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Let $G$ have group order equal to a prime number $p$. Choose any $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$, and it is greater than one. It is therefore $p$, so the cyclic subgroup generated by $g$ is all of $G$. This proves that every nonidentity element generates $G$, and $G$ is isomorphic to the additive group of integers modulo $p$.

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.

Every complex character of a representation of a finite group $G$ is a rational linear combination of characters of induced representations from one-dimensional group representations of cyclic subgroups $C_j$. One-dimensionality follows because the irreducible representations of a finite cyclic group over the complex numbers are one-dimensional. The rational coefficients distinguish this statement from Brauer induction, which permits integer coefficients by allowing more general subgroups. For a rational-valued character, another form of Artin induction uses rational combinations of induced representations of trivial group representations of cyclic subgroups.

Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]

Both constructions are stabilizer subgroups for the conjugation action. The centralizer $C_G(x)$ fixes an individual element $x$; the normalizer $N_G(H)$ preserves a subgroup $H$ as a whole, while possibly moving its elements. The centralizer of a subset $C_G(H)$ fixes every member of $H$ individually, soFor a cyclic subgroup $H=⟨r⟩$, commuting with the group generator $r$ is equivalent to commuting with all its powers, giving $C_G(H)=C_G(r)$.

In the symmetric group $S_3$, let $r=(123)$ and $H=⟨r⟩=\{e,(123),(132)\}$. ThenEvery transposition sends $r$ to $r^{-1}$ under conjugation. In particular, for $t=(12)$,Thus $t$ belongs to the normalizer of $H$ but not to the centralizer of $r$. Preserving the collection of powers allows the two possible directions of its group generator to be exchanged.

This distinction lets one ask separately which operations leave a chosen operation unchanged and which operations preserve the subgroup it generates. The normalizer also supplies a group homomorphism to the automorphism group of $H$, by $g\mapsto (h\mapsto gh{g}^{-1})$; its group kernel is $C_G(H)$. In this example the group homomorphism has two possible outputs, fixing $r$ or replacing it by $r^{-1}$. The quotient group $N_{S_3}(H)/C_{S_3}(H)$ is therefore a cyclic group of group order two. Milne develops the normalizer as a stabilizer subgroup in Group Theory (v4.01) by James S. Milne.

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

A group of order ten is either a cyclic group or the dihedral group of the pentagon. Lagrange's theorem and the fact that a group of exponent two has power-of-two order force an element of order five or ten. In the latter case the group is cyclic. In the former case its order-five cyclic subgroup is normal by index two; an outside involution acts on it either trivially or by inversion. The trivial action gives $C_5\times C_2\cong C_{10}$, and inversion gives $D_{10}$.

Write $D_{2n}=⟨r,s:r^n=s^2=1,srs=r^{-1}⟩$. A subgroup contained in the rotations is a cyclic subgroup $⟨r^d⟩$, with $d\mid n$. If it contains a reflection, every other reflection differs from a chosen $r^{j}s$ by a member of the rotational intersection, so it is $⟨r^d,r^{j}s⟩$, with $j$ specified modulo $d$. These subgroups have orders $n/d$ and $2n/d$ respectively. This gives a complete list, including the trivial and whole subgroups; see Peter Cameron's group-theory exercise solutions, question1.

The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.

For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

In a finite group, a cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function. Each element generates exactly one cyclic subgroup, so dividing the number of elements of order $m$ by $\phi (m)$ counts these subgroups. Include the identity subgroup using $m=1$ and $\phi (1)=1$. In a symmetric group, element counts come from cycle type and order of a group element is the least common multiple of cycle lengths.

For the dicyclic group ${\mathrm{Dic}}_n=⟨a,b\mid a^{2n}=1,b^2=a^n,ba{b}^{-1}=a^{-1}⟩$, with $n\ge 2$, there are four linear characters and $n-1$ irreducible complex characters of degree two.

