Peter Ludvig Sylow
1. The Class Equation
Last post we let \(D_3\) act on itself by conjugation, \(g \cdot x = gxg^{-1}\), and found that the orbits split the group into three pieces:
\[ \{ e \}, \qquad \{ r, r^2 \}, \qquad \{ f, rf, r^2f \} \tag{1} \]
of sizes 1, 2 and 3. They partition the group, and each size divides 6.
Let’s write it down properly.
1.1 Centralisers
The orbit-stabiliser theorem says the size of a conjugacy class is \(|G|\) divided by the size of the stabiliser, so let's work out what the stabiliser is here.
An element \(g\) stabilises \(x\) under conjugation when
\[ gxg^{-1} = x \tag{2} \]
which rearranges to \(gx = xg\). So the stabiliser is everything that commutes with \(x\). It's called the centraliser:
\[ C(x) = \{ g \in G : gx = xg \} \tag{3} \]
It's a subgroup, because stabilisers always are. Plugging it into orbit-stabiliser gives the size of the conjugacy class:
\[ |\operatorname{cl}(x)| = \frac{|G|}{|C(x)|} \tag{4} \]
Check it on \(D_3\). Take \(x = r\). What commutes with \(r\)? The rotations do. Do the reflections? From post 3 we know \(frf^{-1} = r^2 \neq r\), so no. Hence \(C(r) = \{e, r, r^2\}\) and
\[ |\operatorname{cl}(r)| = \frac{6}{3} = 2 \tag{5} \]
which matches the class \(\{r, r^2\}\) we found. Now \(x = f\). The only things commuting with \(f\) are \(e\) and \(f\) itself, so \(|C(f)| = 2\) and the class has \(6/2 = 3\) elements, i.e. the three reflections. Also correct.
1.2 Writing it down
The classes partition \(G\), so their sizes add up to \(|G|\). Picking one representative \(x_i\) from each class:
\[ |G| = \sum_i \frac{|G|}{|C(x_i)|} \tag{6} \]
Now separate out the small classes. A class has one element exactly when \(gxg^{-1} = x\) for every \(g\), which is to say \(x\) commutes with everything. Those elements form the centre:
\[ Z(G) = \{ x \in G : gx = xg \! \forall g \in G \} \tag{7} \]
Every element of the centre contributes a class of size 1, so pulling them out separately gives the class equation:
\[ |G| = |Z(G)| + \sum_i \frac{|G|}{|C(x_i)|} \tag{8} \]
where the sum now runs over representatives of the classes with more than one element only.
For \(D_3\): the centre is trivial, and the two big classes have sizes 2 and 3, so \(6 = 1 + 2 + 3\). Not very exciting on its own, but the equation is cool and works well when you don’t know the answer.
1.3 Groups of prime power order
Here's what it's for. Suppose \(|G| = p^n\) for a prime \(p\), with \(n \geq 1\). Such a group is called a \(p\)-group.
Every term in the sum is \(|G|/|C(x_i)|\), and by Lagrange \(|C(x_i)|\) divides \(p^n\), so each term is a power of \(p\). And each of these classes has more than one element by construction, so the term isn't 1, which means every term in the sum is divisible by \(p\).
Rearrange the class equation:
\[ |Z(G)| = |G| - \sum_i \frac{|G|}{|C(x_i)|} \tag{9} \]
On the right, \(|G| = p^n\) is divisible by \(p\), and every term of the sum is divisible by \(p\). So the whole right-hand side is divisible by \(p\), and therefore
\[ p \text{ divides } |Z(G)| \tag{10} \]
The centre always contains \(e\), so \(|Z(G)| \geq 1\). But now it's a multiple of \(p\), so \(|Z(G)| \geq p\). Which means:
Every group of prime power order has a nontrivial centre.
That's a pretty surprising result. It says nothing about the group's structure, only its size, and yet it guarantees there's something other than the identity commuting with absolutely everything. Compare \(D_3\), whose order 6 isn't a prime power and whose centre is trivial.
1.4 Order \(p^2\) is abelian
Let’s see it in action. Let \(|G| = p^2\).
We just showed \(p\) divides \(|Z(G)|\), and Lagrange says \(|Z(G)|\) divides \(p^2\). So \(|Z(G)|\) is either \(p\) or \(p^2\).
If it's \(p^2\) then \(Z(G) = G\) and everything commutes with everything, so the group is abelian, done.
So suppose \(|Z(G)| = p\). Then the quotient \(G/Z(G)\) has order \(p\), and back in post 2 we showed every group of prime order is generated by a single element. So pick a generator, call its coset \(gZ(G)\). Every element of \(G\) then sits in some coset \(g^kZ(G)\), meaning every element can be written
\[ x = g^k z \tag{11} \]
for some integer \(k\) and some \(z \in Z(G)\). Now take two of them, \(x_1 = g^{k_1}z_1\) and \(x_2 = g^{k_2}z_2\), and multiply:
\[ x_1 x_2 = g^{k_1} z_1 g^{k_2} z_2 = g^{k_1} g^{k_2} z_1 z_2 = g^{k_1 + k_2} z_1 z_2 \tag{12} \]
where the middle step used that \(z_1\) is central and it can rearrange past anything. Doing it the other way round gives \(g^{k_2 + k_1} z_2 z_1\), which is the same thing. So \(x_1x_2 = x_2x_1\), and the group is abelian after all.
But an abelian group has \(Z(G) = G\), contradicting \(|Z(G)| = p\). So that case never happens, and:
Every group of order \(p^2\) is abelian.
Groups of order 4, 9, 25, 49 - all abelian. That’s pretty crazy!
1.5 What just happened?
So what actually just happened? We didn't look at any groups. We took a partition into orbits, converted it into an equation about divisibility, and arithmetic dropped out that answer. A numerical hypothesis gave us a structural conclusion. I haven’t done too much number theory, so that’s pretty rare for me. Cool!
However, we cannot celebrate just yet. Everything here depended on \(|G|\) being finite, so we could count classes and divide. The groups coming next are things like all rotations of a sphere, or all Lorentz transformations, which have infinitely many elements and no meaningful notion of index. Counting arguments are about to stop working entirely, and we'll need to replace them with something continuous.