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    The class equation

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    For a finite group \(G\), each element \(g\) either lies in the center \(Z(G)\), and then its conjugacy class is \(\ts g\), or its conjugacy class has size \([G:C_G(g)] > 1\).

    Class equation

    A finite group is its center plus its noncentral conjugacy classes.Suggest a slogan

    Let \(G\) be a finite group with center \(Z(G)\), and for \(g\in G\) let \(Z(g) = \theset{h\in G \st hgh^{-1} = g}\) be the centralizer of \(g\). Let \(g_1, \ldots, g_r\) be representatives of the conjugacy classes of \(G\) that have more than one element. Then \[ \size G = \size Z(G) + \sum_{i=1}^{r} [G : Z(g_i)], \] where \([G:Z(g_i)]\) is the index of \(Z(g_i)\) in \(G\).

    Class equation

    Let \(G\) be a finite group with center \(Z(G)\), and let \(x_1,\ldots,x_k\) be representatives of the conjugacy classes of \(G\) that contain more than one element. For \(x\in G\), let \(C_G(x) \coloneqq \theset{ g\in G \st gxg^{-1} = x }\) be the centralizer of \(x\). Then \[ \abs{G} = \abs{Z(G)} + \sum_{i=1}^k [G: C_G(x_i)]. \]

    \(G\) acts on itself by conjugation, \(g\cdot x=gxg^{-1}\); its orbits are the conjugacy classes, which partition \(G\), and the stabilizer of \(x\) is \(C_G(x)\). By the orbit-stabilizer theorem, the conjugacy class of \(x\) has \([G:C_G(x)]\) elements. The class of \(x\) is \(\theset{x}\) exactly when \(gxg^{-1}=x\) for all \(g\in G\), that is, when \(x\in Z(G)\). Summing class sizes, the singleton classes contribute \(\abs{Z(G)}\) and the remaining classes contribute \(\sum_{i=1}^k[G:C_G(x_i)]\).

    State the class equation for a finite group $G$.

    (Derivation from orbit-stabilizer)

    For the conjugation action \(\phi\) of \(G\) on itself, the fixed points are the elements of \(Z(G)\) and the orbit of \(g\) is \([g]\), with \(\size{[g]} = [G : C_G(g)]\). If \(g_1,\ldots,g_r\) represent the conjugacy classes of size greater than \(1\), then \[ \begin{aligned} \size G &= \size{\Fix(\phi)} + \sum_{i=1}^r \size{\Orb(g_i)} \\ &= \size{Z(G)} + \sum_{i=1}^r \size{[g_i]} \\ &= \size{Z(G)} + \sum_{i=1}^r [G : C_G(g_i)], \end{aligned} \] which is the class equation.

    \(p\)-groups

    (A nontrivial $p$-group has nontrivial center)

    If \(\size G = p^k\) with \(k\geq 1\), then \(Z(G)\neq 1\).

    Each term \([G : C_G(g_i)]\) in the class equation is a divisor of \(p^k\) greater than \(1\), hence divisible by \(p\), and \(\size G\) is divisible by \(p\). So \(\size{Z(G)}\) is divisible by \(p\), and \(Z(G) \neq 1\).

    Consequences for a group \(G\) of order \(p^k\):

    • If \(k=2\), then \(G\) is abelian.

    • \(G\) has a normal subgroup of every order dividing \(\size G\).

    • \(G\) is nilpotent.

    • \(G\) is simple if and only if \(k=1\).

    Burnside’s lemma

    Burnside's lemma

    Orbit count is the average fixed-point count.Suggest a slogan

    Let \(G\) be a finite group acting on a finite set \(X\). Let \(X/G\) be the set of orbits, and for \(g\in G\) let \(\Fix(g) = \theset{x\in X \st gx=x}\). Then \[ \size (X/G) = \frac{1}{\size G}\sum_{g\in G} \size \Fix(g). \]

    The number of orbits is the average over \(G\) of the number of fixed points.

    Fixed-point set $X^g$ of a group element

    Let a group \(G\) act on a set \(X\), and let \(g\in G\). The fixed-point set of \(g\) is \[ X^g \coloneqq \theset{x\in X \st g\cdot x = x} \subseteq X. \]

    What is the fixed-point set $X^g$ of a group action?

