Torsion-free does not mean torsion-free lower-central factors

The product of (a,pb,c) and (a prime,pb prime,c prime) is $(a+a^{\mathrm{\prime }},p(b+b^{\mathrm{\prime }}),c+c^{\mathrm{\prime }}+pa{b}^{\mathrm{\prime }})$, and the inverse is $(-a,-pb,-c+pab)$. Thus G is a subgroup. Set x=(1,0,0), y=(0,p,0), z=(0,0,1). Then $x^a{y}^b{z}^k=(a,pb,pab+k)$, giving a unique normal form and a finite generating list x,y,z.

Commutators are $(0,0,p(a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b))$, and $[x,y]=z^p$. Thus $[G,G]=⟨z^p⟩$: every commutator lies there and z^p is a commutator. This subgroup is central and nontrivial, so G is nilpotent of class two. The map $\psi (a,pb,c)=(a,b,c<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)$ is an onto homomorphism to ${\mathbb{Z}}^2\oplus \mathbb{Z}\mathrm{/}p\mathbb{Z}$, since the extra product term pab prime vanishes modulo p. Its kernel is exactly $⟨z^p⟩$. The induced quotient map is therefore a bijective homomorphism, with injectivity given by this kernel calculation.

For a positive integer m, repeated multiplication gives $(a,pb,c{)}^m=(ma,mpb,mc+pabm(m-1)\mathrm{/}2)$; induction follows by adding c+p(ma)b at the next multiplication. If this power is the identity, ma=mpb=0 forces a=b=0, and then mc=0 forces c=0. Hence every nonidentity element has infinite order. Nevertheless z[G,G] has order exactly p: z^j belongs to <z^p> exactly when p divides j. Since $p\ge 2$, this is nontrivial torsion in the abelianization.

For the mixed lower-central form choose first-layer lifts x,y,z, with the z exponent reduced to a residue r in {0,...,p-1}, and second-layer generator z^p. For a normal exponent k divide k=pq+r; then $x^a{y}^b{z}^k=x^a{y}^b{z}^r(z^p{)}^q$. The identity is the zero tuple. Thus the p-th power of the finite factor lift is a nontrivial carry into layer two, exactly as retained by F2. Setting p=1 would remove the torsion and is explicitly excluded.