Let $k$ be a field. For each $a \in k^\times$ and each $b \in k$, let $g_{a, b}: k \to k$ be an affine-linear map given by $g_{a, b}(x) = a \cdot x + b$. The transformations $\{g_{a, b}, \text{ }a \in k^\times, b \in k\}$ form a group $G(k)$ with respect to the composition operation.
Question. What is the classification of all (complex) finite dimensional continuous irreducible representations of the group $G(\mathbb{R})$ up to isomorphism?
Assume that $V$ is a finite dimensional irreducible complex representation of the group $G=\{g_{a,b}\}$. Let $H=\{g_{1,b}\}$. Then the subgroup $H$ is commutative. As a result (for this we do not even need continuity), the subgroup $H$ fixes a flag inside $V$. Let $W\subseteq V$ be the subspaces spanned by all vectors $v\in V$ such that $H\cdot v$ is one dimensional. Since $H$ fixes a flag, $W$ is a non-zero subspace. Since $H$ is normal in $G$, $W$ is stable under the action of $G$. Since $V$ is irreducible, it follows that $W=V$. So we can assume that $H$ acts on the space $V$ diagonally. The elements $g_{1,1}$ and $g_{1,n}=g_{1,1}^n$ are conjugate in $G$ (for every $n\in\mathbb{N})$. This means that if $g_{1,1}$ acts as $diag(\lambda_1,\ldots \lambda_d)$, then $g_{1,n}$ acts as $diag(\lambda_1^n,\ldots \lambda_d^n)$. This means that $\{\lambda_1,\ldots \lambda_d\} = \{\lambda_1^n,\ldots,\lambda_d^n\}.$ This is possible only if all the $\lambda_i$ are equal to 1. But this means that the subgroup $H$ acts trivially on $V$.
We are left with an action of the quotient $G/H$. This group is naturally isomorphic with $\mathbb{R}^{\times}$. Since this group is commutative, we conclude as before that every irreducible representation is one dimensional. The $log$ function gives us an isomorphism $\mathbb{R}^{\times}\to\mathbb{R}\times\{\pm 1\}$. So we just need to determine the representations of the group on the right hand side. These will be given by a pair of a representation of $\mathbb{R}$ and a sign.
The one dimensional continuous representations of $\mathbb{R}$ can be classified by first considering the subgroups $$\mathbb{Z}\subseteq \frac{1}{2}\mathbb{Z}\subseteq\ldots\subseteq\frac{1}{2^n}\mathbb{Z}$$ and their union $\mathbb{Z}_{2}$. A representation of $\mathbb{Z}$ will just be given by a complex number $a\in\mathbb{C}^{\times}$. An extension of this representation to $\frac{1}{2}\mathbb{Z}$ will be given by choosing a square root of $a$. An extension to $\frac{1}{2^3}\mathbb{Z}$ will be given by choosing a square root of this square root and so forth. By continuity reasons we can then show that there is a complex number $l$ such that for every $t\in\mathbb{Z}_{2}$ the representation is given by $t\mapsto e^{lt}$. By continuity, it follows that for every $r\in \mathbb{R}$ the representation is given by $r\mapsto e^{lr}$. This already determines all the irreducible representations: Their are one dimensional, and there is a natural correspondence between the irreducible representations and $\mathbb{C}\times\{\pm 1\}$.