    Burnside's orbit-counting formula

    Orbit count is the average fixed-point count.Suggest a slogan

    Let a finite group \(G\) act on a finite set \(X\). Let \(X/G\) be the set of orbits, and for \(g\in G\) let \(X^g=\theset{x\in X\suchthat g\cdot x=x}\) be the set of fixed points of \(g\). Then \[ \abs{G}\,\abs{X/G}=\sum_{g\in G}\abs{X^g}, \] so the number of orbits is the average number of fixed points of an element of \(G\).

    Count the set \(\theset{(g,x)\in G\times X\suchthat g\cdot x=x}\) in two ways. Summing over \(g\) gives \(\sum_{g\in G}\abs{X^g}\). Summing over \(x\) gives \(\sum_{x\in X}\abs{G_x}=\sum_{x\in X}\abs G/\abs{G\cdot x}\) by the orbit-stabilizer theorem, and the \(\abs{G\cdot x}\) points of each orbit contribute \(\abs G\) in total, so this sum is \(\abs G\,\abs{X/G}\).

    State Burnside's orbit-counting formula.

    (of Burnside's lemma)

    Let \(A \da \ts{ (g,x) \in G\cross X \st g\actson x = x }\), and write \(\Stab(x) = \ts{g\in G \st gx=x}\) and \(\Fix(g) = \ts{x\in X\st gx = x}\).

    Partitioning \(A\) according to the first coordinate, \[ A = \Disjoint_{g_0\in G} \ts{ (g_0, x) \st g_0 x = x } \cong \Disjoint_{g_0\in G} \ts{g_0}\cross \Fix(g_0). \] Partitioning \(A\) according to the second coordinate, \[ A = \Disjoint_{x_0\in X} \ts{ (g, x_0) \st gx_0= x_0 } \cong \Disjoint_{x_0\in X} \Stab(x_0) \cross \ts{ x_0 }. \] Taking cardinalities, \[ \sum_{g_0\in G} \size \Fix(g_0) = \size A = \sum_{x_0\in X} \size \Stab(x_0). \] By orbit-stabilizer, \(\size \Stab(x_0) = \size G/ \size \Orb(x_0)\), so \[ {1\over \size G} \sum_{g_0\in G} \size \Fix(g_0) = \sum_{x_0\in X} {1\over \size \Orb(x_0)}. \] Grouping the terms of the right-hand side by orbit, \[ \begin{aligned} \sum_{x_0\in X}{1\over \size \Orb(x_0)} &= \sum_{\Orb(x_0) \in X/G} \qty{1\over \size \Orb(x_0)}\sum_{y\in \Orb(x_0)} 1 \\ &= \sum_{\Orb(x_0) \in X/G} 1 \\ &= \size (X/G). \end{aligned} \]

    (Counting orbits)

    Burnside’s lemma counts orbits, such as colorings of the beads of a necklace or of the faces of a cube up to the action of a rotation group. For \(g,h\in G\), \(x\mapsto hx\) is a bijection \(\Fix(g)\to\Fix(hg\inverseof{h})\), so \(\size{\Fix(g)}\) depends only on the conjugacy class of \(g\), and the sum may be taken over conjugacy classes, each weighted by its size.

    Applications of the class equation

    • Show that \(p\) groups have nontrivial centers.

    • Show that groups of order \(p^2\) are abelian.

    Solution

    \(p\dash\)groups have nontrivial centers:

    • Abusing notation by identifying sets with their cardinalities, the class equation says \(G = Z(G) + \sum_{g}' [G: Z(g)]\) where the terms in the sum are all bigger than 1.

    • Reducing mod \(p\) yields \(0 = Z(G) + 0\), since \(p\) must divide \([G:Z(g)]\) when \([G:Z(g)] > 1\) because \(G = [G:Z(g)]Z(g)\) and \(p\) divides the LHS.

    • So \(p\) divides \(Z(G)\), making \(Z(G)\) nontrivial.

    \(p^2\) groups are abelian:

    • \(Z(G) = 1,p,p^2\), and by the first part \(Z(G)\neq 1\). If \(Z(G) = p^2\) then \(G = Z(G)\) is abelian, so assume \(Z(G) = p\).

    • Then \(G/Z(G) = p\) and groups of order \(p\) are cyclic. If \(G/Z(G)\) is cyclic then \(G\) is abelian, so \(G\) is abelian.

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