Hello again, STEEMathematicians! This is the third addition to your introductory series on group theory - in case you missed them, here's the first and second posts. Last time, we saw how permutation groups reorder a finite number of elements, and we discussed (somewhat concisely) mappings between groups. If you haven't yet had a chance to work through some of the results we have presented with a pen and paper, I suggest you give it a go - it really helps. In today's post, we continue down the rabit hole by thinking about subgroups, how to divide a group, and a few other bits and bobs.
The Standard Model of particle physics is a (gauge) quantum field theory, containing the internal symmetries of the unitary product group SU(3) × SU(2) × U(1) . The pretty picture above is a simulation of the Higgs boson, which is the quantum excitation of the Higgs field.
[ image credit ]
We begin today's post with a definition. It's a rather important one, which we will come to use a lot in the future, so we will try quite hard to understand it's subtleties!
Definition: A subgroup G is a non-empty subset H of G, which under the group operation of G forms a group.
These are just parts of the group which multiply into themselves. A good example for visualisation is the rotations of an equilateral triangle.
An important thing to bear in mind is that any group G contains the "trivial" subgroups G and {I}. No, I'm not being pretentious, that is indeed to how they are referred! All other subgroups are called proper subgroups.
Cast your mind back to the first post, where we talked about Abelian and cyclic groups. It is the case that the set created by {I, X, X2, ...Xm-1} of an element of order m forms an Abelian subgroup. In general; in a cyclic group, this subgroup is the group itself.
Some important details:
Now, lets consider a homomorphism Φ : G → G' (if you're unsure, refer to the second post).
To prove the first of these statements, we will suppose that Z and W are part of H', with Z = X', and W = Y', where X and Y are part of G. We can quite easily demonstrate the closure of H' by observing that
ZW = X'Y' = (XY)',
is part of H'. We can also see that the inverse is present in H':
Z-1 = (X')-1 = (X-1)',
and I' belongs to H'.
Now, in order to prove the second statement, we will suppose that X and Y are part of K. Then, we have closure since
(XY)' = X'Y' = I'I' = I',
and the inverse is present
I' = (XX-1) = X'(X-1) = I'(X-1) = (X-1)',
therefore I is also part of K.
We can gain some understanding of the second statement by considering a homomorphism ℝ → U(1) with Φ : x → eix. We saw this beauty in the last post, so don't be afraid! Its kernel is K = {2πn} with n ∈ ℤ. We see that it forms a subgroup of ℝ under addition! For illustrative purposes, we point out that a real number added to some other real number is also a real number!
Definition: A normal subgroup H of G is defined by the property that for every element X ∈ G and every element Y ∈ G, XYX-1 ∈ H. Or, equivalently, XH = HX.
Normal subgroups are also known as invariant subgroups, or self-conjugate subgroups. The kernel K forms a normal subgroup, since
(XYX-1)' = X'Y'(X-1)' = X'I'(X-1)' = X'(X-1)' = I'.
How on Earth do we go about dividing a group, I hear you ask! That's a very good question because a group is a fairly complicated algebraic object, not just some number! It turns out we have to introduce some more concepts before we can make sense of division (yay more definitions! /s). Now, subgroups are not very well suited to partition the elements of a group, since it is quite possible for elements to be a part of several subgroups. I, for example is part of any subgroup. Equivalence relations to the rescue!
An equivalence relation on a set S is a relationship X~Y between two elements X and Y, part of S, with the properties:
Some examples of equivalence relations include the order of an element, or being part of a specific subgroup. We make the very important note that commutation is not an equivalence relation! We can see this as follows.
A = A · I = I · A and B = B · I = I · B for any elements A, B of a group does not imply that A commutes with B.
An equivalence relation of S divides S into classes C.
Every class defines a subset SX of S labelled by a member X ∈ SX.
Two different subsets SX and SY can have no element in common (by reflexivity), and the collection of all the classes Ci is a partition of S, i.e. every element in S belongs to one, and only one of the classes.
Let's call it a day there! In the next post, we will continue building upon a strong foundation by considering the concept of congruence, along with cosets. We will introduce Lagranges theorem, and prove some rather sleek results using equivalence relations. Before we are able to appreciate the full richness of the subject it is necessary to understand quite a few definitions and results, stick with it! Thanks for reading!
