= Solution
Use the <two-prime bound on rational elliptic torsion> with the counts six at three and eight at five from <2/a/solution>. For a <prime> $\ell\ne3,5$, the $\ell$-primary <torsion subgroup> injects into both reduced groups by <reduction of torsion points on an elliptic curve>. Its order therefore divides the corresponding primary part of $\gcd(6,8)=2$. The three-primary subgroup injects into the group of order eight at five and is trivial; the five-primary subgroup injects into the group of order six at three and is trivial. Hence the entire rational <torsion subgroup> has order at most two.
The factorization $x^3+x^2+x+1=(x+1)(x^2+1)$ supplies $T=(-1,0)$, a nonidentity point of order two under the <elliptic curve group law>. Therefore
$$
\boxed{E(\mathbb Q)_{\rm tors}=\{O,(-1,0)\}\cong\mathbb Z/2\mathbb Z.}
$$
The primary-component argument avoids assuming injectivity on residue-characteristic torsion.
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