= Solution
A <group homomorphism> is a multiplication-preserving map $\phi:G_1\to G_2$, so $\phi(gh)=\phi(g)\phi(h)$. 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 $\det(AB)=\det A\det B\ne0$. Associativity is inherited from matrix multiplication, and every such matrix has an inverse, with $\det A^{-1}=1/\det A\ne0$. Hence the given set $\operatorname{GL}(n,\mathbb R)$ is a group.
For the <special orthogonal group>, the identity is included and
$$
(AB)(AB)^T=A(BB^T)A^T=I,\qquad\det(AB)=1
$$
when both factors are special orthogonal. Moreover $A^{-1}=A^T$ is again orthogonal and has determinant one. Thus $\operatorname{SO}(n)$ 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 $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ of the <special orthogonal group in two dimensions>. Orthogonality and determinant one imply
$$
\begin{pmatrix}a&c\\b&d\end{pmatrix}=A^T=A^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.
$$
Hence $d=a$, $b=-c$, 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 $a=\cos\theta$, $c=\sin\theta$. Therefore
$$
\boxed{A=R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}.}
$$
Define the map from the <circle group> by
$$
\boxed{\phi:U(1)\longrightarrow\operatorname{SO}(2),\qquad a+ic\longmapsto\begin{pmatrix}a&-c\\c&a\end{pmatrix}.}
$$
If $z=a+ic$ and $w=p+iq$, matrix multiplication gives $\phi(z)\phi(w)=\left(\begin{smallmatrix}ap-cq&-(aq+cp)\\aq+cp&ap-cq\end{smallmatrix}\right)=\phi(zw)$. Thus it is a homomorphism. The entries determine $a,c$ uniquely, proving injectivity, and the parametrization proves surjectivity. This establishes
$$
\boxed{\operatorname{SO}(2)\cong U(1).}
$$
The 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.
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