A group records symmetries, but not every symmetry system behaves like addition. If $\sigma$ and $\tau$ are permutations, applying $\sigma$ and then $\tau$ can give a different result from applying $\tau$ and then $\sigma$. The same obstruction appears in matrix multiplication, rotations of space, and symmetries of polygons. Order matters, so the algebra must remember conjugation, normality, and the difference between left and right actions. Abelian groups isolate the opposite situation: the operation is still invertible and associative, but order no longer matters. This does not empty the subject. It changes the questions. Instead of tracking conjugacy classes, we ask how elements decompose into independent pieces, how torsion is organised, how generators and relations describe the group, and why every abelian group behaves like a module over $\mathbb{Z}$. The point of commutativity is not only convenience. It is the reason addition in $\mathbb{Z}$, vector addition, homology groups, divisor class groups, and many algebraic invariants all land in the same category. Abelian groups are often what remains when a complicated object is measured by an additive invariant. The first contrast is visible before any formal definition. Addition of integers has no memory of order, while composition of permutations does. A small permutation computation shows that the ordinary group axioms cannot force commutativity by themselves, so the abelian condition must be added deliberately. [example: Order Matters in $S_3$] Let $S_3$ be the group of permutations of $\{1,2,3\}$ under composition, with the convention that $\sigma\tau$ means "apply $\tau$ first, then apply $\sigma$." Put $\sigma=(12)$ and $\tau=(23)$, so $\sigma$ swaps $1$ and $2$ and fixes $3$, while $\tau$ swaps $2$ and $3$ and fixes $1$. For the product $\sigma\tau$, we compute each value: \begin{align*} (\sigma\tau)(1)=\sigma(\tau(1))=\sigma(1)=2. \end{align*} \begin{align*} (\sigma\tau)(2)=\sigma(\tau(2))=\sigma(3)=3. \end{align*} \begin{align*} (\sigma\tau)(3)=\sigma(\tau(3))=\sigma(2)=1. \end{align*} Thus $\sigma\tau$ sends $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$, so $\sigma\tau=(123)$. For the reversed product $\tau\sigma$, the order of application changes: \begin{align*} (\tau\sigma)(1)=\tau(\sigma(1))=\tau(2)=3. \end{align*} \begin{align*} (\tau\sigma)(2)=\tau(\sigma(2))=\tau(1)=1. \end{align*} \begin{align*} (\tau\sigma)(3)=\tau(\sigma(3))=\tau(3)=2. \end{align*} Thus $\tau\sigma$ sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$, so $\tau\sigma=(132)$. Since $(123)\ne(132)$, we have $\sigma\tau\ne\tau\sigma$. Therefore the group axioms alone do not force commutativity. [/example] The failure in $S_3$ is the obstruction abelian groups remove. Once multiplication commutes, every subgroup is normal, every quotient by a subgroup is available, and conjugation ceases to carry information. The price is that we have restricted the world; the reward is a theory with strong classification theorems and a close relationship to linear algebra. ## Definition The parent notion is a [group](/page/Group): a set with an associative binary operation, an identity element, and inverses. The extra question is whether the operation depends on order. If it does not, the group behaves like addition; if it does, the noncommutative part of group theory enters. This page is the child case in which the parent group structure is already present and commutativity is the additional requirement. [definition: Abelian Group] An abelian group is a set $G$ equipped with a binary operation \begin{align*} \cdot:G\times G\to G \end{align*} such that $(G,\cdot)$ is a group and \begin{align*} gh = hg \end{align*} for all $g,h \in G$. [/definition] The word "abelian" is therefore a property of a group, not a replacement for the group axioms. Every abelian group has an identity element, inverses, and associativity before commutativity is even mentioned. The rest of the chapter studies what this added symmetry buys: additive notation, stable quotients, cyclic decomposition, torsion, and the largest abelian quotient of a general group. ## Notation and Commutators Because many abelian examples behave like addition, we need a notation that exposes sums, zero, negatives, and integer multiples rather than hiding them behind multiplication. This is not a new object; it is a change of language that matches the commutative situation. [definition: Additive Notation for an Abelian Group] Let $G$ be an abelian group. Additive notation writes the group operation as $+$, the identity element as $0$, and the inverse of $g \in G$ as $-g$. For a positive integer $n$ and $g \in G$, write \begin{align*} ng = \underbrace{g+\cdots+g}_{n\text{ times}}. \end{align*} Also write $0g=0$ and $(-n)g=-(ng)$. [/definition] In elementary group theory, additive notation is most natural for abelian groups because $g+h=h+g$ resembles ordinary addition. To compare abelian and nonabelian groups, however, we also need multiplicative notation for the failure of two elements to commute. The expression should vanish exactly when $gh$ and $hg$ agree, and this leads to the commutator. [definition: Commutator] Let $G$ be a group. The commutator map is the function \begin{align*} [-,-]:G\times G\to G \end{align*} defined by \begin{align*} [g,h] = ghg^{-1}h^{-1}. \end{align*} [/definition] The commutator is designed so that $[g,h]=e$ precisely when $gh=hg$. A definition of abelian group asks for commutativity directly, but in practice we often need a test that works inside a larger nonabelian group and can be converted into relations. The next theorem supplies that test: it turns the global phrase "all elements commute" into the vanishing of a family of explicit elements. [quotetheorem:9535] This criterion says that abelian groups are the groups whose internal conjugation noise has