The linear characters are determined by $a\mapsto ϵ\in \{1,-1\}$ and $b\mapsto \beta$, where ${\beta }^2={ϵ}^n$. Their values are ${ϵ}^r$ on $a^r$ and ${ϵ}^r\beta$ on $a^{r}b$. When $n$ is even both choices of $ϵ$ allow $\beta =\pm 1$; when $n$ is odd, $ϵ=-1$ requires $\beta =\pm i$.

Put $\zeta =e^{\pi i/n}$. For $1\le j<n$, take the group representationIts character is $2\cos (\pi jr/n)$ at $a^r$ and zero at every $a^{r}b$. The distinct eigenlines of $a$ are exchanged by $b$, proving that this group representation is an irreducible representation. The degree-square sum $4+4(n-1)=4n$ proves completeness. These formulas specify every row of the character table; the elements outside $⟨a⟩$ form two conjugacy classes, rather than two distinct cosets of that index-two cyclic subgroup.

Write the dihedral group as $D_{2n}=⟨r,s:r^n=s^2=e,srs=r^{-1}⟩$. If $k\mid n$, the cyclic subgroup $⟨r^{n/k}⟩$ has order $k$. For an even divisor $k=2m$ of $2n$, $m\mid n$ and $⟨r^{n/m},s⟩$ has $m$ rotations and $m$ reflections, hence order $k$. An odd divisor of $2n$ always divides $n$. This proves that the necessary divisibility condition from Lagrange's theorem is also sufficient in a dihedral group, although it is not sufficient in arbitrary finite groups.

Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

For finitely generated groups $H\le G$, choose finite symmetric generating sets $T$ for $H$ and $S$ for $G$, and measure subgroup distortion byThe subgroup is exponentially distorted when this function has exponential growth up to the usual coarse comparison. In particular it has both an exponential lower bound and an exponential upper bound after rescaling the argument; an exponential lower bound alone could also occur for more severely distorted subgroups.

For the dyadic affine Baumslag-Solitar groupthe cyclic subgroup $H=⟨a⟩$ is exponentially distorted. The identity $t^{k}a{t}^{-k}=a^{2^k}$ gives ambient word length at most $2k+1$ and intrinsic word length $2^k$.

For the upper bound, use its faithful group action on the real line, given by $a(x)=x+1$, $t(x)=2x$. A word of length at most $n$ representing $a^m$ has at most $n$ translation contributions, each of absolute size at most $2^n$. Thus $|m|\le n{2}^n$. These matching bounds establish the exponential distortion class, independently of the chosen finite generating sets.

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.

A finite group is a group whose underlying set has finitely many elements. Its order of a finite group is the number of elements, counting the identity element. The group operation still has to satisfy all the group axioms: merely having a finite list of states does not make a group.

For example, the six congruence classes modulo six form a finite cyclic group under addition, whereas the permutations of three objects form the symmetric group $S_3$, also with six elements. The first is an abelian group; the second is a non-abelian group. Thus the order of a finite group is a useful first measurement, but the composition rule contains information that the count alone loses.

Finiteness makes group structure accessible to exact calculation. A Cayley table for a finite group of order $n$ has $n^2$ entries, and every group element eventually returns to the identity element when repeated. Lagrange's theorem strengthens this observation: the order of a group element, being the size of its cyclic subgroup, divides $n$. The cyclic group of order four and Klein four-group give a small example of how that constraint can distinguish different composition rules.

A finite subgroup of $O(2)$ is cyclic or dihedral. Its orientation-preserving part consists of rotations and is cyclic: choose the smallest positive rotation angle, divide any other angle by it with remainder, and use minimality to force zero remainder. The smallest angle is therefore $2\pi /n$ for some $n$. A trivial rotational subgroup is included by $n=1$.

If there is no orientation-reversing element, this is the whole group. Otherwise choose a reflection $s$. Every orientation-reversing element is $r^{j}s$, where $r$ generates the rotations; these form the second coset. Since $srs=r^{-1}$, the group is a dihedral group of order $2n$. Its involutions are all $n$ reflections and, when $n$ is even, the extra half-turn $r^{n/2}$. Consequently it has $n$ order-two cyclic subgroups for odd $n$, and $n+1$ for even $n$.

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 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.