# Inverse limit and group completion are isomorphic

It's been a while I get stuck on proof from Lang's Algebra book, which states that the completion and the inverse limit $$\varprojlim G/H_{r}$$ are isomorphic where $$\{H_r \}$$ is a sequence of normal subgroups in $$G$$ with $$H_r \supset H_{r+1}$$ for all $$r$$.

Theorem 10.1: The completion and the inverse limit $$\varprojlim G/H_{r}$$ are isomorphic under natural mappings.

Proof: We give the maps. Let $$x=\{x_n\}$$ be a Cauchy sequence. Given $$r$$, for all $$n$$ sufficiently large, by the definition of Cauchy sequence, the class of $$x_n \mod H_r$$ is independent of $$n$$. Let this class be $$x(r)$$. Then the sequence $$(x(1),x(2),…)$$ defines an element of the inverse limit. Conversely, given an element $$(\overline{x}_1,\overline{x}_2,…)$$ in the inverse limit, with $$\overline{x}_n \in G/H_n$$, let $$x_n$$ be a representative in G. Then the sequence $$\{x_n\}$$ is Cauchy. We leave to the reader to verify that the maps we have defined are inverse isomorphisms between the completion and the inverse limit.

Actually, I don't understand this proof at all. Given $$r$$ and $$x=\{x_n\}$$ be a Cauchy sequence, what is exactly $$x(r)$$? By definition of Cauchy sequence, there is exits $$N$$ such that for all $$m,n \geqslant N$$ we have $$x_nx_m^{-1} \in H_r$$, so is $$x(r)$$ the set $$\{ x_i \mid i \geqslant N\}$$?

In the proof, he said that the sequence $$(x(1),x(2),…)$$ defines an element of the inverse limit. Why does $$(x(1),x(2),…)$$ modulo the null sequences give us a single element?

To see that $$\{x_n \}$$ is a Cauchy sequence from $$(\overline{x}_1,\overline{x}_2,…)$$, by definition of limitt inverse, we have $$f^{n}_{m}(\overline{x}_n)=\overline{x}_m$$, i.e $$x_m, x_n$$ are equal in $$G/H_m$$ which means $$x_m x_n^{-1} \in H_m$$. Therefore, given $$r$$, for any $$m,n \geqslant r$$ we have $$x_m x_n^{-1} \in H_m \subset H_r$$ so $$\{x_n \}$$ is indeed a Cauchy sequence. But how does two representative sequence $$\{x_n \}$$ and $$\{x'_n \}$$ of $$(\overline{x}_1,\overline{x}_2,…)$$ give us a single element in completion?

Any helps would be appreciated! Thanks.

If you have a Cauchy sequence $$x = \{x_n\}$$, then for each $$r$$, the sequence $$\{x_n H_r\}$$ becomes eventually constant and so we can set $$x(r) = x_n H_r$$ where $$n$$ is such that $$x_n H_r = x_m H_r$$ for all $$m \geq n$$. This clearly defines an element $$(x(1),x(2),\dots)$$ in $$\varprojlim G/H_{r}$$.
Now, if $$y = \{y_n\}$$ is another Cauchy sequence differing from $$\{x_n\}$$ by a null sequence, then for each $$r$$, you have $$x_n H_r = y_n H_r$$ for large $$n$$, so the elements $$x(r)$$ and $$y(r)$$ in $$G/H_r$$ coincide.
For the other issue, given two representative sequences $$\{x_n\}$$ and $$\{x_n'\}$$ you have $$x_n H_n = x_n' H_n$$ for all $$n$$. It follows that $$x_n^{-1}x_n' \in H_r$$ for all $$n \geq r$$ and thus $$\{x_n\}$$ and $$\{x_n'\}$$ differ by a null sequence.