disappeared. It also gives a practical way to prove noncommutativity: find a single pair whose commutator is not the identity. The same idea will reappear when we build the largest abelian quotient of a group. ## Examples, Non-Examples, and Centrality ### Additive Models A definition becomes useful only after we know what it includes and what it excludes. The most basic abelian groups are familiar additive systems, and their behaviour explains why abelian groups feel arithmetical. We begin with examples where the operation is already called addition. The integers are the model example. Their operation is addition, and every other cyclic abelian group is built by imposing a periodicity relation on this example. Studying $\mathbb{Z}$ first tells us what one unrestricted generator can do. [example: The Additive Group of Integers] For integers $a,b,c\in\mathbb{Z}$, the sum $a+b$ is again an integer, so $+$ is a binary operation on $\mathbb{Z}$. The group axioms are the standard arithmetic laws: \begin{align*} (a+b)+c=a+(b+c) \end{align*} for associativity, \begin{align*} a+0=a=0+a \end{align*} for the identity element $0$, and \begin{align*} a+(-a)=0=(-a)+a \end{align*} for the inverse $-a$ of $a$. Commutativity is the integer addition law \begin{align*} a+b=b+a. \end{align*} Hence $(\mathbb{Z},+)$ is an abelian group. The group is infinite because the elements $0,1,2,3,\dots$ are pairwise distinct integers. It is generated by $1$: if $m>0$, then \begin{align*} m\cdot 1=\underbrace{1+\cdots+1}_{m\text{ times}}=m, \end{align*} while \begin{align*} 0\cdot 1=0 \end{align*} and, for $m>0$, \begin{align*} (-m)\cdot 1=-(m\cdot 1)=-m. \end{align*} Thus every integer is an integer multiple of $1$, so $\mathbb{Z}=\langle 1\rangle$. [/example] The integer example matters because any group element can be repeatedly added to itself. This process always produces a cyclic subgroup, so $\mathbb{Z}$ is the universal source of one-generator behaviour. Vector spaces give a larger family in which addition remains commutative but extra scalar structure is present. [example: Additive Groups of Vector Spaces] Let $k$ be a field and let $V$ be a [vector space](/page/Vector%20Space) over $k$. Vector addition is a binary operation on $V$, since for any $u,v\in V$ the vector space axioms give $u+v\in V$. The same axioms give associativity: \begin{align*} (u+v)+w=u+(v+w) \end{align*} for all $u,v,w\in V$. The zero vector $0_V$ is the identity because \begin{align*} v+0_V=v=0_V+v. \end{align*} For each $v\in V$, the vector $-v$ is its additive inverse because \begin{align*} v+(-v)=0_V=(-v)+v. \end{align*} Finally, vector addition is commutative: \begin{align*} u+v=v+u. \end{align*} Thus $(V,+)$ is an abelian group. When $V=k^n$, each vector has the form $(a_1,\dots,a_n)$ with $a_i\in k$, and addition is componentwise: \begin{align*} (a_1,\dots,a_n)+(b_1,\dots,b_n)=(a_1+b_1,\dots,a_n+b_n). \end{align*} This is exactly the direct product operation on $k\times\cdots\times k$, where each copy of $k$ is regarded as its additive abelian group. Hence the additive group of $k^n$ is the direct product of $n$ copies of the additive group of $k$. [/example] This example warns that an abelian group may carry extra structure not visible to group theory. A vector space is more than its underlying abelian group, because scalar multiplication by elements of $k$ is additional data. The abelian group remembers addition, but it forgets which field acted on the vectors. ### Noncommutative Boundaries The smallest nonabelian examples show why the abelian condition has content. Symmetry groups of geometric objects often remember order of motion. A rotation followed by a reflection may not be the same as the reflection followed by the rotation, so geometry provides concrete tests for commutativity. [example: The Symmetry Group of a Triangle Is Not Abelian] Label the triangle vertices $1,2,3$ clockwise. Let $r$ be the rotation $1\mapsto 2\mapsto 3\mapsto 1$, and let $s$ be the reflection fixing vertex $1$ and swapping vertices $2$ and $3$. Then \begin{align*} (sr)(1)=s(r(1))=s(2)=3. \end{align*} On the other hand, \begin{align*} (rs)(1)=r(s(1))=r(1)=2. \end{align*} Thus $sr$ and $rs$ send the same vertex to different vertices, so $sr\ne rs$. We can also see the usual relation explicitly. Since $r^{-1}$ sends $1\mapsto 3$, $2\mapsto 1$, and $3\mapsto 2$, we have \begin{align*} (r^{-1}s)(1)=r^{-1}(s(1))=r^{-1}(1)=3. \end{align*} \begin{align*} (r^{-1}s)(2)=r^{-1}(s(2))=r^{-1}(3)=2. \end{align*} \begin{align*} (r^{-1}s)(3)=r^{-1}(s(3))=r^{-1}(2)=1. \end{align*} The product $sr$ has the same values: \begin{align*} (sr)(1)=3. \end{align*} \begin{align*} (sr)(2)=s(r(2))=s(3)=2. \end{align*} \begin{align*} (sr)(3)=s(r(3))=s(1)=1. \end{align*} Therefore $sr=r^{-1}s$, and in particular $sr\ne rs$. Hence the symmetry group $D_6$ of an equilateral triangle is not abelian, even though it has only six elements. [/example] The preceding example raises a useful replacement question. If a whole group is not abelian, which elements still commute with every possible element of the group? The answer is the centre, and it measures the amount of globally compatible abelian behaviour left inside a nonabelian group. [definition: Centre of a Group] Let $G$ be a group. The centre of $G$ is \begin{align*} Z(G)=\{g\in G:gx=xg\text{ for all }x\in G\}. \end{align*} [/definition] The centre is an abelian subgroup of $G$, but it is defined by commuting with every element of the ambient group, not merely with other elements of the centre. This makes it a diagnostic tool. The next result says that a group is abelian exactly when this diagnostic finds every element. [quotetheorem:9536] This theorem translates the abelian condition into a statement about central elements. It is often the fastest way to recognise abelian groups inside larger nonabelian settings. It also