Arthur Cayley
1. What Is a Group?
Group theory is pretty important. Quite a lot in physics rests on symmetry, so groups are naturally some of the most important objects we use to understand our universe. I'm starting group theory, so let’s start talking groups! A group is basically what you get when you strip symmetry down to the only things that are actually needed. Not "a shape that looks nice," but the bare structure underneath: you can do one symmetry after another, you can undo any of them, and doing nothing counts.
1.1 The axioms
Like a vector space, groups are objects that follow a set of axioms; a group is a set \(G\) with an operation combining two elements, satisfying four things.
Closure: \(\forall a, b \in G, \, ab \in G\) (combining two symmetries gives a symmetry).
Associativity: \((ab)c = a(bc)\) (brackets don't matter).
Identity: \(\exists \, e \in G: eg = ge = g \;, \forall g \in G\) (identity).
Inverses: \(\forall g \in G \;, \exists\, g^{-1}: gg^{-1} = g^{-1}g = e\). (every symmetry can be undone).
Now, commutativity is an extra property and is not part of the definition. We’ll talk about it later, commutative groups are called abelian.
1.2 The triangle
Here’s a nice example. Take an equilateral triangle with its corners labelled 1, 2, 3, and ask about every motion that leaves it occupying the same space. There are two obvious ones: let \(r\) be rotation by 120 degrees, which sends the corner in position 1 to position 2, 2 to 3, and 3 to 1. Let \(f\) be the reflection that holds corner 1 still and swaps 2 and 3. So rotating three times gets you back where you started, and reflecting twice does nothing:
\[ r^3 = e, \qquad f^2 = e \tag{1} \]
Now the elements we can write down are \(e\), \(r\), \(r^2\), \(f\), \(rf\), \(r^2f\). That's six elements: a symmetry is determined by where the three corners go, there are \(3! = 6\) ways to rearrange three objects, and we've found six distinct ones. This group is called \(D_3\) (the dihedral group of order 3).
1.3 Does order matter?
Let's check. Convention first: \(ab\) means do \(b\) first, then \(a\), which, kind of sucks, but then you remember how function compositions work and it makes more sense. Compute \(rf\). Start with \(f\): corner 1 stays, 2 and 3 swap. Then apply \(r\), which moves everything one position round.
\[ 1 \xrightarrow{f} 1 \xrightarrow{r} 2, \qquad 2 \xrightarrow{f} 3 \xrightarrow{r} 1, \qquad 3 \xrightarrow{f} 2 \xrightarrow{r} 3 \tag{2} \]
So \(rf\) swaps corners 1 and 2 and leaves 3 alone. Now the other order, \(fr\):
\[ 1 \xrightarrow{r} 2 \xrightarrow{f} 3, \qquad 2 \xrightarrow{r} 3 \xrightarrow{f} 2, \qquad 3 \xrightarrow{r} 1 \xrightarrow{f} 1 \tag{3} \]
That swaps 1 and 3, leaving 2 alone. Different reflection! So
\[ rf \neq fr \tag{4} \]
The group is non-abelian: reflecting then rotating has a different triangle to rotating then reflecting. We can say exactly how they fail to commute. Comparing the two results, \(fr\) is the same motion as \(r^2f\), so
\[ fr = r^2 f = r^{-1} f \tag{5} \]
using \(r^{-1} = r^2\), since \(r^3 = e\). That single relation is enough to push every \(f\) to the right-hand side of any product, which means every element really can be written as \(r^i f^j\), and the six we listed are the whole group.
1.4 The multiplication table
With \(r^3 = e\), \(f^2 = e\) and \(fr = r^2f\) we can compute every product. For instance \(f \cdot rf = (fr)f = (r^2f)f = r^2f^2 = r^2\). Doing all thirty-six:
\[ \begin{array}{c|cccccc} & e & r & r^2 & f & rf & r^2f \\ \hline e & e & r & r^2 & f & rf & r^2f \\ r & r & r^2 & e & rf & r^2f & f \\ r^2 & r^2 & e & r & r^2f & f & rf \\ f & f & r^2f & rf & e & r^2 & r \\ rf & rf & f & r^2f & r & e & r^2 \\ r^2f & r^2f & rf & f & r^2 & r & e \end{array} \tag{6} \]
Two things: first, every row and every column contains all six elements exactly once, with no repeats. That's forced: if \(ab = ac\) then multiplying by \(a^{-1}\) gives \(b = c\), so a row can't repeat itself. Every row is a rearrangement of the group. And the table isn't symmetric about its diagonal, which is the non-commutativity again. Kind of looks like the rotation matrix, no?
1.5 The same group twice
Okay, but who cares about triangles? Everyone, but you know what I mean. Go back and look at what our symmetries actually did. Each one rearranged three labelled corners, and we found all six possible rearrangements. Rearranging three objects doesn’t have to have anything to do with triangles. Shuffle three books, three coordinates, three anything, and you get a set of six operations that compose in exactly this pattern. The group of permutations of three objects is called \(S_3\), and its multiplication table is the one above, symbol for symbol. So \(D_3\) and \(S_3\) aren't two groups that happen to resemble each other, but they're one abstract object with two descriptions. So when we eventually say a particle has spin 2, we won't be saying anything about triangles or shuffling. We'll be saying it carries a particular representation of a particular group.