-to-p torsion lifts uniquely at good reduction>, including uniqueness, rather than merely comparing orders.
For residue-characteristic torsion, use a <deep logarithm subgroup of a formal group>. Take $m=1$ for odd $p$, or $m=2$ at two. The <formal logarithm> identifies $E_m$ with $p^m\mathbb Z_p$, so $E_m$ has no torsion. The finite <quotient group>
$$
G=E(\mathbb Q_p)/E_m,\qquad |G|=N_p p^{m-1},
$$
therefore receives the <torsion subgroup> injectively. There is a complete test for which cosets actually contain torsion. For a coset $g$ of order $n$, choose a lift $P$. Then $[n]P\in E_m$. Its coset has a torsion representative exactly when
$$
\boxed{L([n]P)\in n p^m\mathbb Z_p.}
$$
When this condition holds, let $Q\in E_m$ be the unique point with $L(Q)=L([n]P)/n$. Then $T=P-Q$ is killed by $n$ and represents $g$. Conversely, any torsion representative has order precisely $n$, since an element killed in the quotient would belong to the torsion-free group $E_m$. It must therefore arise from this construction. Changing $P$ by $R\in E_m$ changes $L([n]P)$ by $nL(R)$, so the test is independent of the chosen lift. This proves the <torsion test using a deep formal logarithm subgroup> and gives the full torsion by a finite list of cosets and a logarithmic divisibility test.
For odd $p$, one has $E_m=E_1$, so all <torsion points of an elliptic curve> inject into $\widetilde E(\mathbb F_p)$. Nevertheless a reduced $p$-primary point need not lift to a torsion point: if its order is $p^r$, the test is $L([p^r]P)\in p^{r+1}\mathbb Z_p$. For $p=2$, the subgroup $E_2$ is torsion-free but $E_1/E_2$ has order two. Thus $E_1$ can have a two-torsion point, and one must use $G=E/E_2$ of order $2N_2$ rather than assert that all torsion injects under reduction. The <prime-two torsion bound for good reduction> follows: the reduction kernel contributes at most a subgroup of order two. These distinctions are essential to a correct determination of local <torsion subgroups>.
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