prepares the idea that commutativity can be studied through what conjugation fails to change. ## Subgroups, Quotients, and Homomorphisms ### Subobjects Once commutativity is available, many constructions become less fragile. In a general group, a subgroup may inherit the group law but quotienting by it may fail. In an abelian group, subgroups inherit commutativity and automatically behave well with respect to quotient groups. The first construction is the subgroup. We want to pass to a smaller collection of elements without leaving the category of abelian groups. Since the operation is inherited from the ambient group, commutativity should remain available inside the smaller group. [definition: Subgroup of an Abelian Group] Let $G$ be an abelian group. A subgroup of $G$ is a subset $H \subset G$ such that $H$ is a group under the operation inherited from $G$. [/definition] The notation remains the same as in group theory: write $H \le G$ for a subgroup, and say that $H$ is proper when $H \ne G$. Because $G$ is abelian, the inherited operation on $H$ is commutative. This closure property is the first sign that abelian groups form a stable algebraic setting. [quotetheorem:9537] This result is structurally small but conceptually important: the class of abelian groups is closed under subobjects. The next construction is more delicate in ordinary group theory. We want to identify elements modulo a subgroup, and abelian groups make this possible for every subgroup. ### Quotients A [quotient group](/theorems/790) identifies elements that differ by a chosen subgroup. In a nonabelian group, this only works when the subgroup is normal. In an abelian group, the normality condition is automatic because conjugation does not change subgroup elements. [definition: Quotient of an Abelian Group] Let $G$ be an abelian group and let $H \le G$. The quotient group $G/H$ is the set of cosets \begin{align*} G/H=\{g+H:g\in G\}, \end{align*} equipped with the binary operation $+:(G/H)\times(G/H)\to G/H$ defined by \begin{align*} (g+H)+(g'+H)=(g+g')+H. \end{align*} [/definition] The formula for addition on cosets is only useful if it is independent of the representatives chosen. After that well-definedness issue is settled, there is a second structural question: does passing from $G$ to $G/H$ preserve commutativity, so that quotienting by relations still leaves us inside the world of abelian groups? [quotetheorem:9538] Quotients formalise the act of declaring some elements to be zero. To understand which elements are declared zero by a construction, we need maps that preserve addition. This is where homomorphisms, kernels, and images enter the story. ### Homomorphisms A useful map between abelian groups must respect the additive structure, not merely the underlying sets. Such maps preserve sums, zero, inverses, and integer multiples. They are the morphisms in the category of abelian groups. [definition: Homomorphism of Abelian Groups] Let $G$ and $H$ be abelian groups written additively. A homomorphism of abelian groups is a function $\varphi:G\to H$ such that \begin{align*} \varphi(g+g')=\varphi(g)+\varphi(g') \end{align*} for all $g,g'\in G$. [/definition] A homomorphism may identify different elements of the domain. To understand such a map, the first question is not only where elements go, but which elements become indistinguishable from zero. The subgroup of elements sent to zero records exactly this loss of information; it is the object by which the domain must be quotiented to remove the identifications. [definition: Kernel of a Homomorphism] Let $\varphi:G\to H$ be a homomorphism of abelian groups. The kernel of $\varphi$ is \begin{align*} \ker\varphi=\{g\in G:\varphi(g)=0\}. \end{align*} [/definition] The kernel describes what the map kills. The dual question is what the map reaches. Since a homomorphism may not hit all of its declared codomain, the image records the subgroup that actually appears. [definition: Image of a Homomorphism] Let $\varphi:G\to H$ be a homomorphism of abelian groups. The image of $\varphi$ is \begin{align*} \operatorname{im}\varphi=\{h\in H:h=\varphi(g)\text{ for some }g\in G\}. \end{align*} [/definition] Kernel and image are two halves of the same map. The kernel says which relations are imposed, while the image says what remains after imposing them. The [first isomorphism theorem](/theorems/791) makes this relationship precise by turning a homomorphism into a quotient. [quotetheorem:791] The theorem explains why quotients and homomorphisms are two views of the same process. A map out of $G$ is controlled by the subgroup of $G$ it kills and the subgroup of the target it reaches. [example: Reduction Modulo $n$] Fix a positive integer $n$, and define $\pi:\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ by $\pi(a)=\bar{a}$, the residue class of $a$ modulo $n$. For $a,b\in\mathbb{Z}$, addition in $\mathbb{Z}/n\mathbb{Z}$ is defined by adding representatives, so \begin{align*} \pi(a+b)=\overline{a+b}=\bar{a}+\bar{b}=\pi(a)+\pi(b). \end{align*} Thus $\pi$ is a homomorphism of abelian groups. Its kernel consists exactly of the integers whose residue class is $\bar{0}$: \begin{align*} \ker\pi=\{a\in\mathbb{Z}:\pi(a)=\bar{0}\}. \end{align*} For an integer $a$, the equality $\pi(a)=\bar{0}$ means $\bar{a}=\bar{0}$ in $\mathbb{Z}/n\mathbb{Z}$, which means $n$ divides $a$. Equivalently, there is some $q\in\mathbb{Z}$ such that \begin{align*} a=nq. \end{align*} Therefore \begin{align*} \ker\pi=\{nq:q\in\mathbb{Z}\}=n\mathbb{Z}. \end{align*} The image is all of $\mathbb{Z}/n\mathbb{Z}$, because every element of $\mathbb{Z}/n\mathbb{Z}$ has the form $\bar{r}$ for some $r\in\mathbb{Z}$, and then \begin{align*} \pi(r)=\bar{r}. \end{align*} Hence \begin{align*} \operatorname{im}\pi=\mathbb{Z}/n\mathbb{Z}. \end{align*} By the *First Isomorphism Theorem for Abelian