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For a topological space $X$ and abelian group $A$, singular cohomology is the cohomology of the cochain complex $C^n(X;A)=\mathrm{Hom}(C_n(X;\mathbb{Z}),A)$, where $C_n$ is the singular chain group. Its differential is dual to the singular boundary. This theory has relative cohomology, long exact sequences and the Excision theorem. On spaces that are not locally contractible it can differ from Čech cohomology.

For $A$ in a finite abelian group and normalized Fourier transform $\widehat{1_A}$, the normalized number of sextuples satisfyingisThis follows by inserting orthogonality of complex exponentials for the displayed equation: the three variables on one side contribute $\widehat{1_A}(\gamma {)}^3$ and those on the other side contribute its complex conjugate.

For a point inclusion $i:\{x\}\hookrightarrow X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$ has sections $A$ on opens containing $x$ and zero on other opens. Its restriction maps are identities or maps to zero. If $A\ne 0$, its nonzero stalks occur at every point of the closure $\overline{\{x\}}$: exactly those points whose every neighbourhood contains $x$. Thus the familiar assertion that only the stalk at $x$ is nonzero requires $x$ to be a closed point. It is always flasque, giving the stated skyscraper sheaf cohomology.

The assertion follows using only Sylow theorems. Write the order instead as $p{q}^2$ to separate the prime of multiplicity one. If $p<q$, the number of Sylow $q$-subgroups divides $p$ and is one modulo $q$, so one such subgroup is normal. If $p>q$ and the Sylow $p$-subgroup is not normal, its number must be $q^2$, forcing $p\mid (q-1)(q+1)$. Since $p>q$, this forces $p=q+1$ and hence $p=3,q=2$. In that order-twelve exception, four order-three subgroups account for eight nonidentity elements. The four remaining elements contain every order-four Sylow subgroup, so that subgroup is unique and normal. In every case a normal Sylow subgroup and its quotient have prime or prime-square order, hence are abelian groups. They provide a series with abelian factors and prove solubility. Orders with the squared exponent lowered to one or zero follow by the same simpler counting.

The square-class group identifies two nonzero elements of a field when their quotient is a square. Multiplication makes it an abelian group of exponent two, hence a vector space over ${\mathbb{F}}_2$. It records quadratic extensions and the images of Kummer theory for $n=2$.

Every subgroup of an Abelian group generated by $n$ elements has at most $n$ generators. Lift it under a surjection ${\mathbb{Z}}^n\to A$. A subgroup $L\le {\mathbb{Z}}^n$ has at most $n$ generators by induction: its first-coordinate image is either zero or $d\mathbb{Z}$. In the latter case choose a lift of $d$, and subtract its multiples to reduce to the kernel in ${\mathbb{Z}}^{n-1}$. The images of these generators generate the original subgroup. The Abelian hypothesis matters: $S_{2n+2}$ is generated by a transposition and a full cycle but contains the elementary abelian group generated by $n+1$ disjoint transpositions, requiring $n+1$ generators by its vector-space dimension over ${\mathbb{F}}_2$.

For finite sets $A,B\subseteq G$ in an abelian group $G$, their sumset isEach distinct value occurs only once in this set, even if many ordered pairs produce it. When $B=A$, repetitions are allowed: $a+a$ belongs to $A+A$. For the two integer sets introduced above,For $A$, the value $3$ comes from $(0,3),(1,2),(2,1),(3,0)$. For $B$, two ordered pairs give the same sum only when one reverses the other, apart from a pair being identical to itself. Thus its four pairs $(b,b)$ and six pairs with distinct members, counted without regard to order, produce ten distinct sums.

Every nonempty finite set of integers $S\subseteq \mathbb{Z}$ satisfies $|S+S|\ge 2|S|-1$. To see this, list its members as $s_1<\cdots <s_n$. The $2n-1$ distinct sumsall belong to $S+S$. A nonconstant arithmetic progression attains this bound. The ordered list is essential: in the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the subgroup $H=\{0,2,4\}$ instead satisfies $H+H=H$, of cardinality three. Addition wraps around, for example $4+4=2$ in this group.

For finite sets $A,B\subseteq G$ in an abelian group $G$, the sumset representation function isIt counts ordered pairs in the Cartesian product, so $(a,b)$ and $(b,a)$ are different when $a\ne b$, even though commutativity makes their sums equal. It is zero outside the sumset, andFor $A=\{0,1,2,3\}$, its values at $s=0,1,\dots ,6$ are $1,2,3,4,3,2,1$. For $B=\{0,1,4,10\}$, its values at the increasing elements of $B+B$ are $1,2,1,2,2,1,2,2,2,1$. Both lists total sixteen, but the first concentrates more ordered pairs on fewer sums. This distinction is what additive energy quantifies.

A topological abelian group $A$ with a continuous action of a topological group $G$ by group automorphisms. Continuity means joint continuity of $G\times A\to A$. For a discrete coefficient group this is equivalent to every element of $A$ having an open stabilizer.

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

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.

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.

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