I am trying to read too many books at the moment, and one of them is Keith Devlin’s fascinating, but occasionally infuriating The Language of Mathematics: Making the Invisible Visible.
The book skates through mathematical concepts at a dizzying speed – often pausing only briefly to explain them. It is allowed to do this, it is not a text book. I wish I had read it before because it does at least explain the bare background to some concepts I have come across in the last year while reading other maths books – such as Fermat’s method of infinite descent.
But it is sometimes a little imprecise – at least I think so – in its language. And so it sometimes confuses almost as much as explains – and this is where I come to group theory.
Now, before I read Chapter 5 of the book I knew that groups were like sets, except they were not the same. And I had – while reading Higgs (another one of the too many books right now) come across the (essentially unexplained in that book) concept of the “symmetry group” when discussing sub-atomic particles. Devlin’s book brilliantly and effortless explains what groups are and does so, handily enough, through the question of transformational symmetry.
But this is where the questions begin.
Let us examine the case of a (unmarked) circle. As the book states:
The transformations that leave the circle invariant are rotations round the center (through any angle, in either direction)
Thus there are surely an infinite number of these.
The book then defines three conditions for a group (and later a fourth condition for an abelian group):
G1 For all in , (where is an operation)
G2 There is an element in such that for all
G3 For each element in , there is an element in such that (where is as in G2)
We can see that G1 is a stipulation of associativity, G2 is of the existence of an identity element and G3 of an inverse transformation.
(The condition for an abelian group is that of commutativity, ie., that for all in – but that is not particularly relevant here.)
So back to our circle. The book states that there is precisely one inverse transformation for each element – ie., that in G3 is unique for each . But in our circle, how can this be so? Is not each and every transformation its own inverse as well as the inverse of every other transformation? Because the unmarked circle is, by definition, exactly the same – indeed doesn’t this lead us to conclude that there is only one transformation of the circle – namely the identity transformation (ie., the one where we do nothing) ? At least it feels like a contradiction to me…
Which then takes me on to the precise case of the completely irregular shape. This has a symmetry group with a membership of one, namely the identity transformation, but then the book (and this is, I suspect, a sloppy piece of wording that fails to take account of degenerate cases rather than what looks like a contradiction) states:
Condition G2 asserts the existence of an identity element. Such an element must be unique, for if and have the property expressed by G2 then …
Which is surely precisely what we do have!
Now, I don’t for an instant think I have ripped a hole in the fabric of mathematics – I just think I need this explained to me a bit more clearly: so how many rotational (or reflectional) transformations are in the symmetry group of an unmarked circle and presumably there is no problem with in single member groups? Can anyone help?
Update: Ian Peackcock (@iancpeacock) makes the point to me that, of course, all symmetrical transformations look the same – that is the point. So I guess that knocks down the contradiction between infinite and one – and presumably we can put down the book’s claims about identity being unique to just sloppy wording – after all if a group has a single member, it is of course unique?