Groups*, \begin{align*} \mathbb{Z}/\ker\pi\cong \operatorname{im}\pi. \end{align*} Substituting the computed kernel and image gives \begin{align*} \mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/n\mathbb{Z}. \end{align*} The point of the isomorphism is that reduction modulo $n$ is exactly the quotient map that declares every multiple of $n$ to be zero. [/example] This example is the prototype for presentations by generators and relations: start with a free object, then quotient by the relations you want to impose. ## Cyclic Groups and Torsion ### One Generator The simplest way to build an abelian group is to start with one element and repeatedly add it to itself. The result may continue forever, as in $\mathbb{Z}$, or it may cycle back to zero, as in $\mathbb{Z}/n\mathbb{Z}$. Cyclic groups are the abelian groups controlled by this one-element process. A single element always generates a subgroup. In an abelian group, this generated subgroup is the image of a homomorphism from $\mathbb{Z}$. Naming it lets us talk about the portion of the group controlled by one chosen element. [definition: Subgroup Generated by an Element] Let $G$ be an abelian group and let $g\in G$. The subgroup generated by $g$ is \begin{align*} \langle g\rangle=\{ng:n\in\mathbb{Z}\}. \end{align*} [/definition] The notation $\langle g\rangle$ is standard in group theory. In additive notation it means all integer multiples of $g$. This raises the decisive one-generator question: can every element of the group be reached by repeatedly adding and subtracting a single chosen element? When the answer is yes, the group is controlled by one integer parameter. [definition: Cyclic Abelian Group] An abelian group $G$ is cyclic if there exists $g\in G$ such that \begin{align*} G=\langle g\rangle. \end{align*} The element $g$ is called a generator of $G$. [/definition] Cyclicity is a strong condition. It says the entire group is controlled by the arithmetic of one element. To classify cyclic groups, we need to know whether repeated addition of the generator ever returns to zero, and if so after how many steps. [definition: Order of an Element in an Abelian Group] Let $G$ be an abelian group and let $g\in G$. The order of $g$, denoted $\operatorname{ord}(g)$, is the least positive integer $n$ such that \begin{align*} ng=0, \end{align*} if such an $n$ exists. If no such $n$ exists, then $\operatorname{ord}(g)=\infty$. [/definition] Once the order of a generator is known, the natural map from $\mathbb{Z}$ to the [cyclic group](/page/Cyclic%20Group) has only two possible kernels: either no nonzero integer kills the generator, or the least positive one does. The next theorem turns that observation into a classification, showing that a cyclic abelian group has no hidden structure beyond this order. [quotetheorem:9505] This theorem says there are no mysterious cyclic abelian groups. Each is either the infinite additive group of integers or a finite modular version. The remaining question inside a cyclic group is which elements are generators. [example: Generators of $\mathbb{Z}/12\mathbb{Z}$] In the additive group $\mathbb{Z}/12\mathbb{Z}$, the subgroup generated by $\bar a$ is \begin{align*} \langle \bar a\rangle=\{n\bar a:n\in\mathbb{Z}\}=\{\overline{na}:n\in\mathbb{Z}\}. \end{align*} We show that $\bar a$ generates all of $\mathbb{Z}/12\mathbb{Z}$ exactly when $\gcd(a,12)=1$. If $\gcd(a,12)=1$, Bezout's identity gives integers $r,s$ such that \begin{align*} ra+12s=1. \end{align*} Reducing modulo $12$ gives \begin{align*} \overline{ra+12s}=\bar 1. \end{align*} Since $\overline{12s}=\bar 0$, this becomes \begin{align*} r\bar a=\bar 1. \end{align*} Therefore $\bar 1\in\langle\bar a\rangle$. For any residue class $\bar m$, we then have \begin{align*} \bar m=m\bar 1=m(r\bar a)=(mr)\bar a, \end{align*} so every element lies in $\langle\bar a\rangle$. Conversely, suppose $\bar a$ generates $\mathbb{Z}/12\mathbb{Z}$. Then $\bar 1\in\langle\bar a\rangle$, so there is an integer $n$ such that \begin{align*} n\bar a=\bar 1. \end{align*} Equivalently, \begin{align*} \overline{na}=\bar 1. \end{align*} Thus $na-1$ is divisible by $12$, so for some $q\in\mathbb{Z}$, \begin{align*} na-1=12q. \end{align*} Rearranging gives \begin{align*} na-12q=1. \end{align*} Any common divisor of $a$ and $12$ divides the left-hand side, hence divides $1$, so $\gcd(a,12)=1$. The residue classes between $\bar 0$ and $\bar{11}$ with representatives relatively prime to $12$ are \begin{align*} \bar 1,\bar 5,\bar 7,\bar{11}. \end{align*} These are therefore exactly the generators of $\mathbb{Z}/12\mathbb{Z}$ among those representatives. For $\bar 2$, the successive multiples are \begin{align*} 0\bar 2=\bar 0. \end{align*} \begin{align*} 1\bar 2=\bar 2. \end{align*} \begin{align*} 2\bar 2=\bar 4. \end{align*} \begin{align*} 3\bar 2=\bar 6. \end{align*} \begin{align*} 4\bar 2=\bar 8. \end{align*} \begin{align*} 5\bar 2=\bar{10}. \end{align*} \begin{align*} 6\bar 2=\bar{12}=\bar 0. \end{align*} After this the same six classes repeat, because adding another $\bar 2$ continues the cycle. Hence \begin{align*} \langle\bar 2\rangle=\{\bar 0,\bar 2,\bar 4,\bar 6,\bar 8,\bar{10}\}. \end{align*} Thus $\mathbb{Z}/12\mathbb{Z}$ has several generators, and a non-generator such as $\bar 2$ still generates a proper subgroup with its own cyclic structure. [/example] ### Torsion The cyclic case introduces periodic elements, but periodicity also appears in noncyclic groups. We need a term for elements that are killed by some nonzero integer. This term is torsion, and it is one of the central organising ideas in abelian group theory. [definition: Torsion Element] Let $G$ be an abelian group. An element $g\in G$ is a torsion element if there exists a positive integer $n$ such that \begin{align*} ng=0. \end{align*} [/definition] Torsion is a local periodicity condition on an element. But