Dihedral and Symmetric Groups

Two families of finite nonabelian groups supply nearly every example in the first half of the subject: the dihedral groups, the symmetries of regular polygons, and the symmetric groups, all rearrangements of a finite set. Both admit compact descriptions — the dihedral groups through two generators and three relations, the symmetric groups through cycle notation — and both test the abstract group axioms against noncommutative structure.

The dihedral group

Fix an integer $n\ge 3$. A symmetry of a regular $n$-gon is a rigid motion that carries the polygon onto itself: pick up a copy, move it through space, and set it back so it exactly covers the original.

Label the vertices $1,2,\dots ,n$ clockwise. Each symmetry $s$ is determined by where it sends the vertices, so it corresponds to a permutation of the labels; composition of symmetries corresponds to composition of these permutations, which is associative. The identity motion fixes every vertex, and every motion reverses, so the axioms hold.¹

Counting the symmetries

The count $2n$ comes from a two-step choice. A symmetry can send vertex $1$ to any of the $n$ positions. Once vertex $1$ is placed, vertex $2$ (its neighbor) must go to one of the two positions adjacent to it, and both are achievable — one by a rotation, the other by following with a reflection. After the images of the adjacent pair $1,2$ are fixed, rigidity determines the rest. So there are exactly $n\cdot 2=2n$ symmetries: $n$ rotations and $n$ reflections.

Generators and relations

Let $r$ be the clockwise rotation through $2\pi /n$ radians and let $s$ be the reflection across the axis through vertex $1$ and the center. Direct calculation establishes the following facts.¹

• $1,r,r^2,\dots ,r^{n-1}$ are distinct and $r^n=1$, so $|r|=n$.
• $|s|=2$, and $s$ is not a power of $r$.
• Every element of $D_{2n}$ is $s^k{r}^i$ for a unique $k\in \{0,1\}$ and $i\in \{0,1,\dots ,n-1\}$ — the $n$ rotations $r^i$ and the $n$ reflections $s{r}^i$.
• $rs=s{r}^{-1}$, and more generally $r^{i}s=s{r}^{-i}$.

The last relation is the one used in every computation: it says how to move an $s$ past a power of $r$. Because $r$ and $s$ do not commute, $D_{2n}$ is nonabelian for every $n\ge 3$. All of this compresses into a presentation:

With the presentation in hand, every product reduces to the normal form $s^k{r}^i$ by pushing all $s$'s left and reducing exponents mod $n$. For example, in $D_{24}$ ($n=12$),

A presentation is delicate: relations can interact to collapse the group to something smaller than expected, so a presentation alone does not reveal the order. For $D_{2n}$ the collapse does not happen, because the geometry already exhibits a group of order $2n$ satisfying the relations, pinning the order at exactly $2n$.¹

The same counting principle computes the rotation groups of the Platonic solids: find how many positions an adjacent pair of vertices can be sent to. A tetrahedron gives $12$, a cube and an octahedron each $24$, a dodecahedron and an icosahedron each $60$.

The symmetric group

Where dihedral groups permute the vertices of one polygon, symmetric groups permute an arbitrary finite set with no geometric constraint.

Composition of bijections is a bijection and is associative; the identity map is the identity element; every bijection has an inverse. The elements of $S_n$ are the permutations, not the numbers being permuted.

Cycle notation

Writing $\sigma$ as a table of values is unwieldy. Cycle notation records the orbits of the map instead.

To read $\sigma$, trace each point forward until it returns. Take $n=13$ and

which closes the cycle $(1128104)$; continuing with the smallest unused label gives the full cycle decomposition

a product of pairwise disjoint cycles. The fixed point $3$ would be a $1$-cycle $(3)$; by convention $1$-cycles are omitted, so an unwritten label is understood to be fixed.

Multiplying permutations

Products are computed by tracing a point through both maps, right to left. For $(123)\circ (12)(34)$: the right factor sends $1\mapsto 2$, then the left sends $2\mapsto 3$, so the composite sends $1\mapsto 3$. Continuing yields $(123)\circ (12)(34)=(134)$. Order matters:

so $S_n$ is nonabelian for every $n\ge 3$. Two structural facts make cycle notation efficient: disjoint cycles commute (they move disjoint sets of points), and the entries within a single cycle may be rotated without changing the permutation, e.g. $(1234)=(3412)$.

Order from cycle type

The order of a permutation is legible in its decomposition.

Parity and the alternating group

A single transposition — a $2$-cycle — is the atom of $S_n$. Any $m$-cycle factors into transpositions,

so every permutation is a product of transpositions.³ The factorization is far from unique: the identity is both $(12)(12)$ and $(13)(24)(13)(24)$. One invariant is nonetheless well defined.

The proof rests on a lemma: if the identity equals a product of $r$ transpositions, then $r$ is even. One shows that the leftmost repeated symbol can always be pushed right or cancelled, reducing $r$ by two at a time until an obvious cancellation remains, so $r$ can never be odd.³ Since every permutation has a fixed parity, the assignment $\sigma \mapsto (\text{parity of }\sigma )$ splits $S_n$ cleanly in two.

Comparing the two families

For $n=3$ the two families coincide: $|D_6|=6=3!=|S_3|$, and every permutation of three vertices is realized by a symmetry of the triangle, so $D_6\cong S_3$. For $n\ge 4$ the polygon is too rigid to realize every permutation, and $|D_{2n}|=2n<n!=|S_n|$. The precise sense in which one group sits inside or equals another is homomorphism and isomorphism, and the vertex-permutation correspondence used here is an instance of a group action.