Finite torsion and the torsion-free quotient

In an abelian group, if $a^m=b^n=1$ with $m,n>0$, then $(ab{)}^{mn}=1$; inverses retain finite order. This also covers the trivial group. For a group of class $c\ge 2$ and $b\in G$, put $B=⟨b,{\gamma }_2(G)⟩$. It is normal, being the inverse image of the cyclic subgroup generated by $b{\gamma }_2$ in the abelianization. Modulo ${\gamma }_3$, ${\gamma }_2$ is central, so two elements $b^{r}u,b^{s}v$ commute. Therefore ${\gamma }_2(B)\le {\gamma }_3(G)$; inductively ${\gamma }_j(B)\le {\gamma }_{j+1}(G)$ for $j\ge 2$, using $[B,{\gamma }_{j+1}(G)]\le {\gamma }_{j+2}(G)$. Thus ${\gamma }_c(B)=1$.

Induct on class for torsion closure, with step 1.1 as base. For torsion $a,b$ in class $c$ with $a^m=1$, the subgroup $B$ has smaller class. Its torsion elements form a subgroup $T(B)$ by induction. Automorphisms preserve orders, so $T(B)$ is characteristic in $B$ and normal in $G$. Now $(ab{)}^m=(ab{a}^{-1})(a^{2}b{a}^{-2})\cdots (a^{m}b{a}^{-m})a^m$ is a product of conjugates of $b$ in $T(B)$. It has finite order, hence so does $ab$. Identity and inverses have finite order; thus $T(G)$ is a subgroup. Preservation of orders under every automorphism makes it characteristic.

For the original finitely generated $G$, $T=T(G)$ is finitely generated and nilpotent. Each of its lower-central factors is finitely generated abelian and torsion, since every representative in $T$ has finite order. An abelian group generated by elements of finite orders $d_1,\dots ,d_t$ has at most $\prod d_j$ elements: reduce each exponent modulo $d_j$. Hence all these factors are finite. Lifting their finite sets through the finite series proves $T$ finite (cardinalities multiply in each finite extension).

The quotient is nilpotent. If $(gT{)}^n=T$ for some $n>0$, then $g^n\in T$, so $(g^n{)}^m=1$ for some $m>0$. Thus $g^{nm}=1$, giving $g\in T$ and $gT=T$. This proves torsion-freeness, also when $T=G$ or $T=1$.