classification needs a global object, not just scattered periodic elements. To separate all periodic behaviour from infinite-order behaviour, we gather the torsion elements together and ask whether this collection can be treated as a genuine part of the group. [definition: Torsion Subgroup] Let $G$ be an abelian group. The torsion subgroup of $G$ is \begin{align*} G_{\mathrm{tors}}=\{g\in G:g\text{ is a torsion element}\}. \end{align*} [/definition] The definition names a set, but the theory needs more: if torsion is to be a structural piece of $G$, it must be closed under the group operation. This is where commutativity matters. In a nonabelian group, the product of two finite-order elements need not have finite order, so torsion elements need not form a subgroup. In an abelian group, sums of torsion elements can be controlled by multiplying by a common multiple of their orders, and the next theorem records that the periodic part is really a subgroup. [quotetheorem:9539] The torsion subgroup separates periodic behaviour from infinite-order behaviour. In finite abelian groups, the torsion subgroup is the whole group. In finitely generated abelian groups, it becomes the finite part in the classification theorem. ## Finite Abelian Groups Finite abelian groups are structured enough to classify. The guiding question is: if a finite abelian group is not generated by one element, can it still be decomposed into cyclic components? The answer is yes, and the decomposition is controlled by products and prime powers. To state such decompositions, we need direct products. Externally, a direct product combines separate groups coordinate by coordinate. This construction represents independent additive choices made in several abelian groups at once. [definition: Direct Product of Abelian Groups] Let $G_1,\dots,G_n$ be abelian groups. Their direct product is the set $G_1\times\cdots\times G_n$ equipped with the binary operation $+:(G_1\times\cdots\times G_n)\times(G_1\times\cdots\times G_n)\to G_1\times\cdots\times G_n$ defined by \begin{align*} ((g_1,\dots,g_n),(h_1,\dots,h_n)) \mapsto (g_1+h_1,\dots,g_n+h_n). \end{align*} [/definition] A direct product records independent coordinates. In a finite product of abelian groups, the product and [direct sum](/page/Direct%20Sum) have the same underlying set. For infinite families the distinction becomes important, but finite classification only needs finite products. [example: The Klein Four Group] Let \begin{align*} V=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}. \end{align*} Its elements are the four ordered pairs \begin{align*} (\bar 0,\bar 0),(\bar 1,\bar 0),(\bar 0,\bar 1),(\bar 1,\bar 1). \end{align*} Addition is componentwise, so for $(\bar a,\bar b),(\bar c,\bar d)\in V$, \begin{align*} (\bar a,\bar b)+(\bar c,\bar d)=(\bar a+\bar c,\bar b+\bar d). \end{align*} Since addition in each copy of $\mathbb{Z}/2\mathbb{Z}$ is commutative, we have \begin{align*} (\bar a+\bar c,\bar b+\bar d)=(\bar c+\bar a,\bar d+\bar b). \end{align*} Therefore \begin{align*} (\bar a,\bar b)+(\bar c,\bar d)=(\bar c,\bar d)+(\bar a,\bar b), \end{align*} so $V$ is abelian. Now compute the order of each nonzero element. In $\mathbb{Z}/2\mathbb{Z}$, $\bar 1+\bar 1=\bar 2=\bar 0$, so \begin{align*} 2(\bar 1,\bar 0)=(\bar 1,\bar 0)+(\bar 1,\bar 0)=(\bar 0,\bar 0). \end{align*} Similarly, \begin{align*} 2(\bar 0,\bar 1)=(\bar 0,\bar 1)+(\bar 0,\bar 1)=(\bar 0,\bar 0). \end{align*} Also, \begin{align*} 2(\bar 1,\bar 1)=(\bar 1,\bar 1)+(\bar 1,\bar 1)=(\bar 0,\bar 0). \end{align*} None of these three elements is $(\bar 0,\bar 0)$, so each has order exactly $2$. Thus $V$ has no element of order $4$. If $V$ were cyclic, then some element $g\in V$ would generate all four elements of $V$, and the four multiples \begin{align*} 0g,\ 1g,\ 2g,\ 3g \end{align*} would have to be distinct before returning to $0$ at $4g=0$; equivalently, $g$ would have order $4$. Since every nonzero element of $V$ has order $2$, no such generator exists. Hence $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ is an abelian group of order $4$ that is not cyclic, unlike $\mathbb{Z}/4\mathbb{Z}$. [/example] The Klein four group is the first sign that finite abelian groups need decomposition data. Cardinality alone does not determine the group, so the next question is how to separate the different sources of periodicity inside a finite abelian group. If we try to decompose by arbitrary divisors of $|G|$, behaviour at different primes gets mixed together: an element of order $6$ carries both $2$-torsion and $3$-torsion at once. Prime powers are the scale at which this mixing disappears. For a fixed prime $p$, we therefore want a subgroup that collects precisely the elements whose eventual return to zero is controlled by powers of $p$. This extraction is the group-theoretic analogue of taking the $p$-power part of an integer. It gives a named piece that can later be compared with the corresponding pieces for the other primes, which is why it is the natural next definition before any classification theorem can be stated. [definition: $p$-Primary Component] Let $G$ be an abelian group and let $p$ be a prime. The $p$-primary component of $G$ is \begin{align*} G[p^\infty]=\{g\in G:p^kg=0\text{ for some positive integer }k\}. \end{align*} [/definition] The notation $G[p^\infty]$ records torsion of $p$-power order. After defining these pieces, the necessary question is whether they merely sit inside $G$ or actually reconstruct it. For finite abelian groups they do reconstruct it: prime factorisation of $|G|$ becomes an internal product decomposition of the group. This primary splitting is the reduction step: it separates the different primes before one records the final cyclic factors. The remaining classification problem is to choose cyclic factors in a canonical way. Instead of decomposing the