# How to see that the inverse limit and group completion are isomorphic?

I'm having a bit of a hard time understanding a proof from Lang on why the completion and the inverse limit $\displaystyle \lim_{\longleftarrow}\ G/H_r$ are isomorphic. Here $\{H_r\}$ 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 $\displaystyle \lim_{\longleftarrow}\ G/H_r$ 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 $(x(1),x(2),\dots)$ defines an element of the inverse limit. Conversely, given an element $(\bar{x_1},\bar{x_2},\dots)$ of the inverse limit, with $\bar{x_n}\in G/H_n$, 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.

I want to see if I'm understanding this correctly (unlikely). So $x(r)$ is the class of all elements $y\in G$ such that $x_ny^{-1}\in H_r$? Then $(x(1),x(2),\dots)$ is an element if the inverse limit as for $n\geq m$, we can take $f^n_m\colon G/H_n\to G/H_m$ to be the canonical homomorphism such that $\bar{x_n}$ in $G/H_n$ maps to $\bar{x_n}=\bar{x_m}$ in $G/H_m$ since $H_m\supset H_n$?

Also, to see that $\{x_n\}$ is Cauchy from $(\bar{x_1},\bar{x_2},\dots)$, I know that $f^n_m(\bar{x_n})=\bar{x_m}$ by definition of the inverse limit, so $\bar{x_n}=\bar{x_m}$ in $G/H_m$, so $\overline{x_nx_m^{-1}}=\bar{e}\in G/H_m$, so $x_nx_m^{-1}\in H_m$. Then given any $H_r$, for $n,m\geq r$, $x_nx_m^{-1}\in H_m\subset H_r$, and $\{x_n\}$ is Cauchy?

But then how exactly are the two maps inverses? I didn't quite understand the maps since Lang seems to map a Cauchy sequences $\{x_n\}$ to an element of the inverse limit, but the completion consists of equivalence classes mod the null sequences, so actual Cauchy sequences aren't even elements of the completion $C/C_0$?

Thank you for any clarifying details. I've been struggling to flesh this out for a while.

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Let $C$ be the group of Cauchy sequences in $G$. He's defining a homomorphism $C \to \varprojlim G/H_r$ which kills off the null sequences: if $\{x_n\}$ is a null sequence then for each $r$ there is some $N$ such that $n \geqq N$ implies $x_n \in H_r$, so $x(r)$ is the identity. Hence we get a map on the quotient. –  Dylan Moreland Aug 19 '11 at 0:27
@Dylan, are you saying the isomorphism follows from the first isomorphism theorem? –  yunone Aug 19 '11 at 0:34
I'm just describing how he's getting a map on the completion by defining a map on the Cauchy sequences. –  Dylan Moreland Aug 19 '11 at 0:45
@Dylan, thanks. So when $\{x_n\}$ is a null sequence, $x(r)=\bar{e}$ since $x_ne^{-1}\in H_r$. Sorry, was my interpretation of $x(r)$ in the main question correct? I had a hard time understanding what $x_n\mod H_r$ means, since I'm mostly used to seeing this applied to integers. –  yunone Aug 19 '11 at 0:56

It seems like you want to check both compositions. Let's try to do one of them.

Let $\{x_n\}$ be a Cauchy sequence in $G$. We send this to the sequence $(x(n))$ in the prescribed manner. To get back to the completion, for each $n$ we pick a representative $y_n$ of $x(n)$ in $G$. We need to show that $\{x_n\}$ and $\{y_n\}$ are equal in the completion, i.e. that $\{x_ny_n^{-1}\}$ is a null sequence. To verify this, fix an $r$. There is an $N \geqq r$ such that $n, m \geqq N$ implies $x_nx_m^{-1} \in H_r$.

Fixing an arbitrary $n \geqq N$, there is an $m \geqq N$ such that $x_mH_n = x(n)$. So $$f^n_r(x_mH_n) = x_mH_r = f^n_r(x(n)).$$ From the Cauchy condition, we have $x_mH_r = x_nH_r$. And $y_nH_n = x(n)$ implies $$f^n_r(y_nH_n) = y_nH_r = f^n_r(x(n)).$$ Thus $x_nH_r = y_nH_r$ for $n \geqq N$, so $\{x_ny_n^{-1}\}$ is indeed a null sequence.

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Thanks Dylan, I think I follow this argument. Also, did you mean to say $x_m\equiv x(n)\mod H_r$, in the second to last line? –  yunone Aug 19 '11 at 2:17
Ah, sorry, I didn't mean to second guess you. I was just confused on how an element $x_n$ of $G$ can be equivalent to a whole congruence class $x(n)$ in $G/H_r$. I just thought I would have $x_m\equiv x(n)$ since $x_nx_m^{-1}\in H_r$, which I assumed was how $x(r)$ was characterized. I haven't yet proved how things are well-defined; I was mostly hoping to see how the two maps were inverses, and hence bijective. I'll see if I can see this more clearly in the morning. –  yunone Aug 19 '11 at 6:37
@yunone It's quite alright. I just meant that $x(n)$ is $x_m$ for $m$ very large, so the Cauchy condition is going to make the images of these in $G/H_r$ equal. –  Dylan Moreland Aug 19 '11 at 15:38
I think I see the other direction. So I want to map from the inverse limit to the completion and back. If $(\bar{x_1},\bar{x_2},\dots)$ is in the inverse limit, then it maps to a Cauchy sequence $\{x_n\}$. Mapping back gives $(x(1),x(2),\dots)$. But $x(n)=\bar{x_n}$ since they are both the class of $x_n\mod H_r$. So $(\bar{x_1},\bar{x_2},\dots)=(x(1),x(2),\dots)$ which shows the composition on the inverse limit is the identity? If you have time to expand like you mentioned above, that'd be great. Thanks for your help. –  yunone Aug 19 '11 at 15:57
Oh, I didn't see your most recent comment while writing my own above. I understand your answer better now, but why do we want that $\{x_ny_n^{-1}\}$ is a null sequence? Don't we just want that the composition maps a sequence back to itself to see it is the identity? –  yunone Aug 19 '11 at 16:01