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For $n\ge 3$, the dihedral group is the symmetry group of a regular $n$-gon, including rotations and reflections. Using the convention that the subscript gives the group order,Its $2n$ elements are $r^j$ and $r^{j}s$, with $0\le j<n$: the first are rotations, the second reflections. The relation $sr=r^{-1}s$ reduces every word to one of these forms. Geometrically the image of one vertex and a choice of direction around the polygon distinguish all $2n$ forms.

Another common convention writes $D_n$ for this same group, indexing by polygon vertices. For example the symmetry group of a square has eight elements and may be called either $D_8$ or $D_4$. The group order or group presentation resolves that notation difference. Its rotation subgroup is a cyclic group of index two, and a reflection conjugates each rotation to its inverse.

Write the symmetry group of a square as $D_4=⟨r,s\mid r^4=s^2=e,srs=r^{-1}⟩$, with $r$ a counterclockwise quarter-turn and $s$ reflection across the horizontal axis. The conjugacy classes of the symmetry group of a square areHere $r^2$ is the half-turn and $r^3$ the clockwise quarter-turn. The two members of the fourth class reflect across the horizontal and vertical axes. Those of the fifth reflect across $y=x$ and $y=-x$. The identity and half-turn are central elements, giving singleton classes. The relations $srs=r^3$, $rs{r}^{-1}=r^{2}s$, and $r(rs)r^{-1}=r^{3}s$ show that each displayed pair consists of conjugate elements. More generally, conjugating $r^{k}s$ by $r$ changes $k$ to $k+2$, and conjugating by $s$ changes $k$ to $-k$; parity is preserved, so the two reflection pairs cannot merge.

In a one-dimensional representation, the relation $srs=r^{-1}$ forces the image of $r$ to equal its inverse. Thus $r$ and $s$ can be assigned independent signs $\alpha ,\beta \in \{1,-1\}$. These four choices give four irreducible characters. The ordinary two-dimensional action on the plane gives one more character: a rotation through $\theta$ has trace $2\cos \theta$, and a reflection has trace zero. Therefore the character table of the symmetry group of a square isFor instance the last row has squared character norm $(4+4)/8=1$, so the planar representation is irreducible; its inner product with each one-dimensional row is zero. All five rows obey character orthogonality. Their squared dimensions sum to $1+1+1+1+4=8$, so they exhaust the irreducible representations of this order-eight group.