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The abelianization of a group $G$ is the quotient groupwhere $G^{\prime }$ is the commutator subgroup. Its elements are cosets, and its operation is $(g{G}^{\prime })(h{G}^{\prime })=gh{G}^{\prime }$. The result is an abelian group: the two products $gh$ and $hg$ differ by a group commutator, which disappears in the quotient group.

For the symmetric group $S_3$, the sign of a permutation realizes $S_3^{\mathrm{ab}}\cong C_2$, as shown in commutator subgroup of the symmetric group on three letters. For the dihedral group of a square, abelianization of the square symmetry group gives $C_2\times C_2$: two independent yes-or-no pieces of information survive. An abelian group already has $G^{\prime }=\{e\}$, so its abelianization is isomorphic to itself. At the other extreme, a perfect group has trivial abelianization.

Calling this the largest abelian quotient means that every group homomorphism into any abelian group factors uniquely through $q$. It does not require $G$ to be finite, or compare the sizes of unrelated abelian groups. The universal property of abelianization makes this precise. It lets us solve all such mapping problems in one simpler group: any invariant that respects multiplication and takes values in an abelian group sees only $G^{\mathrm{ab}}$.

For odd $n\ge 3$, the dihedral group $D_{2n}$ of order $2n$ has rotation generator $r$ and reflection generator $s$, with $srs=r^{-1}$. The group commutator $[r,s]=r^{-2}$ generates the rotation subgroup, while the quotient by that subgroup is $C_2$. Hence the commutator subgroup is exactly $⟨r⟩$ and the abelianization is $C_2$. The only abelian quotients are $C_2$ and the trivial group.

Let $D_8$ denote the dihedral group of group order eight acting on a square. Here the subscript specifies group order; the same group is called $D_4$ in the convention that counts vertices of a polygon, as in the group action example. Let $r$ be a counterclockwise quarter-turn rotation and $s$ a reflection in a diagonal. ThenProducts act rightmost first. The group commutator of the generators of a group isIt is a half-turn rotation, so forcing $r$ and $s$ to commute must identify a half-turn with $e$.

More concretely, write each element uniquely as $r^i{s}^j$, with $0\le i<4$ and $j\in \{0,1\}$. The functionis a surjective group homomorphism. Indeed, $srs=r^{-1}$ giveswhere the exponents are reduced modulo four and two. Modulo two, $-k$ and $k$ agree, so $\varphi$ adds the two coordinates separately. Its group kernel is $N=\{e,r^2\}$. Since the target is an abelian group, $D_8^{\prime }\subseteq N$; since $[r,s]=r^2$, also $N\subseteq D_8^{\prime }$. ThusThe four cosets are $\{e,r^2\}$, $\{r,r^3\}$, $\{s,r^{2}s\}$, and $\{rs,r^{3}s\}$. Each coset forgets the difference of a half-turn. The quotient group still distinguishes rotations from reflections, and even from odd numbers of quarter-turns; the two surviving choices are independent. This is the Klein four-group, rather than the cyclic group $C_4$, because each nonidentity element squares to the identity element. For the general dihedral group calculation, see Conrad, Dihedral groups I, theorem A.1; that source uses the vertex-count convention and $xy{x}^{-1}{y}^{-1}$ for individual group commutators.

An opposite-vertex line in the full octahedral symmetry group is an unoriented coordinate axis. Its stabilizer subgroup has the block form $\mathrm{diag}(A,\epsilon )$, where $A$ is a signed two-dimensional permutation matrix and $\epsilon =\pm 1$. The eight possibilities for $A$ form the dihedral group of a square, of order eight; the independent axial sign gives a cyclic group of order two. Thus the stabilizer is their direct product, of order sixteen. It is larger than the stabilizer of an individual vertex, because exchanging the two endpoints still fixes the line.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

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 $D_{10}$.

A coloring bracelet also identifies colorings related by reflection, so it is an orbit under a dihedral group rather than only a cyclic group.

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]

A diagonal of a polygon joins two vertices of a polygon that are not endpoints of the same side of a polygon. A square has two diagonals of a polygon, forming a two-element group orbit under its dihedral group of symmetries.

Automorphisms $\sigma (z)={\zeta }_{n}z$ and $\tau (z)=1/z$ generate a dihedral group of order $2n$. The invariant $w=z^n+z^{-n}$ has rational map degree $2n$, so $[\mathbb{C}(z):\mathbb{C}(w)]=2n$. The fixed-field theorem for a finite automorphism group gives the same degree over its fixed field, proving equality. For odd prime $n$, the nontrivial proper subgroups are the rotation group and the $n$ reflection groups. Their fixed fields are $\mathbb{C}(z^n)$ and $\mathbb{C}(z+{\zeta }_n^{-j}/z)$ for reflections ${\sigma }^j\tau$, respectively.

