Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 3 / 7D (Groups)

The map $H\times K\to HK$, $(h,k)\mapsto hk$, has fibers of size $|H\cap K|$: two pairs have the same product exactly when their quotient is represented by an element of the intersection. Hence

The standard intersection-index inequality gives $[G:H\cap K]\le [G:H][G:K]=ab$. This index is divisible by both coprime numbers $a$ and $b$, so it equals $ab$. The displayed formula then gives $|HK|=|G|$, hence $HK=G$.

The conjugacy class and centralizer areand orbit--stabilizer gives $|{\mathrm{Conj}}_G(x)|=[G:C_G(x)]$. Thus the coprimality hypothesis gives $C_G(x)C_G(y)=G$.

One inclusion in the desired equality follows by conjugating $xy$. Conversely take $g^{-1}xg{h}^{-1}yh$. Write $h{g}^{-1}=ab$ with $a\in C_G(y)$ and $b\in C_G(x)$. The hint then rewrites the product aswhich is conjugate to $xy$. Therefore