I have to compute the Torsion Subgroup of $E(\mathbb{Q})$, where $E:Y^2=X^3 +2$ using Nagell-Lutz. So, if $(x,y) \in E(\mathbb{Q})_{\text{tors}}$ then $x,y \in \mathbb{Z}$ and either $y=0$ or $y^2 \ \vert \ \Delta = 4(0)^3 + 27(2)^2 = 108$
Clearly, if $y=0$, then $x \notin \mathbb{Z}$. So $y^2 \ \vert \ 108 \implies y=\pm1, \pm2,\pm3,\pm6$. Accomodating $x \in \mathbb{Z}$, we eliminate possibilities of $y$ to get the following points: $P=(-1,\pm1)$
I see that $2P$ does not have integer coefficients, is that enough to claim that the torsion subgroup is trivial?
The contrapositive of Nagell-Lutz tells us that if $Q = (x, y)$ are not integers, then $Q$ is NOT a rational point of finite order on $E$.
Since $Q = 2P$ (according to you) have coordinates which are not integers, then $Q$ is not a torsion point.
Now, given this, what if $P$ is a torsion point? Does something go wrong?