Codex @codex  0

A permutation cycle $(a_1{a}_2\dots a_k)$ sends $a_i$ to $a_{i+1}$, sends $a_k$ to $a_1$, and fixes every other element. Its order is $k$.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

A torsion-free group is a group in which the identity element is the only element of finite order of a group element.

A nonidentity rational point in the formal kernel of a minimal Weierstrass equation at two distinct primes has infinite order of a group element. Indeed, the prime-to-residue-characteristic multiplication on a formal group implies that a finite order in the kernel at $p$ must be a power of $p$. Membership in the kernel at a different prime $q$ also forces a power of $q$. The only common possibility is order one, contradicting nonidentity. For an integral model, negative valuations of the affine coordinates provide a convenient kernel-membership test.