Inverse Systems Profinite Groups and Completion - Examples

A compatible tuple is determined uniquely by its $i_0$-coordinate, because for every $j\ge i_0$ the coordinate $g_j$ must be the unique preimage of $g_{i_0}$ under the isomorphism ${\phi }_{i_{0}j}$.

Conversely, let $g\in G_{i_0}$. For each index $i$, choose some $j\ge i,i_0$, possible because the index set is directed, and define $g_i:={\phi }_{ij}({\phi }_{i_{0}j}^{-1}(g))\in G_i\mathrm{.}$ This does not depend on the choice of $j$, because any larger common upper bound gives the same value after applying compatibility of the transition maps. The tuple $(g_i)$ is compatible and has $i_0$-coordinate $g$. Therefore the projection to the $i_0$-coordinate is surjective as well as injective by step 1.1.

The map sending a compatible tuple to its $i_0$-coordinate is therefore a bijective homomorphism, and [L1] identifies it with the inverse-limit comparison map. Hence the inverse limit is isomorphic to $G_{i_0}$.

The finite group $G$ is residually finite because the identity subgroup has finite index. Hence [L2] makes the canonical map ${\iota }_G:G\to \widehat{G}$ injective.

Endow $G$ with the discrete topology, so $G$ itself is profinite. The identity map $G\to G$ is continuous, and [L1] gives a unique continuous homomorphism $\widehat{G}\to G$ left-inverse to ${\iota }_G$. Since ${\iota }_G[G]$ is dense by [L2] and finite subsets of a Hausdorff space are closed, the image of ${\iota }_G$ must be all of $\widehat{G}$. Therefore ${\iota }_G$ is an isomorphism.

Every subgroup of $\mathbb{Z}$ has the form $n\mathbb{Z}$, and it has finite index exactly when $n\ge 1$. Because $\mathbb{Z}$ is abelian, these are all normal.

The quotient by $n\mathbb{Z}$ is the cyclic group $\mathbb{Z}\mathrm{/}n\mathbb{Z}$, and if $m\mid n$ then the map $\mathbb{Z}\mathrm{/}n\mathbb{Z}\to \mathbb{Z}\mathrm{/}m\mathbb{Z}$ is reduction modulo $m$. So [L1] identifies the profinite completion with the displayed inverse limit.

The word $ab$ is nontrivial in the free group $F(a,b)$.

By [L1], there is a finite-index normal subgroup omitting $ab$, equivalently a finite quotient in which the image of $ab$ is nontrivial. This is exactly the residual-finiteness witness promised in the example.

By [L1], $\iota [\mathbb{Z}]$ is dense in $\widehat{\mathbb{Z}}$.

The space $\widehat{\mathbb{Z}}$ is compact and Hausdorff by [L2] together with the general inverse-limit theorem, so any closed dense subset would have to be the whole space. But $\iota [\mathbb{Z}]$ is countable, whereas $\widehat{\mathbb{Z}}$ contains the uncountable subset ${\prod }_p\{0,1\}$ indexed by the primes inside its product model. Therefore $\iota [\mathbb{Z}]$ is a proper dense subset and cannot be closed.

So $\mathbb{Z}$ sits densely but not closedly inside its profinite completion.

For $(m,n)=(2,3)$, the Meskin criterion in the Given clause fails. Therefore $\mathrm{BS}(2,3)$ is not residually finite.

Apply [L1] to $\mathrm{BS}(2,3)$. Since the group is not residually finite, its canonical map into the profinite completion is not injective.

As in FALSE: isomorphic profinite completions force the original groups to be isomorphic, every finite quotient of $H$ factors through the projection onto $\mathbb{Z}$. Hence $G$ and $H$ have the same finite quotients, namely the finite cyclic groups.

Therefore [L1] gives isomorphic profinite completions, both equal to $\widehat{\mathbb{Z}}$. But $G$ and $H$ are not isomorphic because $H$ contains the divisible subgroup $\mathbb{Q}$ and $G$ does not.

This is a concrete counterexample to the statement.