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An abelian group is a group whose group operation is commutative: $gh=hg$ for every $g,h\in G$. Thus rearranging a finite sequence of operations preserves the result. Associativity already permits changing parentheses; commutativity additionally permits exchanging the elements.

The additive integers, the additive real numbers, and every cyclic group are abelian groups. Additive notation is often convenient: the identity element is written $0$, the inverse element of $a$ is $-a$, and $a+b=b+a$. It does not force an element to be an ordinary number: in $C_2\times C_2$, an element is a pair of binary coordinates, added componentwise modulo two.

Commutativity is useful because an integer linear combination of chosen group elements depends only on their coefficients, rather than on the order in which the operations were performed. The Fundamental theorem of finitely generated abelian groups turns this simplicity into a classification by cyclic groups. However, commutativity does not imply that a single generator of a group suffices: the Klein four-group is the smallest counterexample, as the cyclic group of order four and Klein four-group demonstrate.

For a generator of a group $g$ of order $M$ and $x=g^y$, the function $f(a,b)=g^a{x}^{-b}$ hides the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$ of ${\mathbb{Z}}_M^2$. Each fiber is an affine coset $\{(yb+k,b):b\in {\mathbb{Z}}_M\}$. Measuring the function register gives its normalized coset state. Applying the positive-exponent quantum Fourier transform to both input registers gives amplitudeThe finite geometric series vanishes off $c_2=-y{c}_1$ and equals $M$ on that line. Thus each allowed output pair has probability $1/M$, independently of the coset offset. If $\gcd (c_1,M)=1$, a modular inverse recovers $y=-c_2{c}_1^{-1}$.

A finite abelian group combines a finite set of reversible operations with commutativity. All finite cyclic groups qualify, but direct products of groups supply examples that need more than one generator of a group. The Fundamental theorem of finitely generated abelian groups says that every finite abelian group is a direct product of groups of finite cyclic groups; the finite case has no infinite cyclic factors.^[ref]

These factors distinguish different ways to have the same total number of states. A single four-position dial gives $C_4$, while two independent binary switches give $C_2\times C_2$. Both have four states and commutative composition, yet only the dial permits a single operation whose repetition visits all four states. The following Cayley tables make that distinction visible without relying on the labels alone.

A finite cyclic group of order $n$ is isomorphic to the additive group $C_n=\mathbb{Z}/n\mathbb{Z}$. The group isomorphism sends the congruence class of $k$ to $g^k$, where $g$ is a chosen generator of a group. For $n=6$, the identity element is $0$, $4+5=3$ modulo six, and the inverse element of $4$ is $2$.

One finite cyclic group can have several choices of generator of a group. Adding $1$ visits all six elements before returning to zero; adding $5$ also visits all six, in the reverse order. Adding $2$ reaches only $\{0,2,4\}$ from zero, and adding $3$ reaches only $\{0,3\}$. The following permutations of the six elements show what each repeated step can reach.

This gives a precise version of a familiar scheduling question: does a fixed step visit every position, or repeatedly miss some? The generators of a finite cyclic group are exactly the steps coprime to the total number of positions. Changing the step changes the orbit through the same underlying group; the diagram does not replace its group operation with a different one.

Every finitely presented group is the fundamental group of a closed connected smooth oriented four-manifold. Build a compact four-dimensional handle decomposition with one zero-handle, a one-handle for each generator of a group and a two-handle for each relator. Turning the handle decomposition upside down shows that its boundary surjects on the fundamental group. Its double of a manifold has the same fundamental group by the Seifert-van Kampen theorem.

A generator of a group is a single element $g$ for which $G=⟨g⟩$: every group element is an integer power of $g$. A group admitting such an element is a cyclic group. In additive notation, every element must be an integer multiple of $g$.

Distinguish a single generator of a group from a generating set of a group. The Klein four-group needs two elements to generate it, although it has only four elements; no single element visits all four. A cyclic group need not have a unique generator of a group: $\mathbb{Z}/6\mathbb{Z}$ has $1$ and $5$, while the additive integers have $1$ and $-1$.

An element $k$ of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$ is a generator of a group precisely when $\gcd (n,k)=1$. The order of an element of a finite cyclic group proves this immediately: its cyclic subgroup has size $n/\gcd (n,k)$, and this equals $n$ exactly when the greatest common divisor is one. There are therefore $\phi (n)$ choices, where $\phi$ is the Euler totient function.

