Cauchy sequence in group theory

Recall - Cauchy sequence in analysis

Before we go into group theory, let's recall how Cauchy sequence is defined in analysis.

Real/complex number

A sequence \((x_n)_{n=1}^{\infty}\) of real/complex numbers is called a Cauchy sequence if, for every \(\varepsilon>0\), there is a positive integer \(N\) such that for all \(m,n>N\), we have \[ |x_m-x_n|<\varepsilon. \] That said, the distance between two numbers is always 'too close'. Notice that only distance is involved, the definition of Cauchy sequence in metric space comes up in the natural of things.

Metric space

Given a metric space \((X,d)\), a sequence \((x_n)_{n=1}^{\infty}\) is Cauchy if for every real number \(\varepsilon>0\), there is a positive integer \(N\) such that, for all \(m,n>N\), the distance by \[ d(x_m,x_n)<\varepsilon. \]

Topological vector space

By considering the topology induced by metric, we see that \(x_n\) lies in a neighborhood of \(x_m\) with radius \(\varepsilon\). But a topology can be constructed by neighborhood, hence the Cauchy sequence for topological vector space follows.

For a topological vector space \(X\), pick a local base \(\mathcal{B}\), then \((x_n)_{n=1}^{\infty}\) is a Cauchy sequence if for each member \(U \in \mathcal{B}\), there exists some number \(N\) such that for \(m,n>N\), we have \[ x_m-x_n \in U. \] But in a topological space, it's not working. Consider two topological space by \[ X=(0,1)\quad Y=(1,+\infty) \] with usual topology. We have \(X \simeq Y\) since we have the map by \[ \begin{aligned} h:X &\to Y \\ x &\mapsto \frac{1}{x} \end{aligned} \] as a homeomorphism. Consider the Cauchy sequence \((\frac{1}{n+1})_{n=1}^{\infty}\), we see \((h(\frac{1}{n+1}))_{n=1}^{\infty}=(n+1)_{n=1}^{\infty}\) which is not Cauchy. This counterexample shows that being a Cauchy sequence is not preserved by homeomorphism.

Topological group

Similarly, one can have a Cauchy sequence in a topological group (bu considering subtraction as inverse).

A sequence \((x_n)_{n=1}^{\infty}\) in a topological group \(G\) is a Cauchy sequence if for every open neighborhood \(U\) of the identity \(G\), there exists some number \(N\) such that whenever \(m,n>N\), we have \[ x_nx_m^{-1} \in U \]

Completeness

A metric space \((X,d)\) where every Cauchy sequence converges is complete.

Spaces like \(\mathbb{R}\), \(\mathbb{C}\) are complete with Euclid metric. But consider the sequence in \(\mathbb{Q}\) by \[ a_n=\left(1+\frac{1}{n}\right)^{n} \] we have \(a_n\in\mathbb{Q}\) for all \(n\) but the sequence does not converge in \(\mathbb{Q}\). Indeed in \(\mathbb{R}\) we can naturally write \(a_n \to e\) but \(e \notin \mathbb{Q}\) as we all know.

There are several ways to construct \(\mathbb{R}\) from \(\mathbb{Q}\). One of the most famous methods is Dedekind's cut. However you can find no explicit usage of Cauchy sequence. There is another method by using Cauchy sequence explicitly. We are following that way algebraically.

Completing a group

Suppose we are given a group \(G\) with a sequence of normal subgroups \((H_n)_{n=1}^{\infty}\) with \(H_n \supset H_{n+1}\) for all \(n\), all of which has finite index. We are going to complete this group.

Cauchy sequence

A sequence \((x_n)_{n=1}^{\infty}\) in \(G\) will be called Cauchy sequence if given \(H_k\), there exists some \(N>0\) such that for \(m,n>N\), we have \[ x_nx_m^{-1} \in H_k \] Indeed, this looks very similar to what we see in topological group, but we don't want to grant a topology to the group anyway. This definition does not go to far from the original definition of Cauchy sequence in \(\mathbb{R}\) as well. If you treat \(H_k\) as some 'small' thing, it shows that \(x_m\) and \(x_n\) are close enough (by considering \(x_nx_m^{-1}\) as their difference).

Null sequence

A sequence \((x_n)_{n=1}^{\infty}\) in \(G\) will be called null sequence if given \(k\), there exists some \(N>0\) such that for all \(n>N\), we have \[ x_n\in H_k \] or you may write \(x_ne^{-1} \in H_k\). It can be considered as being arbitrarily close to the identity \(e\).

