= Solution
The <kernel of a group homomorphism> found in part (b) is the translation subgroup
$$
\boxed{N=\ker f=\left\{T_c=\begin{pmatrix}1&c\\0&1\end{pmatrix}:c\in\mathbb C\right\}}.
$$
It is a <subgroup> because $T_cT_d=T_{c+d}$, $T_0=I$ 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 $f(gng^{-1})=f(g)e f(g)^{-1}=e$. 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:
$$
M(a,b)T_cM(a,b)^{-1}=T_{ac}.
$$
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> $(\mathbb C,+)$.
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