Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 9

A group homomorphism $f:G\to G^{\prime }$ satisfies $f(gh)=f(g)f(h)$ for all $g,h\in G$. It consequently sends the identity element to the identity and inverses to inverses. Its group homomorphism kernel isA normal subgroup $H$ of $G$ is a subgroup invariant under conjugation:Equivalently, $gH=Hg$ for every $g$, so left and right cosets coincide.

Write $M(a,b)=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$. Direct matrix multiplication givesSince $a,c\ne 0$, the product is again in $G$. Associativity is inherited from matrix multiplication, the identity element is $M(1,0)$, andlies in $G$. These verify all group axioms. This is the Affine group of the complex line, acting by $z\mapsto az+b$.

For the diagonal subset, $M(a,0)M(c,0)=M(ac,0)$, it contains $M(1,0)$, and $M(a,0{)}^{-1}=M(a^{-1},0)$. Thus $G^{\prime }$ is a subgroup. Finally,so $f:G\to G^{\prime }$ is a surjective group homomorphism.

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 $(\mathbb{C},+)$.