Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 3 / 7D / Solution

The first isomorphism theorem for groups states that for a group homomorphism $f:G\to Q$,The map $g\ker f\mapsto f(g)$ is well defined because two representatives differ by an element of the kernel; it is a surjective homomorphism onto the image and is injective because $f(g)=e$ exactly when $g\in \ker f$.

For $n_i\in N$ and $h_i\in H$, normality of $N$ givesand $(nh{)}^{-1}=h^{-1}{n}^{-1}h{h}^{-1}\in NH$. Hence $NH$ is a subgroup. Also $N\cap H⊴H$, since conjugation by $H$ preserves both $N$ and $H$, and $N⊴NH$ because $NH\le G$.

Apply the first isomorphism theorem to $f:H\to NH/N$, $f(h)=hN$. Its kernel is $N\cap H$, and it is surjective because $nhN=hN$. Therefore

If $K,H⊴G$, then $KH$ is a normal subgroup: it is a subgroup by the calculation above, and $g(KH)g^{-1}=(gK{g}^{-1})(gH{g}^{-1})=KH$. Without normality a product need not be a subgroup. In $S_3$, take $H=⟨(12)⟩$ and $K=⟨(23)⟩$. The set $KH$ has four elements, so it cannot be a subgroup of the six-element group by Lagrange's theorem for finite groups.