The Class Equation
The Class Equation
Lemma 1: Let $G$ be a group and let $g \in G$. Then the elements in the conjugacy class of $g$ are in bijection with the left cosets of $C_G(g)$. |
Recall that if $G$ is a group and $H$ is a subgroup of $G$ then for all $a, b \in G$ we have that $aH = bH$ if and only if $a^{-1}b \in H$.
- Proof: Let $g \in G$ and let $[g]$ denote the conjugacy class of $g$, that is:
\begin{align} \quad [g] = \{ h \in G : h = aga^{-1} \: \mathrm{for \: some \:} a \in G \} \end{align}
- Let $h, h' \in [g]$. Then $h = aga^{-1}$ for some $a \in G$ and $h' = bgb^{-1}$ for some $b \in G$. We see that $h = h'$ if and only if $aga^{-1} = bgb^{-1}$, which happens if and only if:
\begin{align} \quad (b^{-1}a)g(a^{-1}b) &= g \\ \quad (b^{-1}a)g(b^{-1}a)^{-1} &= g \end{align}
- But the above equation is true if and only if $b^{-1}a \in C_G(g) = \{ h \in G : hgh^{-1} = g \}$. But $b^{-1}a \in C_G(g)$ if and only if $aC_G(g) = bC_G(g)$.
- So if $\mathcal C$ is the set of all left cosets of $C_G(g)$ and if $\phi : [g] \to \mathcal C$ is defined for each $h \in [g]$ with $h = aga^{-1}$ by $\phi(h) = \phi(aga^{-1}) = aC_G(g)$, then $\phi$ is well-defined above, and is bijective. $\blacksquare$
Theorem 2 (The Class Equation): Let $G$ be a finite group. Then $|G| = |Z(G)| + \sum [G : C_G(g)]$, where the sum on the righthand side runs over one element $g$ in each nontrivial conjugacy class in $G$. |
Recall that $C_G(g)$ is a subgroup of $G$ for each $g \in G$. Also recall from the The Index of a Subgroup page that the index of $C_G(g)$ in $G$ denoted by $[G:C_G(g)]$ is defined to be the number of left (or right) cosets of $C_G(g)$ in $G$.
- Proof: Since conjugacy is an equivalence relation on $G$, we have that $G$ is partitioned by the set of conjugacy classes. So:
\begin{align} \quad |G| = \sum |[g]| \end{align}
- Where the sum above runs over one element $g$ in each conjugacy class in $G$.
- Suppose that $g \in G$ has trivial conjugacy class, i.e., $[g] = \{ g \}$. Then $aga^{-1} = g$ for all $a \in G$, and thus $g \in Z(G)$. Conversely, if $g \in Z(G)$ then $ag = ga$ for all $a \in G$, i.e., $aga^{-1} = g$ for all $a \in G$, so $[g] = \{ g \}$. Therefore:
\begin{align} \quad |G| = |Z(G)| + \sum |[g]| \end{align}
- Where the sum above runs over one element $g$ in each nontrivial conjugacy class in $G$.
- Lastly, by Lemma 1, for each $g \in G$, the set of elements in $[g]$ is in bijection with the number of left cosets of $C_G(g)$ in $G$. The number of left cosets of $C_G(g)$ in $G$ is its index, $[G : C_G(g)]$ and so for each $g \in G$ we have that $|g| = [G : C_G(g)]$. Therefore:
\begin{align} \quad |G| = |Z(G)| + \sum [G:C_G(g)] \end{align}
- Where the sum above runs over one element $g$ in each nontrivial conjugacy class in $G$. $\blacksquare$