Taking abstract algebra was probably one of the best academic decisions I have made in college so far. Although it remains unclear to me where I would ever use abstract algebra given my interests in machine learning and finance, it still is one of the most fun and profound course-sequences I have taken here at UC San Diego. This blog post aims to explain the intuition and discuss some very useful applications of the class equation in group theory. Unfortunately this post assumes some elementary knowledge of abstract algebra.

### The class equation

Here’s the beautiful equation: $|G| = |Z(G)| + \sum_{i}|G:N(a_i)|$. Where $class a_i$ are the non-singleton conjugacy classes in the group $G$. Here $Z(G)$ refers to the center of G, and $N(a) = \{g \in G | ag = ga\}$ is the centralizer of the element $a$. Note that $|class a| = |\{gag^{-1} | g \in G\}|$.

### Intuition / “Proof”

Although this is not the formal proof it should suffice. Firstly it should be fairly easy to see that $|G| = \sum_{i} |class g_{i}|$ where $class g_i$ are the conjugacy classes. This is true because the conjugacy classes of a group partition it (because they induce an equivalence relation). Now, note that if $a \in Z(G)$ then $|class a| = 1$. This is because pick $x \in class a$ then $x = gag^{-1}$ but $a$ commutes with all $g \in G$. So we have $x = gg^{-1}a = a$. Therefore $a$ is the only element in its conjugacy class if $a \in Z(G)$. Therefore we can now separate our conjugacy classes into being singletons and non-singletons. The number of singleton classes is exactly equal to the size of the center of the group, as we just showed, therefore the non-singleton classes remain. Here is where I was at a lost for intuition: $|class a| = [G:N(a)]$. If we can prove this we are clearly done. However, this is rather non-trivial fact. What this equation is saying is that the number of cosets of the normalizer of an element in $G$ is exactly equal to the number of elements in it’s conjugacy class. (How do people think of such stuff?) Anyway, here’s why it makes sense. In plain english, every coset $hN(a), h \in G$ collapses into one exact element in $class a$. Let’s show this: pick $x,y \in hN(a), x \neq y$. Then $x = hg_1, y = hg_2$ such that $g_1, g_2 \in N(a), g_1 \neq g_2$. Now note that $xax^{-1} = hg_1ag_1^{-1}h^{-1} = hag_1g_1^{-1}h^{-1} = hah^{-1} = hg_2g_2^{-1}ah^{-1} = hg_2ag_2^{-1}h^{-1} = yay^{-1}$. Although this seems simple, it is pretty cool. What this proof is telling us that the coset $hN(a)$ is completely determined by $h$ regarding where it sits in $classa$. No matter which element you pick in a given coset, they all map to exactly one element in the conjugacy class of $a$! This proof can be easily formalized by showing that this is indeed an isomorphism.

### Applications

#### Centers of p-groups

It follows directly from the class equation that p-groups can not have trivial center since no prime $p$ divides 1.

#### Orbit decomposition theorem

Given a G-Set $X$, we have the following equation: $X = X_{F} + \sum_i |G:Stab(x_i)|$. Here $X_{F}$ is the fixed set (analogous to the center of a group) and $Stab(x_i)$ is the stablizer of the element (analogous to the normalizer). Instead of classes of an element we have the orbits of an element (which also partition X). Using the same idea as the class equation we have a much stronger equation! In fact note that the class equation is just a special case of $G$ acting upon itself via conjugation.

### Conclusion

Abstract algebra is full of such profound, strong theorems and equations, and that’s what makes it so beautiful to study.