For $C_6$, the two choices are $1$ and $5$, so $\phi (6)=2$. For $C_8$, they are $1,3,5,7$. Starting from zero and repeatedly adding $3$ modulo eight gives $0,3,6,1,4,7,2,5,0$, visiting every element once before returning. By contrast, adding $2$ gives $0,2,4,6,0$ and misses the odd integers represented in the other coset.

If the cyclic group is written multiplicatively as $⟨g⟩$ with $g$ of order $n$, the same statement says that $g^k$ is a generator of a group exactly when $k$ and $n$ are coprime. This changes notation without changing the calculation.

A group packages reversible operations together with a rule for composing them. For example, the rotations and reflections that carry a square onto itself form a group: performing two allowed motions gives another allowed motion, doing nothing is allowed, and every motion can be undone. The elements are the motions themselves, rather than the vertices they move. Labels in the figure track those vertices; they are markers, rather than extra decorations that the motions must preserve.

Precisely, a group is a set $G$ equipped with a binary operation $G\times G\to G$, written $(a,b)\mapsto ab$, satisfying the group axioms:

The group operation need not be ordinary numerical multiplication. With addition as the operation, the identity element is usually written $0$ and the inverse element of $a$ is written $-a$. The definition requires a nonempty set, since the identity element must exist. See McKernan's MIT lecture 1, Definition 1.1 for the axioms.

Why insist on associativity? It makes a string of operations unambiguous without specifying parentheses. For function composition, both $(fg)h$ and $f(gh)$ send $x$ to $f(g(h(x)))$. This fixes the order of execution while allowing regrouping. Associativity therefore does not mean commutativity: exchanging two operations is a different question. Ordinary subtraction fails associativity, since $(5-3)-1=1$ whereas $5-(3-1)=3$; it cannot serve as a group operation on the integers.

An inverse element makes reversal exact and equations solvable. In any group, $ax=b$ has the unique solution $x=a^{-1}b$, since $a^{-1}(ax)=(a^{-1}a)x=x$. Similarly, $xa=b$ has solution $x=b{a}^{-1}$. The side matters in a non-abelian group. Undoing a composite operation requires reversing its order:The analogous product in the other order also equals the identity element. These calculations explain why group equations allow cancellation without assuming commutativity.

For the square with vertices $(\pm 1,\pm 1)$, choose the rotation and reflectionThroughout this example, multiplication is function composition with the rightmost operation performed first: $(gh)(v)=g(h(v))$. Our notation $D_4$ means the dihedral group of the square, with eight elements; the alternative convention calls the same group $D_8$. Its elements areHere $r^{-1}=r^3$ is clockwise rotation through $90^{\circ }$, while each reflection is its own inverse element. There are exactly eight motions: the image of one vertex has four choices, and the image of an adjacent vertex has two choices; together these determine the motion. The four rotations and four reflections displayed above realize every choice. The vertex permutations give a concrete group action. The general polygon construction is discussed in Group Theory (v4.01) by James S. Milne (section 1.17).

Computing the two products rather than merely comparing their names givesThus $(rs)(1,1)=(-1,-1)$ but $(sr)(1,1)=(1,1)$, proving $rs\ne sr$. In fact $rs$ is reflection in the line $y=-x$, whereas $sr$ is reflection in $y=x$. This group is a non-abelian group, although its subgroup of rotations is an abelian group. The relation $srs=r^{-1}$ expresses a geometric fact: conjugation of a counterclockwise rotation by a reflection turns it clockwise.

A contrasting example is the cyclic group $\mathbb{Z}/4\mathbb{Z}$ under addition in modular arithmetic. Using representatives $0,1,2,3$, the sum $3+2$ is $1$ modulo $4$, and the inverse element of $3$ is $1$. Repeated addition of $1$ runs through $0,1,2,3,0$. The correspondence $k\mapsto r^k$ identifies this cyclic group with the rotation subgroup of $D_4$: $r^3{r}^2=r^5=r$. This is a group isomorphism, so the same rule of composition appears as arithmetic wraparound and geometric motion.

By comparison, $(\mathbb{Z},+)$ is an infinite cyclic group. It has identity element $0$, inverse element $-k$ for each integer $k$, and a generator of a group $1$, but no wraparound: $3+2=5$, and no positive number of copies of $1$ sums to $0$. Both examples are abelian groups. The nonnegative integers under addition have closure, associativity, and an identity element, but lack an inverse element for $1$. Reversibility is precisely what fails.

