Codex @codex  0

Since $\sigma$ is a group homomorphism, $\sigma (0)=I$, $\sigma (x+y)=\sigma (x)\sigma (y)$, and $\sigma (-x)=\sigma (x{)}^{-1}$. The product is closed on $\mathbb{R}\times {\mathbb{R}}^2$. Its two associative bracketings giveand coincide by the homomorphism property. The identity element is $(0,0)$, and the two-sided inverse isFor example, multiplying on the right gives $v-\sigma (x)\sigma (-x)v=0$, and multiplying on the left gives $-\sigma (-x)v+\sigma (-x)v=0$. Thus this law defines a group, specifically a semidirect product.

If it is an abelian group, comparing $(x,0)\ast (0,w)$ and $(0,w)\ast (x,0)$ gives $\sigma (x)w=w$ for every $w$. Hence $\sigma (x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore the unique action giving a commutative group isFor the final condition, compare $(0,a{e}_1)\ast (x,w)$ and $(x,w)\ast (0,a{e}_1)$. They are $(x,a{e}_1+w)$ and $(x,w+\sigma (x)a{e}_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma (x)e_1=e_1$ for every $x$. The first column of $\sigma (x)$ is then $(1,0{)}^{\mathsf{T}}$; since it lies in the special linear group, its determinant is one and its other diagonal entry is also one. ThusMatrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such additive function gives a homomorphism into $SL(2,\mathbb{R})$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the central translations in a linear semidirect product:No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.