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Every Euclidean isometry of ${\mathbb{R}}^n$ is a product of at most $n+1$ operations of reflection in a hyperplane. An isometry fixing zero is orthogonal: the polarization identity preserves inner products, and its values on an orthonormal basis determine its linear action. An orthogonal transformation is a product of at most $n$ linear reflections, by reflecting the image of the first basis vector back to that vector and inducting on its orthogonal complement. For an arbitrary isometry, first reflect its image of zero back to zero, if necessary, and apply the orthogonal result. Affine reflection hyperplanes, rather than only hyperplanes through zero, are essential here.

A finite group of Euclidean isometries has a common fixed point: average any orbit. Each isometry is an affine map and therefore preserves this finite affine average; applying a group element merely permutes the orbit terms. To see that an isometry is affine, subtract its value at zero and use the polarization identity to show that the resulting map preserves inner products. Images of a standard orthonormal basis then determine a linear orthogonal map.

A discrete subgroup of Euclidean isometries is a frieze group when its translation subgroup is infinite cyclic and has finite index. Translations along a line alone give $\mathbb{Z}$; adding perpendicular mirrors gives the infinite dihedral group, while adding a mirror along the translation direction gives $\mathbb{Z}\times C_2$. These groups describe patterns periodic in one planar direction.

The map $A$ associated with an affine map $x\mapsto Ax+b$; it is independent of the translation $b$. An affine Euclidean isometry has an orthogonal linear part, and an isometry between flat tori has a lift whose linear part carries one Euclidean lattice onto the other.

Take the scalene triangle vertex set $T=\{(0,0),(1,0),(0,2)\}$. Its edge lengths are $1,2,\sqrt{5}$. Each vertex has a different unordered pair of incident edge lengths. A Euclidean isometry preserving the set must preserve these pairs, and therefore fixes every vertex.

To justify rigidity of a scalene triangle fully, Euclidean distances from an arbitrary plane point to these three fixed points determine that point uniquely: subtracting its squared Euclidean distance to the origin from the other two squared Euclidean distances determines its two coordinates. Thus an isometry fixing the three vertices fixes every plane point. Only the identity preserves $T$.

For the general union of group translates, let $h\in G$. Left multiplication by $h$ permutes $G$, soTherefore $G\subseteq H$. This containment uses only the group action; the absence of symmetries of $T$ is not needed for the containment itself.

Let $b=f(0)$ and $g(x)=f(x)-b$. Distance preservation and polarization give $⟨g(x),g(y)⟩=⟨x,y⟩$. The vectors $g(e_1),g(e_2),g(e_3)$ form an orthonormal basis, and $⟨g(x),g(e_i)⟩=x_i$. Hence $g(x)={\sum }_i{x}_{i}g(e_i)=Qx$ for an orthogonal matrix $Q$. Every Euclidean isometry is therefore $f(x)=Qx+b$.

To decompose $Q$, if $Q{e}_1\ne e_1$ reflect in the origin plane normal to $Q{e}_1-e_1$. This reflection sends $Q{e}_1$ to $e_1$. The remaining orthogonal transformation fixes $e_1$ and acts on its two-dimensional orthogonal complement; repeat with $e_2$, and then on the last one-dimensional complement. At most three plane reflections are needed. A translation $x\mapsto x+b$ is itself a product of two reflections in parallel planes: for $\mathbf{n}=b/|b|$, reflecting first in $\mathbf{n}\cdot x=0$ and then in $\mathbf{n}\cdot x=|b|/2$ gives that translation. Thus every isometry is a finite composition of plane reflections, the finite reflection decomposition of a Euclidean isometry.

For an isometry fixing zero, the preceding construction gives $N\le 3$. A plane reflection's linear part has $\mathrm{rank}(I-R)=1$. The identity $I-AB=(I-A)+A(I-B)$ implies that a product of $m$ such linear parts has rank of $I-Q$ at most $m$, even if the reflecting planes are affine. For central inversion $Q=-I$, this rank is three. The reflection length of a Euclidean orthogonal map therefore gives

For the regular tetrahedron, center it at zero. Every permutation of its four vertices determines an isometry, and the orientation-reversing ones correspond to odd permutations in $S_4$: six transpositions and six four-cycles. Each transposition is a plane reflection fixing the two unswapped vertices; its plane contains their edge and the midpoint of the opposite edge. These give six distinct mirror planes.

The other six are rotoreflections: rotate by $+\pi /2$ or $-\pi /2$ about an axis through the midpoints of a pair of opposite edges, and then reflect in the plane through the center perpendicular to that axis. There are three axes and two senses. For vertices $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, the map $(x,y,z)\mapsto (-y,x,-z)$ is one such four-cycle; the other axes and inverse rotations give all six. Hence the complete orientation-reversing tetrahedral symmetry list is six mirror reflections and six quarter-turn rotoreflections.

