# A distraction: infinite groups

This is the first in the distraction series. In these posts (put in the Group Theory-distractions category) I will explain something that I think is interesting and compliments the main material, but would be less likely to be seen in an early course on group theory. You will need to know about the cyclic and dihedral groups, which are in the previous two posts.

Revisiting: the infinite cylic group. In the first post we saw $\mathbb{Z}$, the group of integers (a.k.a. whole numbers) with addition. Recall that this notation is not ambiguous since $(\mathbb{Z}, \times)$ is not a group. We will first try and build some geometric intuition for $\mathbb{Z}$, and then develop this for an infinite dihedral group.

So what pictures can we have in mind? The first one we saw was as the group generated by a rotation by an irrational amount (which was a symmetry of the circle) in part 1.

The blue dot is there to keep track of rotations. This is useful to see that the element had infinite order and to see our group as cyclic: the circle can be thought of as the limit of the sequence of regular polygons. But we’re getting off track. Although polygons seem natural objects for the groups we’ve seen so far (the families of Cn and Dn), there is a more natural object for the infinite case in my opinion.

A number line…really? We could also view this with a slight curve so that the two ‘ends’ of this line meet. To help visualise this point, think of it as the anitpodal point to the blue point on the circle. Hopefully you’ve now got the picture I have in my head…a very zoomed in version of the circle where you have labelled the integers and where there is a rotation of infinite order which moves the circle to itself and moves all of the whole numbers up by a certain amount (so that on the number line we would call this ‘move’ a translation). For convenience, it’s worth thinking that we’re moving all of the numbers on the solid black line (the real numbers) by an integer translation. Roughly the picture is:

We will come back to this picture soon, where it will help us visualise direct products. Before continuing: write down all of the ways we have viewed the infinite cyclic group.

Spoilers! The different ways we’ve seen are:

1) The structure of the integers with addition.
2) Those symmetries of the circle generated by an irrational rotation (which does not generate all of the rotational symmetries of a circle, see part 1).
3) The group presentation $\langle r \mid \_ \rangle$.
4) The group of translations by $\mathbb{Z}$ of the real numbers (the number line above).

An infinite dihedral group. So how does this generalise for the family of dihedral groups; is there an infinite dihedral group? We’ve seen an infinite cyclic group. So can we extend this infinite cyclic group to include reflections as we did for Dn? Yes, we can! We could do this either geometrically or algebraically.

D, the geometric way. We have seen many ways to view $\mathbb{Z}$. I will use (4) from above: the group of translations (of an integer amount) of the real line. The ‘symmetry’ that we included to go from Cn to Dn was a reflection. The natural reflection to include for $\mathbb{Z}$ is the reflection of the number line (the real numbers) about 0. This reflection can be realised as swapping $x$ with $-x$ for all $x \in \mathbb{R}$ (which fixes 0).

This reflection can also be thought of as multiplying by minus 1. Thus D consists of the transformations of $\mathbb{Z}$ which are either a reflection about any integer (so that it leaves one point fixed), a reflection about the midpoint of two integers (having no fixed points), or a translation (again having no fixed points). Notice how these fit with the types of elements seen in Dn.

Before continuing: play with these symmetries. Can D be generated by 2 elements? (Think about the picture we have made and what we know about the translations.)

D, the algebraic way. We have our group presentation for Dn. In the same way as going from Cn to $\mathbb{Z}$ by removing the relationship that made the generator have finite order, we may go from Dto D and obtain the group presentation:

$\langle r, s \mid s^2=1, srs=r^{-1} \rangle$.

Question: decide whether the algebraic rule srs=r-1 which was satisfied by Dn holds for the geometric intuition of the infinite dihedral group. Do our two different approaches produce the same group? Answering this is the last aim of this post.

Do the two approaches agree? First of all we need to check whether the geometric way of defining D can be generated by 2 elements. If it cannot, then the two approaches we have taken cannot produce the same group (since the algebraic definition only uses 2 generators). Viewing the geometric transformations algebraically we have

• the translation which sends $x$ to $x+1$ for all $x \in \mathbb{R}$
• the reflection which sends $x$ to $-x$ for all $x \in \mathbb{R}$

which we will denote by r (for rotation) and s respectively. Now,

• r generates all possible translation by an integer amount
• r-ksrk is a reflection with fixed point k for any integer k
• r-ksrk+1 is a reflection which fixes the line between k and k+1 for any integer k

and so the group produced by the geometric approach can be generated by 2 elements.

We can next check that this r and s satisfy the relations we have in the presentation for D. In order for these to actually be the same group, we also need to check that there are no other relationships between our translation r and reflection s. This requires checking that no other ‘words’ formed from the letters r and s produce the ‘do nothing’ symmetry (the identity). First we consider what our current relations provide. We know that no powers of s other than 0 and 1 are needed since s2 is the identity meaning that s2k is also the identity for all integers k and finally that s-1=s1. Using our other relation srs=r-1 to imply that sr=r-1s, we have a way of shuffling all of the s letters to the right (or to the left). This means that any ‘word’ formed from r and s is of the form

rk or rks, where k is an integer.

We wish to conclude that no two of these words produce the same symmetry. Firstly, rk ≠ rjs for any integers j and k: the right hand side permutes two integers whilst the left hand side does not. We could also see that these are different by considering the order of these group elements, noticing that -since they are not of the same order- that they cannot be the same group element. Secondly if rk = rj, then j=k by comparing where they send 0. Similarly if rks = rjs, then j=k either by again comparing where they send 0, or by noting that if rks = rjs, then multiplying on the right by s would imply that rk = rj (and so if j were not equal to k this would contradict our previous statement). This means that there are no other relations -we have listed all of the rules for cancelling- and that our two approaches have produced the ‘same’ group.

I hope you enjoyed this excursion…it’s made me realise that we should see direct products before our final families of finite groups. Hopefully see you there.