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The group *I*(**R**) of isometries of the line **R** is an interesting and (for such an apparently easy case) complicated group.

Since the group *O*(1) = {±1} we can divide the elements of the group into two subsets: those with orthogonal part +1 and those with the orthogonal part -1. The former are just *translations* of the form *x* ↦ *a* + *x* while the latter are maps of the form *x* ↦ *a* - *x*.

These latter maps have the effect of "reversing the direction" on the line. To see more clearly what they do, observe that the point -*a*/2 is mapped to itself and so you can think of these maps as *reflection in this fixed point*.

Reflection in the point *b* is the map *x* ↦ 2*b* - *x*.

Observe that composing two of these reflections gives a translation. Calculation should show you that composing reflections in points *b* and *c* gives translation by twice the distace between *b* and *c*. The direction of the translation depends on the order in which you do the reflections and so the group is *non-abelian*.

Notice that translation by any *a* ≠ 0 has infinite order, while all the reflections have order 2.

We can think about this group *I*(**R**) in another way. Since the linear part *L* ∈ *O*(1) = {±1} we can represent the element *T*_{a}∘ *L* by the pair (*a*, ±1).

This gives *I*(**R**) = {(*a*, *b*) ∈ **R**^{2} | *b* = ±1 } and composing the maps gives a multiplication * on these pairs defined by

(

a_{1},b_{1})*(a_{2},b_{2}) = (a_{1}+b_{1}a_{2},b_{1}b_{2}).

It is now no longer obvious that the multiplication is associative. The identity element (which is, of course, the pair representing the map *x* ↦ *x*) is the pair (0,1). The inverse of an element (*a*, 1) is (-*a*, 1) (translation in the opposite direction) while an element (*a*, -1) is its own inverse.

The group of matrices { | *a*, *b* ∈ **R**, *b* = ±1 } is isomorphic to *I*(**R**). See Exercises 3 Question 3.

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