# An elementary introduction to profinite groups

## Introduction

This blog serves as an introduction to profinite groups without touching anything other than elementary group theory (no ring, field, Galois theory, topological group, etc.), though we may not be able to go for further application.

### Number system

We begin with an easy-to-understand motivation by introducing \(\mathbb{Z}_p\). Consider the binary expansion of an integer \[
a=\sum_{k=0}^{n}a_k2^k
\] where \(a_k=0,1\). For example we may have \[
\begin{aligned}
88&=8+16+64 \\
&=0\cdot2^0+0\cdot2^1+0\cdot2^2+1\cdot2^3+1\cdot2^4+0\cdot2^5+1\cdot2^6 \\
&=(1011000)_2.
\end{aligned}
\] You must be familiar with binary expansion if you write codes. As a topology exercise, show that the set containing all such \(a\) is uncountable. In the octal number system you may also write \[
\begin{aligned}
88&=24+64 \\
&=0\cdot8^0+3\cdot8^1+1\cdot8^2 \\
&=(130)_8.
\end{aligned}
\] This notation is pretty useful in some real life occasions but not here. We are looking for connections between number systems and **prime** numbers (you will see why later)), but number systems with bases like \(8,10,16\) definitely won't work.

### \(p\)-adic integer

Fix a prime number \(p\), a \(p\)-adic integer \(\alpha\) is defined by a sequence of integers \(x_k\) for which we writes \[ \alpha=(x_k)_{k=1}^{\infty} \] satisfying \[ x_{k+1}\equiv x_k\mod{p^k}\text{ for all $k \geq 1$} \] For example, we write \(88\) as a \(2\)-adic number by \[ 88=(88,88,\cdots)=(0,0,0,8,24,24,88,88,\cdots,88,\cdots) \] As you may realize, \(x_k\) can be written by \[ x_k=a_0+a_1p+\cdots+a_{k-1}p^{k-1} \] where \(a_i=0,1,\cdots,p-1\) for \(i \leq k-1\), where \(a_i\) is called the \(p\)-adic digits.

In fact, if we define the addition componentwise, i.e. \[ (x_k)+(y_k)=(x_k+y_k) \] then it's a group. Further, if we define the multiplication componentwise, we get a ring. The group of all \(p\)-adic numbers is denoted by \(\mathbb{Z}_p\). But this blog won't touch anything other than group theory.

As you may wonder, it doesn't seems to work for 'negative' one. For example if we have \[ \alpha=(1,1,\cdots,1,\cdots), \] how do we get \(-\alpha\)? In fact we have \[ -\alpha=(p-1,p^2-1,\cdots,p^n-1,\cdots) \] which suggests the limit value of \(x_k\) associated to \(-\alpha\) as \(k\to\infty\) is \[ (p-1)(1+p+p^2+\cdots). \] It doesn't converge in the usual sense. But if it does, we have \[ \lim_{k\to\infty}x_k=(p-1)/(1-p)=-1. \] But this is valid under such circumstance. We can check this using \(p\)-adic digits. In fact, the \(p\)-adic digits of \(1\) is \[ (1,0,\cdots,0,\cdots) \] if we add \(p-1\) to each component, we get \[ (1,0,\cdots)+(p-1,p-1,\cdots)=(0,0,\cdots,0) \] (there are infinitely many \(p-1\)!).

With all these stuff being said, you can treat \(\alpha=(x_1,x_2,\cdots)\) as a **limit**: \[
\alpha=\lim_{k\to\infty}x_k
\] which makes everything natural. We are not digging into \(\mathbb{Z}_p\) further. But keep two words in mind: limit and group.

### \(\mathbb{Z}_p\) and \(\mathbb{Z}/p^k\mathbb{Z}\)

The definition of \(\mathbb{Z}_p\) by \(x_{k+1} \equiv x_k\mod{p^k}\) might remind you of \(\mathbb{Z}/p^k\mathbb{Z}\). Let's give a review of \(\mathbb{Z}/p^k\mathbb{Z}\).

For integers \(x,y\), we have \[
x \equiv 0 \mod p^k\mathbb{Z}
\] if \(x \in p^k\mathbb{Z}\). Further we have \[
x \equiv y \mod p^k\mathbb{Z}
\] if \((x-y)\in p^k\mathbb{Z}\). We also may write \(x \equiv y \mod p^k\). So there are infinitely many \(x_{k+1} \equiv x_k \mod p^k\mathbb{Z}\), shall we associate infinitely many \(\mathbb{Z}/p^k\mathbb{Z}\)? If it works, we may treat \(\mathbb{Z}_p\) as the 'limit' of \(\mathbb{Z}/p^k\mathbb{Z}\). But we need some proper **operation** to do that.

