FALSE: isomorphic profinite completions force the original groups to be isomorphic

Every homomorphism from the divisible group $\mathbb{Q}$ to a finite group is trivial: if $F$ is finite of order $m$ and $\varphi :\mathbb{Q}\to F$, then for any $q\in \mathbb{Q}$ one has $q=mr$ for some $r\in \mathbb{Q}$, so $\varphi (q)=m\varphi (r)=0$. Therefore every finite quotient of $H=\mathbb{Z}\oplus \mathbb{Q}$ factors through the projection onto $\mathbb{Z}$.

The finite quotients of both $G$ and $H$ are therefore exactly the finite cyclic groups $\mathbb{Z}\mathrm{/}n\mathbb{Z}$, with the same transition maps. By [L1], both profinite completions are the inverse limit of that same system, namely $\widehat{\mathbb{Z}}$. But $G$ and $H$ are not isomorphic because $H$ contains a nontrivial divisible subgroup and $G$ does not.

Thus nonisomorphic groups can have isomorphic profinite completions. The statement is false.