same torsion repeatedly, the invariant-factor form packages the finite abelian group as a product of cyclic groups whose orders divide one another. That divisibility condition is what makes the list unique. [quotetheorem:850] This theorem is one of the main reasons finite abelian groups are tractable. It converts an abstract finite abelian group into a finite list of cyclic orders. Concrete classification problems then become arithmetic exercises in comparing the primary pieces and assembling the corresponding invariant factors. [example: Abelian Groups of Order $12$] Since $12=2^2\cdot 3$, the *Primary Decomposition of Finite Abelian Groups* says that any finite abelian group $G$ of order $12$ splits into its $2$-primary part and its $3$-primary part: \begin{align*} G\cong G[2^\infty]\times G[3^\infty]. \end{align*} The first factor has order $4$, and the second has order $3$. By the *Fundamental Theorem of Finite Abelian Groups*, the abelian groups of order $4$ are \begin{align*} \mathbb{Z}/4\mathbb{Z} \end{align*} and \begin{align*} \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}. \end{align*} The only abelian group of order $3$ is \begin{align*} \mathbb{Z}/3\mathbb{Z}. \end{align*} Thus the two possibilities are \begin{align*} \mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z} \end{align*} and \begin{align*} \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}. \end{align*} The first product is cyclic. Let \begin{align*} g=(\bar 1,\bar 1)\in \mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}. \end{align*} For an integer $k$, \begin{align*} kg=(\bar k,\bar k). \end{align*} This equals $(\bar 0,\bar 0)$ exactly when $\bar k=\bar 0$ in $\mathbb{Z}/4\mathbb{Z}$ and $\bar k=\bar 0$ in $\mathbb{Z}/3\mathbb{Z}$, which means $4\mid k$ and $3\mid k$. Since $4$ and $3$ are relatively prime, this is equivalent to $12\mid k$. Hence the least positive $k$ with $kg=0$ is $12$, so $g$ has order $12$. A group of order $12$ containing an element of order $12$ is generated by that element, and therefore \begin{align*} \mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\cong \mathbb{Z}/12\mathbb{Z}. \end{align*} The second product has no element of order $12$. Take \begin{align*} h=(\bar a,\bar b,\bar c)\in \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}. \end{align*} Then \begin{align*} 6h=(6\bar a,6\bar b,6\bar c). \end{align*} In $\mathbb{Z}/2\mathbb{Z}$, $6\bar a=\bar 0$ and $6\bar b=\bar 0$ because $2\mid 6$. In $\mathbb{Z}/3\mathbb{Z}$, $6\bar c=\bar 0$ because $3\mid 6$. Therefore \begin{align*} 6h=(\bar 0,\bar 0,\bar 0) \end{align*} for every element $h$ of the second product. Since every element is killed by $6$, no element can have order $12$. Thus the two abelian groups of order $12$ are distinguished by cyclicity: one is cyclic and the other is not. [/example] The example illustrates how classification turns group theory into arithmetic. The prime factorisation of the order gives the primary components, and the partitions of exponents describe the possible cyclic factors. ## Abelian Groups as $\mathbb{Z}$-Modules ### Integer Scalars The deeper reason abelian groups behave linearly is that every abelian group admits scalar multiplication by integers. Repeated addition by an element gives multiplication by positive integers, inverses give multiplication by negative integers, and the identity gives multiplication by zero. This turns abelian groups into modules over the ring $\mathbb{Z}$. The module viewpoint is not extra structure; it is another language for the same structure. It lets us use generators, relations, free objects, rank, and normal forms. It also explains why abelian groups sit naturally beside linear algebra. [definition: $\mathbb{Z}$-Module Structure on an Abelian Group] Let $G$ be an abelian group written additively. The associated $\mathbb{Z}$-module structure is the scalar multiplication \begin{align*} \mathbb{Z}\times G\to G \end{align*} given by \begin{align*} (n,g)\mapsto ng. \end{align*} [/definition] The module axioms are exactly the familiar laws of repeated addition. This is why methods from modules belong naturally to abelian group theory. To use generators and relations, we first need the analogue of a vector space with a basis: an abelian group in which every element has a unique expression as an integer linear combination of chosen generators. [definition: Free Abelian Group] An abelian group $F$ is free abelian on a set $S$ if every element of $F$ can be written uniquely as a finite sum \begin{align*} \sum_{s\in S} n_s s \end{align*} where $n_s\in\mathbb{Z}$ and all but finitely many $n_s$ are zero. [/definition] The uniqueness of integer coefficients means there are no hidden relations among the basis elements. Once a free abelian group has a basis, we need an invariant that records how many independent integer directions it has. For finite bases this is a [natural number](/page/Natural%20Number); for infinite bases it is a cardinal. This number plays the role of dimension, but only for the free part rather than any torsion part. [definition: Rank of a Free Abelian Group] Let $F$ be a free abelian group. The rank of $F$ is the cardinality of a basis of $F$: \begin{align*} \operatorname{rank}F=|S|, \end{align*} where $S$ is any basis of $F$. [/definition] Rank plays the role of dimension for free abelian groups, but quotients over $\mathbb{Z}$ can create torsion. This is a major difference from vector spaces over a field. The next example shows that even a quotient of a free abelian group can become finite. [example: A Quotient of a Free Abelian Group with Torsion] The group $\mathbb{Z}$ is free abelian on the generator $1$, since every integer $m$ has the unique form \begin{align*} m=m\cdot 1. \end{align*} It is torsion-free: if $a\in\mathbb{Z}$ and $n>0$ satisfy $na=0$, then the ordinary integer product $na$ is