Torsion element

A torsion element has finite order.

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.

Enumerate all words $w_i$ on two generators in a free pro-p group. Choose powers $p^{e_i}$ so large that ${\sum }_i{\tau }^{p^{e_i}}<2\tau -1$ for some $1/2<\tau <1$, and impose the closed normal relations $w_i^{p^{e_i}}=1$. The Golod-Shafarevich infinitude criterion for pro-p groups makes the resulting pro-$p$ group infinite. Its abstract subgroup generated by the two generator images is dense and infinite; each element is represented by an enumerated word and has finite order. The ambient finite quotients separate its elements, so it is also a residually finite group. The abstract subgroup, rather than the whole topological group, is the desired finitely generated torsion group.

A group is locally finite when every subgroup that is a finitely generated group is finite. It is a torsion group, since the cyclic subgroup generated by one element must be finite. A restricted direct sum of groups with finite factors is locally finite. A locally finite group can be infinite, while a locally finite finitely generated group is finite by definition.

For $1\to N\to G\to Q\to 1$ with $N,Q$ locally finite, take a finitely generated subgroup $L\le G$. Its image in $Q$ is finite, so $L\cap N$ has finite index in $L$ and is finitely generated by Schreier's lemma. Local finiteness of $N$ makes that intersection finite. Thus $L$ itself is finite, proving local finiteness of $G$.

// Target: algebra.bigb

The torsion subgroup of an abelian group consists of all elements of finite order. For a finitely generated abelian group, it is the finite direct summand complementary to the free abelian part.

For an abelian group $G$ and an integer $m\ge 1$, its $m$-torsion subgroup is $G[m]=\{x\in G:mx=0\}$, the kernel of multiplication by $m$. In a finite abelian group, every nonempty fiber of this homomorphism has cardinality $|G[m]|$, so $|G:mG|=|G[m]|$. This follows directly: if $mx=my$, then $x-y\in G[m]$, and conversely translating by an element of $G[m]$ leaves the image unchanged. On the cyclic group ${\mathbb{Z}}_N$, the solutions of $mx=0$ number $\gcd (m,N)$; divide both integers by their greatest common divisor to see that $x$ must be a multiple of $N/\gcd (m,N)$.

For an abelian group $A$, its $p$-primary torsion subgroup is $A\{p\}=\{a:p^{r}a=0\text{ for some }r\ge 0\}$. It is zero exactly when $A[p]=0$, since a nonzero element of $p$-power order has a multiple of order $p$. Multiplication by an integer prime to $p$ is an automorphism on it.