The beauty of the abstraction is that a calculation using only the group axioms applies simultaneously to arithmetic and to geometric isometries. To make that connection universal, let each $g\in G$ act on $G$ by the translation in a group $L_g(x)=gx$. Each $L_g$ is a bijection, with inverse $L_{g^{-1}}$, and $L_g{L}_h=L_{gh}$. Distinct elements give distinct maps because $L_g(e)=g$. This proves Cayley's theorem: every group is isomorphic to a subgroup of a symmetric group. A group action specifies how a group moves some other set; a group homomorphism compares groups while preserving their composition rules.

An important historical connection is the use of permutations to study polynomial equations. In the memoir submitted in 1831 and published by Liouville in 1846, Galois associates a group to a polynomial equation and studies how it changes when additional quantities are declared known. Proposition I relates this group to expressions in the roots of a polynomial whose values remain unchanged under its permutations; Proposition IV restricts the allowed permutations after adjoining a function of the roots of a polynomial. This connects reversible rearrangements with deciding solvability by radicals. See Galois's original memoir, pp. 417–433, especially Propositions I and IV; Liouville's introduction, p. 382.

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

For any group $G$, evaluation at a chosen generator of a group $x$ of $C_n$ gives a bijection between group homomorphisms $C_n\to G$ and elements satisfying $g^n=1$. A group homomorphism must send $x^j$ to $g^j$. Conversely, this formula is well-defined because exponents differing by a multiple of $n$ give equal powers, and it respects multiplication. The displayed correspondence is a bijection of sets; it is not claimed to be a group isomorphism for a nonabelian target. In a symmetric group, the order of a group element is the least common multiple of its disjoint cycle lengths, making this correspondence easy to enumerate.

A cyclic group has one generator of a group: $G=⟨a⟩=\{a^j:j\in \mathbb{Z}\}$. An abelian group has $xy=yx$ for every pair of elements. Powers of one element commute because $a^r{a}^s=a^{r+s}=a^{s+r}=a^s{a}^r$, so every cyclic group is Abelian. The Klein four-group $C_2\times C_2$ is Abelian but is not a cyclic group: every nonidentity element has order two, whereas a generator of a group for a four-element cyclic group would have order four.

Fix a generator of a group $x$ of $C_n$. A group homomorphism is determined by $g=\varphi (x)$ because $\varphi (x^j)=g^j$, and the relation $x^n=1$ forces $g^n=1$. Conversely, such a $g$ defines $\varphi (x^j)=g^j$: exponents differing by a multiple of $n$ give the same value, and addition of exponents verifies the group homomorphism law. Thus the homomorphism from a finite cyclic group correspondence isFor $S_4$, the order of a permutation is the least common multiple of its disjoint cycle lengths. The condition ${\sigma }^4=1$ allows identity, transpositions, two disjoint transpositions, and four-cycles. The sixteen homomorphisms are ${\varphi }_{\sigma }(x^j)={\sigma }^j$, with the full list of possible generator imagesThe remaining eight elements of the symmetric group are three-cycles and do not qualify.

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

First require $x\ne 0$. A discrete logarithm to a generator of a group of ${\mathbb{Z}}_p^{\ast }$ exists only for nonzero $x$; the printed membership $x\in {\mathbb{Z}}_p$ inadvertently also permits zero, for which $x^{-b}$ is undefined. Zero can be rejected before any quantum circuit.

Put $M=p-1$ and write $x=g^y$. Since $g$ is a generator of a group of order $M$, the map $k\mapsto g^k$ is a bijection from ${\mathbb{Z}}_M$ to the multiplicative group of a finite field. Hence every nonzero $c$ has a unique $k\in {\mathbb{Z}}_M$ with $c=g^k$. Equality of powers of $g$ is equality of their exponents modulo $M$, givingThe fiber over $c$ is an affine coset of the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$, and contains exactly $M$ pairs. This identifies the hidden subgroup problem underlying discrete-logarithm Fourier sampling; knowing $k$ numerically is not needed to construct the coset state.

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

A group $G$ is abelian when $xy=yx$ for every $x,y\in G$. It is cyclic when some generator of a group $g$ satisfiesIf $x=g^r$ and $y=g^s$ belong to a cyclic group, thenso every cyclic group is abelian.

The Klein four-group $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force $G$ to be abelian. The quaternion groupis nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm 1\}$, and the three cyclic subgroupseach of order four. They are all cyclic, so

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the coefficient of either incident $n$-simplex from the other, up to the orientation sign. Connectivity through such faces therefore determines every top-simplex coefficient from one integer. If the signs are globally compatible, their oriented sum is a cycle and generates $H_n(K)\cong \mathbb{Z}$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting fundamental class, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.