Random thoughts on isometries on lines

Let \(L\) be a Euclidean line. This means \(L\) is an affine space over \(V\), where \(V\) is a vector space of dimension one with a euclidean norm, and \(L\) has the metric inherited from \(V\).

Now assume \(A\) is an isometry on \(L\). What could \(A\) be?

There are two cases:

• \(A\) has a fixed point.
• \(A\) does not have a fixed point.

Let's consider the case where \(A\) has a fixed point \(P\) first. Consider the mapping \[ B\colon V \ni v \mapsto (A(P + v) - P) \in V \] We have, since \(A\) is an isometry and \(A(P) = P \), \[ |B(v)| = |(A(P+v) - A(P)| = | (P + v) - P| = |v|. \] So \(B\) is an isometry.

There are only two such things! Either \(B\) is the identity, or \(B\) is \(v \mapsto -v\). In the first case, \(A\) is the identity on \(L\). In the second case, \(A\) is reflection in the point \(P\) on \(L\).

Now consider the case where \(A\) has no fixed point. Take any point \(P\) on \(L\), and let \(S\) be reflection in the point midway between \(P\) and \(A(P)\). The map \(S\circ A\) clearly has \(P\) as a fixed point. Call it \(A'\). According to the reasoning above, \(A'\) is either the identity or a reflection in some point. But \(A'\) cannot be the identity, because then \(A\) would be \(S\), and \(P\) would be a fixed point, but \(A\) had no fixed points. So \(A'\) must be a reflection in some other point. Call it \(S'\). We have \[ \begin{align} S\circ A = S' \\ A = S\circ S' \end{align} \] So \(A\) is the composition of two reflections. Some simple considerations will show that \(A\) is a translation on \(L\).

In conclusion, the kinds of isometries on a euclidean line are:

• The identity
• Reflections
• Compositions of two reflections (these are translations)

Definition, and introductory examples

So, we make a definition. A group is a set \(G\) and an operation \(\cdot\) on \(G\) such that

• \((a \cdot b ) \cdot c = a \cdot (b \cdot c)\) for all \(a,b,c \in G\)
• There exists an element \(e\) in \(G\) such that \(a \cdot e = e \cdot a = a\) for all \(a \in G\)
• For every \(a\) in \(G\) there exists an element \(a^{-1} \in G\) such that \(a \cdot a^{-1} = a^{-1} \cdot a = e\)

The element \(e\) is called the neutral element of the group. \(a^{-1}\) is called the inverse of \(a\).

We can find some obvious examples of groups:

• The set of all integers, with the operation of addition. The neutral element is 0, and the inverse of an integer \(n\) is \(-n\).
• The set of all real numbers, with the same operation
• The set of all non-zero rational numbers with the operation of multiplication. The neutral element is 0, and the inverse of a rational number \(r\) is \(\frac{1}{r}\).

We note that all of these examples have the set \(G\) being infinite. However, there are many finite groups.

Let \(n\) be a positive integer. We say that two integers \(x\) and \(y\) are congruent modulo \(n\) if \(n\) divides their difference, that is, if there exists an integer \(k\) such that \( x-y = kn\). For example, 3 and 13 are congruent modulo 5, since \(13 - 3 = 2 \cdot 5\). It is obvious that being congruent modulo a fixed positive integer is an equivalence relation on the set of integers. We can define addition of two equivalence classes by taking representatives, and it is clear that this will constitute a group with \(n\) elements. This group is called the cyclic group of order \(n\).

This shows us that for each positive integer n there exists at least one group of size n. For now on we will refer to the size of a group as the order of the group. However, there are other finite groups.

Let n be a positive integer. Consider the set of permutations of n elements, that is, invertible functions from the set \(\{1,2,\ldots,n\}\) to itself. This set, with function composition as an operation, is a group. It has order \(n!\), and we will realize soon that it is not "the same" as the cyclic group of order \(n!\).

Introduction

I decided that I want to see what I remember of the theory of groups, and particularly their representations. And also see if it is bearable to type mathematics on blogger. So here we go!