Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 8D (Groups)

A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

Take $G=C_2\times C_2$ and $N=C_2\times \{1\}$. Both are nontrivial, and $N$ is normal because $G$ is abelian. The second-coordinate projection identifies the quotient group $G/N$ with $C_2$. Swapping the two coordinates identifies $(G/N)\times N$ with $G$, soThis is a direct product of groups with its first factor as the specified normal subgroup.

Use additive notation and take $G=\mathbb{Z}/4\mathbb{Z}$, $N=\{0,2\}$. Its quotient group is $C_2$. A group homomorphism from $C_2$ sends its generator to an element annihilated by $2$, so the image must be either $0$ or $2$. Both project to the identity coset $N$, and neither can project to the nonidentity generator of $G/N$. Therefore this quotient projection has no homomorphic section. It is a nonsplit group extension, despite both $G$ and $N$ being nontrivial finite groups.

Take $G=S_3$ and its normal subgroup $N=A_3=⟨r⟩$, where $r=(123)$. Normality follows either from the sign of a permutation or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The quotient group has order two, and $t=(12)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.

However, the map $\Phi (q,n)=i(q)n$ from the direct product of groups is not a group homomorphism. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $tr{t}^{-1}=r^{-1}\ne r$. This is a split group extension giving a genuine semidirect product instead of a direct product. The pair $S_3,A_3$ supplies the required split but non-direct example.

For the general bijectivity assertion, let $\pi :G\to G/N$ be the quotient projection and assume $\pi \circ i$ is the identity. Given $g\in G$, put $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Then $\pi (n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q^{\prime })n^{\prime }$, applying $\pi$ gives $q=q^{\prime }$, after which cancellation gives $n=n^{\prime }$. ThusThis section normal form for a split group extension does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.