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An abelian group in which no nonzero element has finite order. Free abelian groups are torsion-free, but torsion-free groups need not be free, as the additive rationals show. This algebraic meaning differs from the torsion-free connection in differential geometry.

For an abelian group, the scheme $\forall x(nx=0\to x=0)$ for every positive integer $n$ expresses absence of nonzero elements of finite order. Each fixed $n$ gives one sentence, while the entire scheme expresses the torsion-free abelian group property.

A divisible Abelian group is torsion-free precisely when division by each nonzero integer is unique. Defining rational scalar multiplication by $n((m/n)a)=ma$ identifies these groups with vector spaces over the rational numbers, including the zero space.

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 $N/\gcd (m,N)$.

Integer row vectors forming a square unimodular matrix are a basis of the corresponding free abelian group: the adjugate formula makes the inverse integral. The subgroup spanned by any subset of that basis is therefore a direct summand, complemented by the remaining basis vectors. In contrast, a proper finite-index subgroup of a free abelian group cannot be a direct summand: its nonzero finite quotient would have to embed as a subgroup of a torsion-free group.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.

For every abelian group $A$ and group homomorphism $f:G\to A$, there is a unique group homomorphism $\bar{f}:G^{\mathrm{ab}}\to A$ such thatThis is a universal property. To see why it holds, compute $f([x,y])=[f(x),f(y)]=e_A$, using commutativity in $A$. Hence $G^{\prime }\subseteq \ker f$, and $f$ is constant on each coset of $G^{\prime }$. The displayed formula is therefore well-defined and preserves multiplication. It is unique because $q$ is surjective. This is an instance of factoring a group homomorphism through a quotient group; see Conrad, Universal mapping properties, theorem 3.4.

In particular, every abelian quotient $G/N$ comes from a surjective group homomorphism $G^{\mathrm{ab}}\to G/N$, $g{G}^{\prime }\mapsto gN$. For $S_3$, any group homomorphism to an abelian group is determined by the common image $a$ of its three transpositions, with $a^2=e_A$. Every even permutation must map to $e_A$. Thus sign of a permutation captures all possible information of this kind, including maps with infinite abelian groups as their targets.

Every balanced map $b:M\times N\to P$ factors uniquely through an $R$-module homomorphism $\tilde{b}:M{\otimes }_{R}N\to P$ satisfying $\tilde{b}(m\otimes n)=b(m,n)$.

The value group of a valued field is the ordered abelian group formed by the values of its nonzero elements. A valuation is discrete when its value group is infinite cyclic with the order inherited from $\mathbb{Z}$.

Addition makes $V$ an Abelian group: for all $u,v,w\in V$, $(u+v)+w=u+(v+w)$ and $u+v=v+u$, there is a zero vector $0$, and every $v$ has an additive inverse $-v$. Scalar multiplication satisfies, for $a,b\in \mathbb{F}$,Both operations take values in $V$. These requirements make manipulating linear combinations independent of how sums are grouped or expanded.

Every abelian group is amenable, and amenability is preserved by finite extensions. Hence every group with an abelian finite-index subgroup is an amenable group.

A group is virtually cyclic when it has a cyclic subgroup of finite index. A finitely generated abelian group is virtually cyclic exactly when its free abelian rank is one.

The word metric measures how many allowed moves are needed to change one group element into another. For a finite symmetric generating set of a group $S$ with $1\notin S$, it is the graph distance between vertices of the Cayley graph, equivalentlyHere word length counts letters in a shortest representative of $g^{-1}h$. The path metric on the entire Cayley graph also permits points inside edges; the word metric is its restriction to the group itself.^[ref]

The word metric is left invariant:This corresponds to the label-preserving group action on the Cayley graph. Symmetry follows by reversing a shortest graph path; concatenating routes gives the triangle inequality.^[ref]

Right invariance can fail in a non-abelian group, because right multiplication replaces $g^{-1}h$ by $a^{-1}{g}^{-1}ha$ through conjugation. In the free group $F(x,y)$ with $S=\{x,x^{-1},y,y^{-1}\}$, we have $d_S(1,y)=1$, whereas $d_S(x,yx)=|x^{-1}yx{|}_S=3$: the latter freely reduced word has three letters. In an abelian group, conjugation fixes each element, so right invariance does hold. Cayley graphs of the integers and word metric balls in the integer lattice give examples where the word metric can be calculated explicitly.

Any abelian group becomes a nonunital ring by declaring every product zero. Associativity and the distributive property hold because both sides are zero. When the group has more than one element, no multiplicative identity element exists: an identity would have to satisfy $ex=x$, whereas every such product is zero. This differs from the zero ring, which has just one element.

The zeroth homology group is the free abelian group on the path components of $X$. In particular, if $X$ has finitely many path components, then