Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 302 / 1 / d

The matrix is invertible precisely when $a\ne 0$, andThe largest group is thereforethe real affine group $\mathbb{R}⋊{\mathbb{R}}^{\ast }$ under $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$.

As a smooth manifold, $G\cong {\mathbb{R}}^{\ast }\times \mathbb{R}$. It has two connected components, distinguished by the sign of $a$: each of $a>0$ and $a<0$ is connected, and a continuous path in $G$ cannot cross $a=0$.

The defining two-dimensional group representation is reducible, because the line $\mathbb{R}{e}_1$ is an invariant subspace:For $b\ne 0$, the complementary line $\mathbb{R}{e}_2$ is not invariant, so this representation need not split as a direct sum of one-dimensional representations.

The map $(a,b)\mapsto \mathrm{sgn}(a)$ is a surjective group homomorphism with kernel $H_0$, so $H_0$ is a normal subgroup and $G/H_0\cong C_2$. The subgroup $H_1$ is not normal: for a translation $t=(1,c)$,which leaves $H_1$ when $a\ne 1$ and $c\ne 0$. Finally, $(a,b)\mapsto a$ is a surjective homomorphism onto ${\mathbb{R}}^{\ast }$ with kernel $H_2$, so $H_2◃G$ and the quotient group $G/H_2\cong {\mathbb{R}}^{\ast }$.