zero, so $a=0$. Now take the subgroup \begin{align*} 2\mathbb{Z}=\{2q:q\in\mathbb{Z}\}. \end{align*} In the quotient $\mathbb{Z}/2\mathbb{Z}$, the class $\bar 1=1+2\mathbb{Z}$ is not the zero class, because $1\notin 2\mathbb{Z}$. But twice this class is zero: \begin{align*} 2\bar 1=\bar 1+\bar 1=\overline{1+1}=\bar 2=\bar 0, \end{align*} where $\bar 2=\bar 0$ because $2\in 2\mathbb{Z}$. Thus $\bar 1$ is a nonzero torsion element in $\mathbb{Z}/2\mathbb{Z}$. Quotienting the torsion-free free abelian group $\mathbb{Z}$ by $2\mathbb{Z}$ has created torsion, which is a basic difference between abelian groups and vector spaces. [/example] ### Finite Generation Many abelian groups in algebra and topology are described by finitely many generators and finitely many relations. Before classifying them, we need a definition that says a finite list of elements generates the whole group. This is the abelian-group analogue of finite spanning sets in linear algebra. [definition: Finitely Generated Abelian Group] An abelian group $G$ is finitely generated if there exist $g_1,\dots,g_n\in G$ such that every $g\in G$ can be written as \begin{align*} g=a_1g_1+\cdots+a_ng_n \end{align*} for some $a_1,\dots,a_n\in\mathbb{Z}$. [/definition] Finitely generated abelian groups are described by a finite list of generators, but generators alone do not show which combinations are forced to be zero. The classification theorem says that the only remaining data are a free rank and a finite torsion decomposition. In elementary abelian-group notation, this means that every finitely generated abelian group is isomorphic to a group of the form \begin{align*} \mathbb{Z}^r\oplus \mathbb{Z}/n_1\mathbb{Z}\oplus\cdots\oplus \mathbb{Z}/n_k\mathbb{Z}, \end{align*} with $n_i>1$ and with the integers chosen in a standard divisibility pattern. Here $\mathbb{Z}^r$ means $r$ independent infinite cyclic directions, $\oplus$ is the direct sum of abelian groups, and $\mathbb{Z}/n\mathbb{Z}$ is the cyclic group of residues modulo $n$. This classification is the abelian-group version of diagonalising a matrix over $\mathbb{Z}$. It explains why finitely generated abelian groups are computable from generators and relations. The computation is governed by integer row and column operations, often organised as Smith normal form. [example: Computing from a Relation] Let $G=\langle x,y:2x+4y=0\rangle$. Start with the free abelian group on $x,y$, and impose the single relation $2x+4y=0$. Define new generators \begin{align*} u=x+2y \end{align*} and \begin{align*} v=y. \end{align*} These determine the old generators because \begin{align*} y=v \end{align*} and \begin{align*} x=u-2v. \end{align*} Thus the change of generators is invertible over $\mathbb{Z}$. In the new generators, the relation is \begin{align*} 2x+4y=2(u-2v)+4v=2u-4v+4v=2u. \end{align*} So the presentation becomes \begin{align*} G=\langle u,v:2u=0\rangle. \end{align*} Every element of the free abelian group on $u,v$ has the form $au+bv$ with $a,b\in\mathbb{Z}$. Write $a=2q+\epsilon$, where $q\in\mathbb{Z}$ and $\epsilon\in\{0,1\}$. Then \begin{align*} au+bv=(2q+\epsilon)u+bv=q(2u)+\epsilon u+bv. \end{align*} Since $2u=0$ in $G$, this element represents the same element of $G$ as \begin{align*} \epsilon u+bv. \end{align*} This representative is unique. If \begin{align*} \epsilon u+bv=0 \end{align*} in $G$, then in the free abelian group on $u,v$ there is some $k\in\mathbb{Z}$ such that \begin{align*} \epsilon u+bv=k(2u). \end{align*} Comparing the $v$-coefficients gives $b=0$, and comparing the $u$-coefficients gives $\epsilon=2k$. Since $\epsilon\in\{0,1\}$, this forces $\epsilon=0$. Therefore every element of $G$ is uniquely expressible as \begin{align*} \epsilon u+bv \end{align*} with $\epsilon\in\{0,1\}$ and $b\in\mathbb{Z}$. The map \begin{align*} \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}\to G \end{align*} given by \begin{align*} (\bar\epsilon,b)\mapsto \epsilon u+bv \end{align*} is a well-defined homomorphism, because changing $\epsilon$ by $2$ changes the image by $2u=0$. The uniqueness just proved shows that it is bijective, so \begin{align*} G\cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}. \end{align*} Here $u$ generates the torsion factor and $v$ generates the free factor; the computation is the one-relation prototype for Smith normal form. [/example] The module viewpoint also explains why abelian groups appear everywhere in algebraic topology and algebraic geometry. Chain groups, homology groups, divisor class groups, and many cohomology groups are abelian because their operations are additive and relations are imposed by quotienting. ## Abelianisation A nonabelian group can be forced to become abelian by killing all commutators. This construction is important because many invariants naturally ignore noncommutative information. Abelianisation is the best abelian approximation to a group. The commutator subgroup is the subgroup generated by all commutators. In a general group it measures the obstruction to being abelian. Naming it allows us to quotient by exactly the relations needed to make every pair of elements commute. [definition: Commutator Subgroup] Let $G$ be a group. The commutator subgroup of $G$ is \begin{align*} [G,G]=\langle [g,h]:g,h\in G\rangle. \end{align*} [/definition] Before quotienting by $[G,G]$, we must know that this subgroup is normal. This is not automatic for arbitrary generated subgroups, but commutators are stable under conjugation in exactly the way needed. The next result supplies the local group-theoretic fact that makes the quotient construction legitimate. [quotetheorem:9540] With normality in place, the quotient by the commutator subgroup is available. The reason to single out this quotient is that every attempt to map $G$ into an abelian group must make each commutator invisible: in an abelian