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For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

A group homomorphism is a multiplication-preserving map $\varphi :G_1\to G_2$, so $\varphi (gh)=\varphi (g)\varphi (h)$. A group isomorphism is a bijective homomorphism; its inverse is automatically multiplication-preserving as well.

For the real general linear group, the identity matrix has nonzero determinant. If $A,B$ have nonzero determinants, multiplicativity of the determinant gives $\det (AB)=\det A\det B\ne 0$. Associativity is inherited from matrix multiplication, and every such matrix has an inverse, with $\det A^{-1}=1/\det A\ne 0$. Hence the given set $\mathrm{GL}(n,\mathbb{R})$ is a group.

For the special orthogonal group, the identity is included andwhen both factors are special orthogonal. Moreover $A^{-1}=A^T$ is again orthogonal and has determinant one. Thus $\mathrm{SO}(n)$ is a subgroup. The orientation-preserving general linear group is also a subgroup: the identity has positive determinant, products preserve positivity, and inversion replaces the determinant by its positive reciprocal.

Now take an element $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ of the special orthogonal group in two dimensions. Orthogonality and determinant one implyHence $d=a$, $b=-c$, and the unit length of the first column gives $a^2+c^2=1$. There is an angle $\theta$, unique modulo $2\pi$, such that $a=\cos \theta$, $c=\sin \theta$. ThereforeDefine the map from the circle group byIf $z=a+ic$ and $w=p+iq$, matrix multiplication gives $\varphi (z)\varphi (w)=\left(\begin{array}{cc}ap-cq& -(aq+cp)\\ aq+cp& ap-cq\end{array}\right)=\varphi (zw)$. Thus it is a homomorphism. The entries determine $a,c$ uniquely, proving injectivity, and the parametrization proves surjectivity. This establishesThe use of the unit complex number, rather than an unrestricted real angle, removes the $2\pi$ ambiguity: angles add modulo $2\pi$, exactly as unit complex numbers multiply.

A parity symmetry in quantum field theory reverses spatial orientation, so its spatial matrix must lie in the determinant-$-1$ component of the orthogonal group. In two spatial dimensions, reversing both coordinates gives $-I_2$, which has determinant $+1$ and is a rotation through $\pi$ in the special orthogonal group in two dimensions. It cannot test a discrete orientation-reversing symmetry independent of ordinary rotations. Reflecting one coordinate, for example $\mathrm{diag}(-1,1)$, has determinant $-1$ and supplies the required parity symmetry in quantum field theory.

In three spatial dimensions, $-I_3$ has determinant $-1$, so simultaneous inversion of all three coordinates is the standard choice. A single-coordinate reflection also reverses orientation there, and differs from $-I_3$ by a proper rotation through $\pi$. Thus, in a rotationally invariant theory, it tests the same discrete symmetry. The word is in the source specifies conventional representatives: the essential distinction is orientation reversal, rather than a unique choice of axis. Both transformations preserve the ordinary spatial integration measure through the absolute value of their Jacobian determinants.