 Solution

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.