Let G be a compact group, suppose $G=AB$ where $A$,$B$ are profinite subgroups of $G$. Is it true that G is profinite group?



Set product of profinite subgroups of a compact group is profinite
Yes. Let $f$ be a finite-dimensional continuous unitary representation of $G$. Then $f(G)=f(A)f(B)$. Since $A$, $B$ are profinite and $f(G)$ is a compact Lie group, $f(A)$ and $f(B)$ are both finite. Hence $f(G)$ is finite; in particular the unit component $G^0$ is contained in the kernel of $f$. Since this holds for every $f$, it follows from the Peter-Weyl Theorem that $G^0=\{1\}$, that is, $G$ is profinite. 


More generally, a compact group $G$ generated by countably many profinite subgroups $(G\_n)$ is profinite. Indeed, if $f$ is above, then $f(G\_n)$ is finite, so they generate a countable group. Since $f(G)$ is a countable compact Lie group, it has to be finite, and we conclude again using the Peter-Weyl theorem


(On the other hand, the compact connected group $\mathrm{SO}(3)$ is *topologically* generated by finitely many finite subgroups, say by the group of motions of the icosahedron and a nontrivial conjugate.) 