target, $\varphi(ghg^{-1}h^{-1})=e$. Therefore the quotient by $[G,G]$ is the smallest forced collapse that removes exactly the obstruction to commutativity. That universal quotient is called the abelianisation. [definition: Abelianisation] Let $G$ be a group. The abelianisation of $G$ is the quotient group \begin{align*} G^{\mathrm{ab}}=G/[G,G]. \end{align*} [/definition] Abelianisation turns a general group into an abelian group by imposing exactly the relations needed to make all elements commute. It does not preserve all group-theoretic information; it preserves the part visible to abelian targets. The following universal property expresses that statement without reference to a chosen presentation. [quotetheorem:9541] The universal property says that any map from $G$ into an abelian group must factor through $G^{\mathrm{ab}}$. In this sense, abelianisation is the canonical answer to the question of what abelian group a general group determines. [example: Abelianisation of $S_3$] Let $A_3=\{e,(123),(132)\}$ be the alternating subgroup of $S_3$. The [sign homomorphism](/theorems/778) $\operatorname{sgn}:S_3\to\{\pm1\}$ has kernel $A_3$. For any $\alpha,\beta\in S_3$, \begin{align*} \operatorname{sgn}([\alpha,\beta])=\operatorname{sgn}(\alpha\beta\alpha^{-1}\beta^{-1})=\operatorname{sgn}(\alpha)\operatorname{sgn}(\beta)\operatorname{sgn}(\alpha^{-1})\operatorname{sgn}(\beta^{-1}). \end{align*} Since $\operatorname{sgn}(\alpha^{-1})=\operatorname{sgn}(\alpha)$ and $\operatorname{sgn}(\beta^{-1})=\operatorname{sgn}(\beta)$, this becomes \begin{align*} \operatorname{sgn}([\alpha,\beta])=\operatorname{sgn}(\alpha)^2\operatorname{sgn}(\beta)^2=1. \end{align*} Thus every commutator lies in $A_3$, and since $A_3$ is a subgroup, the subgroup generated by all commutators satisfies \begin{align*} [S_3,S_3]\le A_3. \end{align*} Now take $\sigma=(12)$ and $\tau=(23)$. Since $\sigma^{-1}=\sigma$ and $\tau^{-1}=\tau$, their commutator is \begin{align*} [\sigma,\tau]=\sigma\tau\sigma\tau. \end{align*} With the convention that products act from right to left, \begin{align*} [\sigma,\tau](1)=\sigma(\tau(\sigma(\tau(1))))=\sigma(\tau(\sigma(1)))=\sigma(\tau(2))=\sigma(3)=3. \end{align*} Also, \begin{align*} [\sigma,\tau](2)=\sigma(\tau(\sigma(\tau(2))))=\sigma(\tau(\sigma(3)))=\sigma(\tau(3))=\sigma(2)=1. \end{align*} And \begin{align*} [\sigma,\tau](3)=\sigma(\tau(\sigma(\tau(3))))=\sigma(\tau(\sigma(2)))=\sigma(\tau(1))=\sigma(1)=2. \end{align*} Therefore $[\sigma,\tau]$ sends $1\mapsto3$, $3\mapsto2$, and $2\mapsto1$, so \begin{align*} [\sigma,\tau]=(132). \end{align*} Hence $(132)\in[S_3,S_3]$. Since $[S_3,S_3]$ is a subgroup, it also contains \begin{align*} (132)^2=(123) \end{align*} and \begin{align*} (132)^3=e. \end{align*} Thus $A_3=\{e,(123),(132)\}\le[S_3,S_3]$. Combining both inclusions gives \begin{align*} [S_3,S_3]=A_3. \end{align*} Therefore \begin{align*} S_3^{\mathrm{ab}}=S_3/[S_3,S_3]=S_3/A_3. \end{align*} The quotient $S_3/A_3$ has the two cosets $A_3$ and $(12)A_3$, because $A_3$ has three elements and $S_3$ has six elements. The coset $(12)A_3$ has order $2$, since \begin{align*} (12)A_3+(12)A_3 \end{align*} in multiplicative quotient notation means \begin{align*} (12)A_3\cdot(12)A_3=(12)^2A_3=eA_3=A_3. \end{align*} Thus $S_3/A_3$ is the cyclic group with two elements, so \begin{align*} S_3^{\mathrm{ab}}\cong\mathbb{Z}/2\mathbb{Z}. \end{align*} This computation shows that abelianisation keeps only the parity information in $S_3$ and kills the alternating subgroup. [/example] Abelianisation is especially important in topology, where the first homology group is the abelianisation of the fundamental group. It turns path composition, which may be noncommutative, into an additive invariant. ## Beyond and Connected Topics Abelian groups are the meeting point of elementary group theory, module theory, and invariant-building constructions. The next natural step is the general theory of [groups](/page/Group), where abelian groups appear as the commutative child case and where conjugation, normal subgroups, and group actions explain what is missing in the abelian setting. The module viewpoint leads directly to modules over a ring. Abelian groups are precisely $\mathbb{Z}$-modules, so many ideas from [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) generalise the present chapter by replacing $\mathbb{Z}$ with an arbitrary ring $R$. Free modules, quotient modules, exact sequences, tensor products, and structure theorems all refine the same additive language. Finite abelian groups are also a gateway to representation theory and character theory. Here $\mathbb{C}^\times$ denotes the multiplicative group of nonzero complex numbers. Homomorphisms from a finite abelian group into $\mathbb{C}^\times$ form the character group, and the classification theorem makes these characters explicitly computable. This is one reason finite Fourier analysis works so cleanly on abelian groups. In algebraic topology, abelian groups appear as homology groups. The operation is addition of cycles modulo boundaries, and quotienting by boundaries is exactly the kind of construction abelian groups are built to support. The passage from the fundamental group to first homology is abelianisation in a geometric disguise. Commutative algebra uses abelian groups both as underlying additive groups of rings and as modules. Ideals, quotient rings, divisor class groups, and localisations all rely on additive abelian structure before multiplicative structure enters. The course-level continuation [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra) develops this module-theoretic world in a deeper direction. ## References Androma, [Group](/page/Group). Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Dummit and Foote, *Abstract Algebra* (2004). Lang, *Algebra* (2002). Rotman, *An Introduction to the Theory of Groups* (1995).