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Every discrete additive subgroup of ${\mathbb{R}}^2$ has the form $\mathbb{Z}{v}_1+\cdots +\mathbb{Z}{v}_r$ for linearly independent vectors, with $r=0,1,$ or $2$. A full translation lattice has $r=2$, and conversely every such integer span is a discrete cocompact subgroup. This classifies planar lattices up to a choice of basis.

A frieze group is a discrete group of planar Euclidean isometries whose translation subgroup is infinite cyclic and has finite index. Consider the glide reflection $g(x,y)=(x+1,-y)$. Its square is translation by $(2,0)$, whereas every odd power reverses orientation. Therefore $⟨g⟩\cong \mathbb{Z}$, but none of its translation elements generates the whole group: they form the index-two subgroup $⟨g^2⟩$.

To realize exactly this frieze group, repeat a small scalene triangular motif at the points $(n,(-1{)}^n)$, reflecting the motif when $n$ is odd. Choose the motif with three distinct side lengths and no internal symmetry. The displayed pattern has glide reflection symmetry $g$ and translation period $2$, but no pure horizontal reflection or vertical mirror. Distinct side lengths force any symmetry to preserve corresponding vertices, which rules out extra motif permutations; the sequence of heights then leaves exactly the powers of $g$.

For a line through the origin making angle $\theta$ with the positive $x$-axis, its reflection matrix isIts determinant is $-1$. An orthogonal matrix has determinant $\pm 1$. If its determinant is $-1$, write its first column as $(\cos \phi ,\sin \phi {)}^T$; orthonormality and the determinant then force the second column to be $(\sin \phi ,-\cos \phi {)}^T$. The matrix is $S_{\phi /2}$, hence is one Euclidean reflection. If its determinant is $+1$, the second column is $(-\sin \phi ,\cos \phi {)}^T$, so it is a rotation matrix. Direct multiplication givesThe identity may be taken as an empty product or as two identical reflections. Thus at most two origin-line reflections suffice in every case.

For a nonzero vector $b=dn$ with $d=|b|$ and unit normal $n$, reflection in the affine line $n\cdot x=c$ is $s_c(x)=x+2(c-n\cdot x)n$. Consequently $s_{d/2}{s}_0(x)=x+dn=x+b$. A translation is therefore a product of two parallel-line reflections. Combining this with the given decomposition of a Euclidean isometry as translation followed by an orthogonal map proves that every plane isometry is a finite product of affine-line reflections. The affine lines need not pass through the origin. In fact three suffice: if an isometry $f$ sends zero to $b\ne 0$, reflection in the perpendicular bisector of $0,b$ takes $b$ to zero. Composing that reflection with $f$ gives an isometry fixing zero, to which the two-reflection linear result applies. This is the planar finite reflection decomposition of a Euclidean isometry.

For an example that genuinely needs three, take the glide reflectionIt has no fixed point, so it is not one reflection. Its orthogonal part has determinant $-1$, whereas the product of two reflections has determinant $+1$, so it is not two reflections; it is not the identity either. Explicitly it is reflection in $y=0$ followed by reflection in $x=0$ and then in $x=1/2$. Thus its minimum reflection count is exactly three.

Let $v=T(0)$ and $B(x)=T(x)-v$. The Euclidean isometry $B$ fixes the origin and preserves norms. The polarization identity therefore gives $⟨Bx,By⟩=⟨x,y⟩$. In particular, $B{e}_1,B{e}_2$ are an orthonormal basis. The coefficients of $Bx$ in this basis are $⟨Bx,B{e}_j⟩=x_j$, so $Bx=x_{1}B{e}_1+x_{2}B{e}_2$. Thus $B$ is a linear map represented by an orthogonal matrix, and $T(x)=Bx+v$. Conversely, every such map preserves distances because $\Vert B(x-y)\Vert =\Vert x-y\Vert$.

When $\det B=-1$, the real eigenvalues of the two-dimensional orthogonal matrix are $1,-1$. Choose corresponding unit eigenvectors $e,n$, and write $v=ae+bn$. In these coordinates $T(s,t)=(s+a,-t+b)$. This is reflection in the line $t=b/2$, followed by translation $ae$ along that line. It is a Euclidean reflection if $a=0$, and a glide reflection if $a\ne 0$.

For unit $u$, put $A=I-2u{u}^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an orthogonal matrix. The affine reflection in a hyperplane is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the Euclidean norm and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an Euclidean isometry fixing the hyperplane pointwise.

For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:Changing both signs of $u,c$ leaves the same reflection. It belongs to the orthogonal group $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\Vert p{\Vert }^2-\Vert q{\Vert }^2)/(2\Vert p-q\Vert )$,To prove the finite reflection decomposition of a Euclidean isometry, first note that a distance-preserving map $g$ fixing zero preserves inner products by the polarization identity. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)={\sum }_i{x}_{i}g(e_i)$. Thus $g$ is orthogonal.

Every orthogonal transformation is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^{\perp }$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.

For an arbitrary isometry $f$, if $f(0)\ne 0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. ThereforeThe glide reflection $(x,y)\mapsto (x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.