Permutation and dihedral groups

The symmetries of a regular polygon combine a cyclic rotation group with reflections that reverse rotation direction. Permutations make these operations exact and finite. We develop cycle computations, prove the dihedral normal form, and connect the triangle's symmetries with the full permutation group on three vertices.

1 Permutations and disjoint cycles

The group $S_n$ consists of all bijections of $\{1,\ldots,n\}$ under composition. Throughout, $\sigma\tau$ means apply $\tau$ first. A cycle $(a_1\ \cdots\ a_k)$ sends each entry to the next, the last to the first, and fixes all other elements.

Every permutation decomposes into disjoint cycles. Starting at one element, follow repeated images. Finiteness forces a repeat, and bijectivity makes the first repeat return to the starting element; otherwise a repeated value would have two different predecessors. The resulting orbit is a cycle. Repeat with an unused element. Distinct orbits are disjoint, so their cycles commute and together describe the whole permutation. The decomposition is unique apart from ordering the disjoint cycles and rotating the written entries of each cycle.

A $k$-cycle returns every affected element after exactly multiples of $k$ iterations. Hence the order of disjoint cycles of lengths $k_1,\ldots,k_r$ is $\operatorname{lcm}(k_1,\ldots,k_r)$. For $(1\ 3\ 5)(2\ 4)$ the order is 6, not the sum 5 and not necessarily the product in other examples.

Worked composition. Let $\sigma=(1\ 2\ 3)$ and $\tau=(1\ 2)$. Tracing each input gives $\tau\sigma=(2\ 3)$: 1 returns to 1, 2 goes to 3, and 3 goes to 2. In contrast, $\sigma\tau=(1\ 3)$. Changing the convention changes calculations, not the group itself, so state the convention before computing.

2 Building the dihedral group explicitly

Fix $n\ge3$ and label polygon vertices by $\mathbb Z/n\mathbb Z$. Let $r(k)=k+1$ and $s(k)=-k$. These are permutations satisfying

$$r^n=e,\qquad s^2=e,\qquad sr=r^{-1}s.$$

The group they generate is denoted $D_n$ here, with order $2n$. Some books use the subscript $2n$ instead; the stated order removes the notational ambiguity.

Using $sr^a=r^{-a}s$, move every $s$ in a product to the right and reduce powers. Every element has form $r^as^\varepsilon$, where $0\le a<n$ and $\varepsilon\in\{0,1\}$. These $2n$ permutations are distinct. Within either type, evaluation at zero distinguishes $a$. A rotation cannot equal a reflection: if $k+a=b-k$ for every vertex, then comparing $k=0$ and $k=1$ would force $2=0$ modulo $n$, impossible for $n\ge3$.

Thus the normal forms do not merely give an upper bound; they give every element exactly once. The multiplication rule is

$$(r^as^\varepsilon)(r^bs^\delta)=r^{a+(-1)^\varepsilon b}s^{\varepsilon+\delta},$$

with rotation exponents modulo $n$ and reflection exponents modulo 2. This formula is often faster and safer than drawing repeated polygon moves.

3 Orders normality and the triangle

Every reflection $r^as$ has order 2:

$$(r^as)^2=r^a(sr^a)s=r^ar^{-a}s^2=e,$$

and no reflection is identity. A rotation $r^a$ has order $n/\gcd(n,a)$. The rotations form a normal cyclic subgroup $R=\langle r\rangle$: conjugation by a rotation preserves it, and conjugation by $s$ takes $r^a$ to $r^{-a}$. The quotient has two cosets and is isomorphic to $\mathbb Z/2\mathbb Z$, recording whether the symmetry is a rotation or reflection.

For $n=3$, label vertices by 1,2,3 and take $r=(1\ 2\ 3)$, $s=(2\ 3)$. The relation $srs=r^{-1}$ holds by tracing vertices. The six normal forms are six distinct permutations, and $S_3$ has exactly $3!=6$ elements, so $D_3\cong S_3$. This is an isomorphism of operations, not just a coincidence of sizes: both are the same transformations of the labeled vertex set under the indicated generators.

For every $n\ge4$, $D_n$ is a proper subgroup of $S_n$, because $2n<n!$. A polygon symmetry preserves adjacency; a general permutation need not. Already $D_4$ has 8 elements while $S_4$ has 24.

4 Graded practice

E1 Foundation. Decompose the permutation sending $1\mapsto4$, $4\mapsto2$, $2\mapsto1$, $3\mapsto5$, $5\mapsto3$, $6\mapsto6$ into disjoint cycles. Find its order.

E2 Dihedral calculation. In $D_5$, simplify $(r^2s)(r^3s)$ and find the orders of $r^2$ and $r^2s$.

E3 Structure. Prove $D_n$ is nonabelian for every $n\ge3$ under the convention above.

E4 Center. Determine the rotations that commute with a reflection. Use this to find the center of $D_n$ for odd and even $n$.

5 Full solutions

S1. The decomposition is $(1\ 4\ 2)(3\ 5)$, with 6 fixed. Its order is $\operatorname{lcm}(3,2)=6$. Fixed points contribute cycle length 1 and do not change the least common multiple.

S2. The product is $r^{2-3}=r^{-1}=r^4$. The rotation $r^2$ has order $5/\gcd(5,2)=5$, while $r^2s$ has order 2 by the reflection calculation.

S3. If $sr=rs$, the relation $sr=r^{-1}s$ would imply $r=r^{-1}$, hence $r^2=e$. But $r$ has order $n\ge3$, a contradiction. Preserving order of factors matters even when each reflection is its own inverse.

S4. $sr^a=r^{-a}s$ equals $r^as$ exactly when $2a=0$ modulo $n$. For odd $n$, this gives only $a=0$; for even $n$, $a=0,n/2$. No reflection commutes with $r$, by the same contradiction as E3. Therefore the center is $\{e\}$ for odd $n$ and $\{e,r^{n/2}\}$ for even $n$.

Summary

Disjoint cycles organize permutations into independent orbits. A dihedral group has a unique rotation-or-reflection normal form and an explicit multiplication rule. Its cyclic rotation subgroup is normal, while reflections reverse rotations. These structural arguments explain the group without relying on pictures or equal-size guesses.