The Class Equation for Groups Acting on a Set

# The Class Equation for Groups Acting on a Set

Recall from The Orbit and Stabilizer of a Point in a Group Acting on a Set page that if $(G, \cdot)$ is a group acting on a (nonempty) set $A$, then for each $a \in A$ we defined the orbit of $a$ in $G$ to be the set:

(1)\begin{align} \quad Ga = \{ b \in A : b = ga \: \mathrm{for \: some \:} g \in G \} \end{align}

And for each $a \in A$ we defined the stabilizer of $a$ under $G$ to be the set:

(2)\begin{align} \quad G_a = \{ g \in G : ga = a \} \end{align}

We then proved that for each $a \in A$, $G_a$ is a subgroup of $G$. We are now about to prove an extension of The Class Equation for groups acting on a set. We first need the following lemma.

Lemma 1: Let $(G, \cdot)$ be a group acting on a (nonempty) set $A$. Then for each $a \in A$, $Ga$ is in bijection with the set of left cosets of $G_a$ in $G$, and so, for each $a \in A$, $|Ga| = [G:G_a]$ |

**Proof:**Fix $a \in A$.

- Let $g_1, g_2 \in G$. Observe that $g_1a = g_2a$ if and only if:

\begin{align} \quad g_2^{-1}(g_1a) = g_2^{-1}(g_2a) \end{align}

- Or equivalently, if and only if:

\begin{align} \quad (g_2^{-1} \cdot g_1)a = a \end{align}

- This happens if and only if $g_2^{-1} \cdot g_1 \in G_a$. But this happens if and only if $g_1G_a = g_2G_a$.

- Let $\mathrm{cosets}(G_a)$ be the set of all left cosets of $G_a$. Let $f : \mathrm{cosets}(G_a) \to Ga$ be defined for all $gG_a \in \mathrm{cosets}(G_a)$ by:

\begin{align} \quad f(gG_a) = ga \end{align}

- From the discussion above, we see that $g_1G_a = g_2G_a$ if and only if $g_1a = g_2a$, and so $f$ is well-defined and injective. Furthermore, $f$ is surjective since for all $ga \in Ga$ we have that $gG_a \in \mathrm{cosets}(G_a)$ is such that $f(gG_a) = ga$.

- So indeed, $Ga$ is in bijection with the set of left cosets of $G_a$ in $G$, and for each $a \in A$:

\begin{align} |Ga| = |\mathrm{cosets}(G_a)| = [G:G_a] \quad \blacksquare \end{align}

Lemma 2: Let $(G, \cdot)$ be a group acting on a (nonempty) set $A$. Then the set of orbits $\{ Ga : a \in A \}$ partition $A$. |

**Proof:**Define an equivalence relation $\sim$ on $A$ for all $a, b \in A$ by $a \sim b$ if and only if there exists a $g \in G$ such that $a = gb$. We now check that $\sim$ is indeed an equivalence relation on $A$.

**Reflexivity:**By the second axiom of a group action we have that $a = ea$ for all $a \in A$, where $e \in G$ denotes the identity. Thus $a \sim a$ for every $a \in A$.

**Symmetry:**Suppose that $a \sim b$. Then there exists a $g \in G$ such that $a = gb$. Then we have that $g^{-1}a = g^{-1}(gb) = b$. Since $g \in G$ implies $g^{-1} \in G$ and $b = g^{-1}a$ we see that $b \sim a$. Thus, for all $a, b \in A$, if $a \sim b$ then $b \sim a$.

**Transitivity:**Suppose that $a \sim b$ and $b \sim c$. Then there exists a $g \in G$ such that $a = gb$, and there exists a $g' \in G$ such that $b = g'c$. So by the first axiom of a group action we have that $a = gb = g(g'b) = (g \cdot g')c$. Since $g, g' \in G$ we have that $g \cdot g' \in G$. Thus $a \sim c$. So for all $a, b, c \in A$, if $a \sim b$ and $b \sim c$ then $a \sim c$.

- So indeed, $\sim$ is an equivalence relation on $A$, and thus, the equivalence classes partition $A$. But for each $a \in A$, the equivalence class $[a]$ of $a$ is:

\begin{align} \quad [a] = \{ b \in A : b \sim a \} = \{ b \in A : b = ga \: \mathrm{for \: some \:} g \in G \} = Ga \end{align}

- So the set of orbits $\{ Ga : a \in A \}$ partitions $A$. $\blacksquare$

Definition: Let $(G, \cdot)$ be a group acting on a (nonempty) set $A$. The Subset of $A$ Fixed by $G$ is defined to be the set $A^G = \{ a \in A : ga = a \: \mathrm{for \: all \:} g \in G \}$. |

We are now ready to state and prove the generalized class equation for groups acting on a set.

Theorem 3 (The Class Equation for Groups Acting on a Set): Let $(G, \cdot)$ be a group acting on a finite (nonempty) set $A$. Then $\displaystyle{|A| = |A^G| + \sum [G : G_a]}$ where the sum runs over one representative of each orbit $Ga$ with $|Ga| > 1$. |

**Proof:**By Lemma 2 we have that $\{ G_a : a \in A \}$ partitions $A$, and so:

\begin{align} \quad |A| &= \sum |[a]| \\ &= \sum |Ga| \end{align}

- Where the above sum runs over one representative for each orbit. Suppose that $a \in A$ is such that $|Ga| = 1$. Then $Ga = \{ b \in A : b = ga \: \mathrm{for \: some \:} g \in G \} = \{ a \}$. That is, for all $g \in G$ we have that $ga = a$. So:

\begin{align} \quad |A| = |A^G| + \sum|Ga| \end{align}

- Where the above sum runs over one representative for each orbit $Ga$ with $|Ga| > 1$. By Lemma 1, we have that:

\begin{align} \quad |A| = |A^G| + \sum [G:G_a] \end{align}

- Where the above sum runs over one representative for each orbit $G_a$ with $|Ga| > 1$. $\blacksquare$