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A glide reflection of the plane composes reflection in a line with a nonzero translation parallel to that line. It reverses orientation and has no fixed point. It cannot be one reflection, because a reflection fixes a line, or a product of two reflections, because such a product preserves orientation. The finite reflection decomposition of a Euclidean isometry therefore shows that it requires exactly three reflections.

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.

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.

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 $y=0$.