Cauchy sequence and null sequence as groups

The Cauchy sequences (of \(G\)) form a group under termwise product

Proof. Let \(C\) be the set of Cauchy sequences, we shall show that \(C\) forms a group. For \((x_1,x_2,\cdots),(y_1,y_2,\cdots)\in C\), the product is defined by \[ (x_1,x_2,\cdots)(y_1,y_2,\cdots)=(x_1y_1,x_2y_2,\cdots) \] The associativity follows naturally from the associativity of \(G\). To show that \((x_1y_1,x_2y_2,\cdots)\) is still a Cauchy sequence, notice that for big enough \(m\), \(n\) and some \(k\), we have \[ x_nx_m^{-1}\in H_k \quad y_ny_m^{-1}\in H_k. \] But \((x_ny_n)(x_my_m)^{-1}=x_ny_ny_m^{-1}x_m^{-1}\). To show that this is an element of \(H_k\), notice that \[ x_ny_ny_m^{-1}x_m^{-1}=(x_ny_ny_m^{-1}{\color{red}{x_n^{-1}}})({\color{red}{x_n}}x_m^{-1}) \] Since \(y_ny_m^{-1}\in H_k\), \(H_k\) is normal, we have \(x_ny_ny_mx_n^{-1} \in H_k\). Since \(x_nx_m^{-1} \in H_k\), \((x_ny_n)(x_my_m)^{-1}\) can be viewed as a product of two elements of \(H_k\), therefore is an element of \(H_k\).

Obviously, if we define \(e_C=(e_G,e_G,\cdots)\), where \(e_G\) is the identity of \(G\), \(e_C\) becomes the identity of \(C\), since \[ (x_1,x_2,\cdots)(e_G,e_G,\cdots)=(x_1,x_2,\cdots)=(e_G,e_G,\cdots)(x_1,x_2,\cdots). \] Finally the inverse. We need to show that \[ (x_1,x_2,\cdots)^{-1}=(x_1^{-1},x_2^{-1},\cdots) \] is still an element of \(C\). This is trivial since if we have \[ x_nx_m^{-1} \in H_k \] then \[ (x_n^{-1})(x_m^{-1})^{-1}=x_n^{-1}x_m \in H_k \] as \(H_k\) is a group.


The null sequences (of \(G\)) form a group, further, it's a normal subgroup of \(C\), that is, the group of Cauchy sequences.

Let \(N\) be the set of null sequences of \(G\). Still, the identity is defined by \((e_G,e_G,\cdots)\), and there is no need to duplicate the validation. And the associativity still follows from \(G\). To show that \(N\) is closed under termwise product, namely if \((x_n),(y_n) \in N\), then \((x_ny_n)\in N\), one only need to notice that, for big \(n\), we already have \[ x_n,y_n \in H_k. \] Therefore \(x_ny_n \in H_k\) since \(x_n\) and \(y_n\) are two elements of \(H_k\).

To show that \((x_n^{-1})\), which should be treated as the inverse of \((x_n)\), is still in \(N\), notice that if \(x_n \in H_k\), then \(x_n^{-1} \in H_k\).

Next, we shall show that \(N\) is a subgroup of \(C\), which is equivalent to show that every null sequence is Cauchy. Given \(H_p \supset H_q\), for \((x_n)\in{N}\), there are some big enough \(m\) and \(n\) such that \[ x_n \in H_p \quad x_m \in H_q \] therefore \[ x_nx_m^{-1} \in H_p \] as desired. Finally, pick \((p_n) \in N\) and \((q_n) \in C\), we shall show that \((q_n)(p_n)(q_n)^{-1} \in N\). That is, the sequence \((q_np_nq_n^{-1})\) is a null sequence. Given \(H_k\), we have some big \(n\) such that \[ p_n \in H_k \] therefore \[ q_np_nq_n^{-1} \in H_k \] since \(H_k\) is normal. Our statement is proved.

The factor group

The factor group \(C/N\) is called the completion of \(G\) (with respect to \((H_n)\)).

As we know, the elements of \(C/N\) are cosets. A coset can be considered as an element of \(G\)'s completion. Let's head back to some properties of factor group. Pick \(x,y \in C\), then \(xN=yN\) if and only if \(x^{-1}y \in N\). With that being said, two Cauchy sequences are equivalent if their 'difference' is a null sequence.

Informally, consider the addictive group \(\mathbb{Q}\). There are two Cauchy sequence by \[ (1,1,1,\cdots,1,\cdots) \\ (0.9,0.99,0.999,0.9999,\cdots). \] They are equivalent since \[ (0.1,0.01,0.001,\cdots) \] is a null sequence. That's why people say \(0.99999... = 1\) (in analysis, the difference is convergent to \(0\); but in algebra, we say the two sequences are equivalent). Another example, \(\ln{2}\) can be represented by the equivalent class of \[ (0.6,0.69,0.693,\cdots). \]


Remarks

We made our completion using Cauchy sequences. The completion is filled with some Cauchy sequence and some additions of 'nothing', whence the gap disappears.

Again, the sequence of normal subgroups does not have to be indexed by \(\mathbb{N}\). It can be indexed by any directed partially ordered set, or simply partially ordered set. Removing the restriction of index set gives us a great variety of implementation.

However, can we finished everything about completing \(\mathbb{Q}\) using this? The answer is, no - the multiplication is not verified! To finish this, field theory have to be taken into consideration.