A discrete subgroup of the Euclidean isometry group has the identity isolated in the group topology. Equivalently, no sequence of distinct group elements converges to the identity; translating neighbourhoods then isolates every element. Such a subgroup is closed, and its intersection with any compact subset is finite: otherwise two elements of a convergent sequence would have quotient tending to the identity.

Write each Euclidean isometry uniquely as $x\mapsto Qx+b$, $Q\in O(2)$. Its linear part gives the group homomorphism $G\to O(2)$, since the linear part of a composite is the product of the linear parts. Its kernel is exactly the translation subgroup $G_T$. The translation vectors form a discrete additive subgroup $L\subset {\mathbb{R}}^2$.

If $L\ne \{0\}$, choose a shortest nonzero vector $a$, possible by discreteness and finiteness in compact balls. On its line, Euclidean division of a scalar coefficient by one proves $L\cap \mathbb{R}a=\mathbb{Z}a$: a nonzero remainder would be shorter than $a$. If all vectors are collinear, this finishes the proof. Otherwise orient the perpendicular coordinate so some vector has positive height. Reduce every vector's component along $a$ modulo $a$ into a bounded interval. Among vectors with positive height at most that of a chosen vector, these reduced representatives lie in a compact rectangle and are finite. Choose $b$ with the smallest such positive height. Subtracting an integer multiple of $b$ from any vector of $L$ reduces its height into $[0,b_{\perp })$; minimality makes the remainder's height zero, so it is an integer multiple of $a$. ThusThis derives the rank-at-most-two translation lattice rather than assuming every discrete subgroup of the plane is a full lattice.

Two examples with $G_T\cong \mathbb{Z}$ are the group generated by $T(x,y)=(x+1,y)$ alone, and the infinite dihedral group generated by $T$ together with $R(x,y)=(-x,y)$. The latter has $RTR=T^{-1}$; its elements are translations $T^k$ and reflections $T^{k}R$, whose mirror lines are $x=k/2$. The first group preserves the orientation of every translated motif; the second also reverses it across these mirrors.

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.

A frieze group is a discrete subgroup of Euclidean isometries whose translation subgroup is generated by one nonzero vector and has finite index. A plane crystallographic group is a discrete subgroup of Euclidean isometries with a rank-two translation lattice of finite index; equivalently its action has a compact fundamental region.

Here are three frieze groups, with explicit generators. The translation $T(x,y)=(x+1,y)$ alone gives $\mathbb{Z}$. Adding the vertical reflection in mathematics $S(x,y)=(-x,y)$ gives $\mathbb{Z}⋊C_2$, the infinite dihedral group, since $STS=T^{-1}$. Instead adding the horizontal reflection in mathematics $H(x,y)=(x,-y)$ gives $\mathbb{Z}\times C_2$, since $HT=TH$. Their translation parts are exactly $⟨T⟩$ and their point parts have respectively one, two and two elements.

For plane crystallographic groups, let $T_1(x,y)=(x+1,y)$ and $T_2(x,y)=(x,y+1)$. These generate ${\mathbb{Z}}^2$. Adding the half-turn $R_2(x,y)=(-x,-y)$ gives ${\mathbb{Z}}^2⋊C_2$, with inversion on both translation generators. Adding the quarter-turn $R_4(x,y)=(-y,x)$ gives ${\mathbb{Z}}^2⋊C_4$, with $R_4{T}_1{R}_4^{-1}=T_2$ and $R_4{T}_2{R}_4^{-1}=T_1^{-1}$. These are the translation, half-turn and square-rotation wallpaper groups. Integer translation vectors and a finite point part make each group a discrete subgroup; the unit square meets every orbit, establishing a cocompact group action.

For the final claim, suppose a Euclidean isometry $R$ has order five. An orientation-reversing planar isometry has finite order only when it is a reflection in mathematics, so $R$ is a rotation through $\theta =2\pi j/5$, $j=1,2,3,4$, about some centre. Write $A$ for its linear rotation part. If $t_v\in G$ is a nonzero translation, conjugation gives $R{t}_v{R}^{-1}=t_{Av}$. Thus $G$ contains translations by $v$ and by $(A+A^{-1})v=cv$, where $c=2\cos \theta$.

For a primitive fifth root $\zeta$, dividing $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4=0$ by ${\zeta }^2$ gives $c^2+c-1=0$. Hence $c$ is irrational. The pigeonhole principle, applied to $N+1$ fractional parts of $0,c,\dots ,Nc$ in $N$ equal intervals, gives integers $m,n$ with $0<|nc-m|\le 1/N$. Consequently $G$ contains nonidentity translations by $(nc-m)v$ tending to the identity. This contradicts the discrete subgroup hypothesis. There is no element of order five. This argument needs only the given nonzero translation, not a rank-two lattice.

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 $a\ne 0$.

Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector $z$ has squared norm ${\sum }_i{a}_i{z}_i^2$. These four numbers are linearly independent over $\mathbb{Q}$: the four sign changes of $\sqrt{2},\sqrt{3}$ in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.

LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives $S^T=-S$, $S^2=-3I$, andIn particular both $L_{\pm }$ are full-rank Euclidean lattices. Denote their four generator columns by $v_i^{\pm }$. Modulo $3$ the two matrices have the same columns, and the nonzero words of the given ternary tetracode $T$ are exactlyThese are the reductions of $\pm v_i^{\pm }$, respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo $3$ maps either Euclidean lattice onto precisely $T$.

To construct a length-preserving correspondence, find the kernels $M_{\pm }=L_{\pm }\cap 3{\mathbb{Z}}^4$ of this reduction. The inverse matrix is $A_{\pm }^{-1}=A_{\pm }^T/12$. Hence $3w\in L_{\pm }$ exactly when $A_{\pm }^{T}w\equiv 0(\mathrm{mod}4)$. These congruences are equivalent toFor clarity, the first $+$ congruence is $-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; subtracting the others gives evenness of each required pair sum. For the $-$ matrix the first congruence is $2w_1-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.

Let $R_i$ be the orthogonal reflection changing only coordinate $i$. If the coordinates of $w$ are even, this changes their sum by a multiple of $4$; if all are odd, it changes the sum by $2$ modulo $4$. ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets $\pm v_i^{\pm }+M_{\pm }$, since $T$ has nine words. Define a map $\Phi :L_{+}\to L_{-}$ by using $R_1$ on $M_{+}$ and $R_i$ on each of $v_i^{+}+M_{+}$ and $-v_i^{+}+M_{+}$. Each piece maps bijectively onto the corresponding $-$ coset. Thus $\Phi$ is a bijection satisfyingIt also commutes with $z\mapsto -z$. The closed geodesics of a flat torus have lengths $\Vert z\Vert$ for nonzero translation vectors $z$ in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at $\ell$ is the sum of primitive multiplicities at $\ell /k$, and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.

It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry $L_{+}$ onto $L_{-}$. In integer coordinates write this part as $B$. For every $z\in L_{+}$, both $z$ and $Bz$ have integer coordinates and have equal squared norms. Rational independence of the $a_i$ impliesApply this to $z=A_{+}n$ for all $n\in {\mathbb{Z}}^4$. Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since $A_{+}$ is invertible, $B$ is consequently a diagonal sign change $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_4)$.

Modulo $3$, such a $D$ must preserve $T$. Each nonzero word of $T$ has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4$. Thus only $D=I$ and $D=-I$ are possible. Neither changes $L_{+}$.

Finally $v_1^{+}\notin L_{-}$: it has the same ternary word as $v_1^{-}$, but their difference is $6e_1=3(2,0,0,0)$, which fails the congruence defining $M_{-}$. Hence $L_{+}\ne L_{-}$ and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.

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 $y=0$.

Under the Euclidean isometry $\tilde{\mathbf{r}}=\mathbf{a}+R\mathbf{r}$, where $R^{T}R=I$,ThereforeSince the integration parameter is unchanged,Both the induced metric and arc length are invariant.

The vertex set of a triangle with three distinct side lengths has trivial setwise stabilizer subgroup in the plane isometry group. Each vertex is characterized by the unordered pair of lengths of its incident edges, so a setwise-preserving Euclidean isometry fixes every vertex. An isometry fixing three noncollinear plane points is the identity: squared Euclidean distances to them determine every other point, since subtracting two of the Euclidean distance equations gives two independent linear equations. For example, $(0,0),(1,0),(0,2)$ have side lengths $1,2,\sqrt{5}$.

Let ${\Lambda }_1,{\Lambda }_2\subset {\mathbb{R}}^d$ be full-rank Euclidean lattices, and choose $\epsilon$ smaller than every nonzero vector length in either. The shortest vectors of ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$ are exactly the new coordinate vectors and their negatives. Any isometry between the associated flat tori lifts to an affine Euclidean isometry, whose orthogonal part maps the lattices. It must preserve the span of the shortest vectors, hence also its orthogonal complement, and therefore maps ${\Lambda }_1$ to ${\Lambda }_2$. Consequently a nonisometric pair remains nonisometric after adding sufficiently short equal circle factors. The spectrum of a flat torus shows that adding the same factors preserves isospectral manifolds.

For a group of Euclidean isometries, every element preserves the displayed union of images of a set, because left multiplication permutes its group indices. Its full setwise isometry group can be larger even when $T$ itself has no nonidentity symmetry. Take $G$ to be the integer translations of the plane and $T=\{(0,0),(1,0),(0,2)\}$. By rigidity of a scalene triangle, $T$ has trivial stabilizer subgroup. Yet $S={\mathbb{Z}}^2$, whose quarter-turn about the origin is a symmetry outside the translation group.

The translation subgroup of a group of Euclidean isometries consists of its translations. It is a normal subgroup, because conjugation by $x\mapsto Ax+b$ takes translation by $v$ to translation by $Av$. Its vectors form an additive subgroup of Euclidean space; for a plane crystallographic group they form a rank-two translation lattice.