 Solution

The kernel of a group homomorphism found in part (b) is the translation subgroupIt is a subgroup because $T_c{T}_d=T_{c+d}$, $T_0=I$ and $T_c^{-1}=T_{-c}$. It is nontrivial since $T_1\ne I$, and proper since, for example, $M(2,0)\notin N$.

Every kernel is a normal subgroup: if $n\in \ker f$, then $f(gn{g}^{-1})=f(g)ef(g{)}^{-1}=e$. Applying the same argument with $g^{-1}$ gives equality of the conjugate subgroup and the kernel. Here one can also check invariance under conjugation directly:Multiplication by nonzero $a$ permutes all $c\in \mathbb{C}$, so $M(a,b)NM(a,b{)}^{-1}=N$. Thus $N$ is a nontrivial proper normal subgroup, isomorphic to the additive group $(\mathbb{C},+)$.