On $K=\mathbb{Q}(x,y)$, let $\sigma (x,y)=(y,-x)$ and $\tau (x,y)=(x,-y)$. They generate a dihedral group of order eight, andThe inclusion from right to left is immediate. In the other direction, $x^2,y^2$ solve a quadratic over the displayed field and adjoining their square roots has degree at most four; the resulting total degree is at most eight, while the automorphism-count bound for a finite field extension gives the reverse bound.

For the natural plane action of the dihedral group, rotations constrain monomial weights and reflection makes coefficients real. The reflection axes support the bifurcating equilibria. For $n=3$, the quadratic anisotropic term gives three transcritical-like branches that are saddles in the full plane. For $n=4$, cubic isotropic and anisotropic terms determine two reflection-isotropy classes. For $n=5$, the cubic isotropic term sets the leading radial amplitude while quartic anisotropy separates stable and saddle angular classes.

// Target: dynamical-systems.bigb

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.

For the order-eight dihedral group generated by $r$ of order four and $s$ of order two with $srs=r^{-1}$, let $H=K=⟨s⟩$. There are exactly three double cosets: $H$, $H{r}^{2}H$, and $HrH$, of sizes two, two, and four. Their number does not divide eight. This differs from the number of ordinary cosets, which divides group order by Lagrange's theorem.

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 $n$.

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 left group action of a group $G$ on a set $X$ is a function $(g,x)\mapsto g\cdot x$ satisfyingwhere $e$ is the identity element of $G$. The second rule means that $h$ moves the point first and $g$ moves it next. These rules let group multiplication describe consistent rearrangements of a set; see Ackerman's algebra notes, definition 0.4.1.1.

For each $g$, the function $x\mapsto g\cdot x$ is a bijection: its inverse is $x\mapsto g^{-1}\cdot x$, using the inverse element of $g$. Equivalently, a group action is a group homomorphism $\rho :G\to \mathrm{Sym}(X)$ into the symmetric group of all permutations of $X$, with $\rho (g)(x)=g\cdot x$ and $\rho (gh)=\rho (g)\circ \rho (h)$. See MIT's group-action notes, definition 23.1.

For a concrete group action, put the vertices of a polygon of a square at $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$ and $D=(1,-1)$. Let $r(x,y)=(-y,x)$ be a counterclockwise quarter-turn rotation and let $s(x,y)=(y,x)$ be a reflection. Their eight permutations of the vertices of a polygon form the dihedral groupHere $D_4$ means the symmetry group of a four-sided polygon, with group order eight; the alternative convention calls this same group $D_8$. Products apply the rightmost operation first. For example, $s\cdot B=D$ and $(rs)\cdot B=r\cdot D=A$. The same group also has separate group actions on the sides of a polygon, the diagonals of a polygon, and the singleton set containing the centroid of the square. Specifying $X$ matters: changing what is moved changes the orbits of a group action and the stabilizer subgroups.

A group action need not be a faithful group action. For example, the cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ acts on $X=\{0,1\}$ by $k\cdot x=x+k(\mathrm{mod}2)$, using modular arithmetic. Only the parity of $k$ matters: $0,2$ fix both points and $1,3$ exchange them. The associated group homomorphism $C_4\to \mathrm{Sym}(X)$ has group kernel $\{0,2\}$, of group order two. Either point has a group orbit of size two and a stabilizer subgroup of group order two, so the orbit-stabilizer theorem gives $4=2\cdot 2$.

The useful questions are which positions can be exchanged, and which transformations survive after imposing a constraint such as keeping one marked point fixed. An orbit of a group action answers the first question; a stabilizer subgroup answers the second. This turns a list of transformations into a count of distinct configurations, and explains why several transformations can produce the same configuration.

Up to group isomorphism, the groups of order eight are $C_8$, $C_4\times C_2$, $C_2^3$, $D_8$ and $Q_8$, with $D_8$ here denoting the eight-element dihedral group. An element of order eight gives the first case. If every nonidentity element has order two the group is an elementary abelian group. Otherwise, after excluding elements of order eight, a cyclic subgroup of order four has index two; an element outside it either commutes with its generator or inverts it. In the latter case its square is either the identity or the central involution, giving the dihedral group or quaternion group.

These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

A polygon with six sides and six vertices. A regular hexagon has equal sides and angles, with each interior angle equal to $2\pi /3$. Its symmetry group consists of six rotations and six reflections, forming a dihedral group of order twelve.