Let \(G_n=\mathbb{Z}/p^{n+1}\mathbb{Z}\) for each \(n \geq 0\). Let \[
f_n: G_n \to G_{n-1}
\] be the canonical homomorphism. Notice that \(f_n\) is surjective. Now consider a \(p\)-adic integer \[
\alpha=(x_1,x_2,\cdots)
\] we have \[
f_n(x_n)=x_{n-1}
\] Therefore we got a expression of \(\mathbb{Z}_p\) by \[
\mathbb{Z}_p=\{(x_n)_{n=0}^{\infty}\in\prod_{n\geq0}\mathbb{Z}/p^n\mathbb{Z}:f_n(x_n)=x_{n-1}\}
\] We will write \(\mathbb{Z}_p=\varprojlim\mathbb{Z}/p^n\mathbb{Z}\) since it's an example of a inverse limit. It's *inverse* since \(f_n\) goes 'back' by associating each \(x_n\) to \(x_{n-1}\). Since \(f_n\) is **surjective**, we can always raise \(x_{n-1}\) to \(G_{n}\) via \(f_{n}\). We treated one group as a limit of a sequence of groups. We don't want to limit ourself in number theory. In the following section we are offering a much more generalized definition where even numbers are generalized.

## Definition of profinite group

We are going to give a generalized definition for profinite group. Notice that in the example of \(\mathbb{Z}_p\), the sequence is indexed by \(\mathbb{N}\). It's easy to understand but this index set prevents profinite group from being further applied. Of course, the index \(\mathbb{N}\) is not excluded.

### Directed partially ordered set

A set \(I\) is **directed partially ordered** if it's associated with a partial order \(\geq\) such that for any two elements \(i,j \in I,\) there exists a \(k \in I\) such that \(k \geq i\) and \(k \geq j\).

\(\mathbb{Z}\) with the natural inequality is of course directed partially ordered. However we can define another partial order by division. If we define \(n \geq m\) if \(m|n\), then we have \(\operatorname{lcm}(m,n) \geq m,n\).

As another example, consider the family of all subgroups \(\mathcal{F}\) of a group \(G\). The partial order is defined by inclusion. i.e. for \(M,N \in \mathcal{F}\), we have \(M \geq N\) if \(M \supset N\). In this case \(M \cup N \geq M,N\).

### Projective system

A **projective system** is a collection of groups \(G_i\) (\(i \in I\)), together with group homomorphisms \(f^{j}_i: G_j \to G_i\) for \(i,j\in {I}\) with \(j \geq i\) such that

- \(f_{i}^{i}=\operatorname{id}_{G_i}\) for every \(i \in I\).
- \(f_{i}^{j}\circ f_{j}^{k}=f_{i}^{k}\) for \(k \geq j \geq i\).

### Inverse limit and profinite group

Given any such projective system with a directed partially ordered index set, we has the **inverse limit** (or projective limit) defined by \[
\varprojlim_{i}G_i=\left\{(x_i)_{i\in{I}}\in\prod_{i\in{I}}G_i:f_i^j(x_j)=x_i,j\geq i\right\}
\] It's easy to see that \(\mathbb{Z}_p\) can be defined with the same manner, although we have \(I = \mathbb{N}\). It also can be verified that the inverse limit forms a group (also topological group, but we are not discussing that here).

A group is **profinite** if it is a **pro**jective limit of **finite** groups (up to isomorphism).

### Examples

For any \(g \in \mathbb{N}_+\), it would be interesting to consider the following projective limit by \[ \mathbb{Z}_g=\varprojlim_{n}\mathbb{Z}/g^n\mathbb{Z}. \] It can be verified that we have \[ \mathbb{Z}_g \simeq \prod_{p|g}\mathbb{Z}_p. \] That said, base \(8\) number system is 'useless' since it's isomorphic to base \(2\) number system. That's why we focus on prime first. We will give another 'generalization' of \(p\)-adic numbers.

Suppose we have a sequence of normal subgroups \((H_n)\) of \(G\) such that \(H_n \supset H_{n+1}\) for all \(n\). It doesn't matter whether \(G\) is finite. Let \[ f_n:G/H_n \to G/H_{n-1} \] be the canonical homomorphisms. Then the inverse limit follows: \[ \varprojlim_n G/H_n=\left\{(x_i)_{i\in\mathbb{N}}\in\prod_{i\in\mathbb{N}}G/H_i:f_n(x_n)=x_{n-1}\right\}. \] We also have a natural homomorphism \[ g:G \to \varprojlim_{n}G/H_n \] by sending \(x\) to the sequence \((x_i)\), where \(x_n\) is the image of \(x\) in \(G/H_n\). Notice that we don't have to use \(\mathbb{N}\) as the index. This inverse limit can also be indexed by the set containing all \(H_n\).

## Further

You may think this is like, algebraists stole something from analysts and made it up with the magic of algebra. There are many other applications that I want to show you in the future (not beyond elementary group theory). If you learned functional analysis you may know that \(L^p\) space for \(1 \leq p < \infty\) is not a Banach space due to the functions equal to \(0\) a.e.. But \(L^p/N\) can be a Banach space where \(N\) contains all functions equal to \(0\) a.e.. Both \(L^p\) and \(N\) are groups, and we 'completed' \(L^p\) by defining a factor space which still is a group. In fact, in algebra, we also have **Cauchy sequence** and **completion** of a group, which are associated with inverse limit still.

## References / Further readings

- Luis Ribes,
*Introduction to Profinite Groups* - Hendrik Lenstra,
*Profinite Groups* - Serge Lang,
*Algebra Revised Third Edition*