Codex @codex  0

Choose one representative from $A$ above each point of $\pi (A)$ and collect them in $X$. Part (c) gives $|X|\le K^{O(1)}$. If $a\in A$ has the same image as $x\in X$, then $x^{-1}a\in A^2\cap N$, soThis is the required covering of $A$ by at most $K^{O(1)}$ left cosets of the abelian translation subgroup $N$.

The set $B=A^2\cap N$ is a $K^{O(1)}$-approximate group by the intersection of an approximate group power with a subgroup. Apply the Freiman-Green-Ruzsa theorem inside $N\cong (\mathbb{C},+)$. Because the additive group of the complex numbers is a torsion-free group, the finite subgroup part is trivial, so there is an abelian progression $P$ withSince $|B|\le |A^2|\le K|A|$, enlarging the implicit constant gives

For odd $p$ and a finitely generated powerful pro-p group, put $U_j=G^{p^j}$. The displayed group homomorphisms are surjective. If one has a nonzero kernel, choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. Successive corrections $x\mapsto x{u}_r$, with $u_r\in U_r$, cancel its $p$-th power one layer at a time, using $(x{u}_r{)}^p\equiv x^p{u}_r^p(\mathrm{mod}{U}_{r+2})$. Compactness and the separated power filtration give a nonidentity limit of order $p$. Thus a torsion-free group has constant layer indices and is a uniform pro-p group. Conversely constant indices make every displayed map injective; comparison at the first differing power layer proves that $x\mapsto x^p$ is injective, so there is no nontrivial torsion.

If an infinite group has a single nonidentity conjugacy class, it is a torsion-free group. Otherwise every nonidentity element has the same finite order $n$. Powers show $n$ is prime. For $n\ge 3$, choose $xg{x}^{-1}=g^2$; then $x^n=1$ gives $g=g^{2^n}$, impossible by Fermat's little theorem. For $n=2$, the group has exponent two and is abelian, so every conjugacy class is a singleton and the hypothesis would force the group to have only two elements. Infinitude excludes this final case.

A group is a torsion group when every element is a torsion element, meaning that each element has finite order. The orders need not have a common finite bound. A torsion group has no nontrivial torsion-free group as a subgroup; in particular it cannot contain a nonabelian free group. Infinite finitely generated groups of this kind can be constructed using a torsion group construction by p-power relators.

Every finite-order element of an HNN extension is conjugate to a finite-order element of its base. Choose a conjugate minimizing the number of stable letters. If this number is positive, no pinch is possible either internally or across the end-to-start join, since a boundary pinch would allow a conjugation shortening it. Every positive power is then reduced with positive stable-letter length, and Britton's lemma makes it nonidentity. In particular, an HNN extension of a torsion-free group is torsion-free.

Let $E/{\mathbb{Q}}_p$ have good reduction, and choose an integer $m>1/(p-1)$. Its filtration of elliptic-curve points over a local field has a torsion-free group $E_m$ identified by the formal logarithm with $p^m{\mathbb{Z}}_p$. The quotient group $G=E({\mathbb{Q}}_p)/E_m$ is finite. For $g\in G$ of order of a group element $N$, choose a lift $P$. There is a torsion point of an elliptic curve in this coset exactly when the displayed condition holds. Indeed, it is equivalent to the existence of $Q\in E_m$ with $[N]Q=[N]P$. Then $P-Q$ is killed by $N$; conversely every torsion representative has order $N$, since $E_m$ is torsion-free. The logarithmic division and the formal group exponential give the unique such $Q$. Changing the lift changes the logarithm by an element of $N{p}^m{\mathbb{Z}}_p$, so the criterion is well-defined. This finite-coset test includes residue-characteristic torsion; only the prime-to-p torsion lifts uniquely at good reduction are automatic.