Orbits and Stabilizers

Let \(G\) be a group and \(X\) a set. Let \(\alpha\) be an action of \(G\) on \(X\). Define the relation \(\sim\) on \(X\) such that \(x \sim y \iff \exists g \in G \Big(\alpha_{g}(x) = y\Big)\). It is easy to see that \(\sim\) is an equivalence relation.

Definition (Orbit)

Let \(G\) be a group and \(X\) a set such that \(G\) acts on \(X\) by the action \(\alpha\). Given the equivalence relation \(\sim\) defined above, we define the orbit of \(x \in X\) as

\[\mathcal{O}_{x} := \{ y \in X:y \sim x \} = \{ y \in X: \exists g \in G \big( \alpha_{g}(x) = y \big) \}.\]

Since \(\sim\) is an equivalence relation, the orbits form a partition of \(X\), i.e., \(X = \dot\bigcup_{x \in X} \mathcal{O}_{x}\).

Definition (Transitive action)

Let \(G\) be a group and \(X\) a set such that \(G\) acts on \(X\). If there exists \(x \in X\) such that \(\mathcal{O}_{x} =X\) we say that the action is transitive. Equivalently, the action is transitive if for all \(x,y \in X\) there exists \(g \in G\) such that \(x = \alpha_{g}(y)\).

Definition (Stabilizer)

Let \(G\) be a group and \(X\) a set such that \(G\) acts on \(X\) by the action \(\alpha\). Given \(x \in X\), we define the stabilizer of \(x\) by \(G_{x}:= \{ g \in G:\alpha_{g}(x)=x \}\).

Lemma (Stabilizer is a subgroup)

Proof: \(G_{x} \neq \emptyset\) since \(1_{G} \in G_{x}\) by definition. Now, let \(g,h \in G_{x}\). Note that \(\alpha_{h^{-1}}(x) = (\alpha_{h})^{-1}(x) = x\). Finally, we have that

\[\alpha_{gh^{-1}}(x) = \alpha_{g}(\alpha_{h^{-1}}(x)) = x,\]

so that \(g \cdot h^{-1} \in G_{x}\), from which we conclude that \(G_{x} \leq G\).

Examples (Orbits and stabilizers)

Let \(n\geq2\). Consider the action of \(S_{n}\) on \(X = \{ 1,\dots,n \}\) such that \(\alpha_{\sigma}(i) = \sigma(i)\). In this case there is a single orbit. Let \(i,j \in X\) and consider \(\tau = \begin{pmatrix}i & j\end{pmatrix}\). Note that \(\alpha_{\tau}(i) = j\), so that \(i \sim j\), i.e., any two elements of \(X\) are related. Hence the action is transitive. Now, note that if \(1 \leq i\leq n\), \(G_{i} = \{ \sigma \in S_{n}: \sigma(i) = i \} \cong S_{n-1}\).
Consider \(G = (\mathbb{Z}, + )\) and \(X=\mathbb{R}\) under the action \(\tau(n,x) = x+n\). Note that if \(x,y \in \mathbb{R}\),
\[x \sim y \iff \exists n \in \mathbb{Z} \Big(\tau_{n}(x) = y\Big) \iff \exists n \in \mathbb{Z} \Big(y=x+n\Big) \iff x-y \in \mathbb{Z},\]

from which we have that \(x\) and \(y\) have the same decimal part. There is an uncountable number of orders. It is evident that for every \(x \in \mathbb{R}\), \(G_{x} = \{ 0 \}\).

Consider \(G = \mathbb{S}^{1} =\{ e^{i\theta}:\theta \in [0, 2\pi) \}\) and \(X = \mathbb{C}\). Define the action \(\rho_{\theta}(z) = z e^{i \theta}\) (checking that this is an action is an exercise). Let \(z \in \mathbb{C}\). Then,
\[\mathcal{O}_{z} = \{ \rho_{\theta}(z): \theta \in [0, 2\pi) \} = \{ ze^{i \theta}: \theta \in [0,2\pi) \} = \{ w \in \mathbb{C}: \lvert w \rvert = \lvert z \rvert \}.\]

Now, let us see what happens with the stabilizer. If \(z = 0\), we have that

\[G_{0} = \{ \theta \in \mathbb{S}^{1}: \rho_{\theta}(0) = 0 \} = \{ g \in \mathbb{S}^{1}: 0 e^{i \theta} = 0 \} = \mathbb{S}^{1}.\]

On the other hand, if \(z \neq 0\),

\[\rho_{\theta}(z) = z \iff z e^{i \theta} = z \iff \theta = 0 \iff e^{i \theta } = 1,\]

which implies that \(G_{Z} = \{ 1 \}\).

Let \(G\) be a group and \(H \leq G\). Consider \(G / H = \{ gH : g \in G\}\). Define an action of \(G\) on \(G / H\) such that \(\mu:G \times G / H\to G /H\) such that \(\mu_{g}(hH) = (g \cdot h) H\). It is left as an exercise to see that \(\mu\) is a transitive action. Now, note that
\[g \in G_{xH} \iff (g \cdot x) H = xH \iff x ^{-1} \cdot g \cdot x \in H \iff g \in x H x ^{-1},\]

so that \(G_{xH} = xH x ^{-1}\).

Let \(G = \mathrm{GL}(n,\mathbb{R})\) and \(X = \mathbb{R}^n\). Define the action \(\alpha_{A}(v)=Av\). Note that
\[\mathcal{O}_{0} = \{ v \in \mathbb{R}: \exists A \in G (Av = w) \} = \{ 0 \}.\]

Hence this action is not transitive.

Let \(G\) be a group. Define \(\mu:G\to G\) such that \(\mu_{g}(h) = gh\). Let \(h,g \in G\). Note that \(\mu_{gh^{-1}}(h) = (gh^{-1})h=g\), so that \(h \sim g\). Hence the action is transitive.
Let \(G\) be a group and consider the action of \(G\) on \(G\) given by conjugation \(\mu_{g}(h) = ghg^{-1}\). Let \(x \in G\). We have that
\[x^G := O_{x} = \{ y \in G: \exists g \in G (g xg^{-1} = y) \} = \{ gxg^{-1}:g \in G \}.\]

On the other hand, \(G_{x}:= \{ g \in G: gxg^{-1} = x \} = C_{G}(x)\) (the centralizer).

Theorem (Orbit and quotient of the stabilizer)

Let \(G\) be a group and \(X\) a set such that \(G\) acts on \(X\). Let \(x \in X\). Then \(\lvert \mathcal{O}_{x} \rvert = [G:G_{x}]\).

Proof: We shall construct a bijection \(\Phi:\mathcal{O}_{x} \to G / G_{x}\). Let \(y \in \mathcal{O}_{x}\). Then there exists \(g \in G\) such that \(\alpha_{g}(x) = y\). Define \(\Phi(y) = g \cdot G_{x}\). \(\Phi\) is well defined, since if \(g,h\) are such that \(\alpha_{g}(x) = \alpha_{h}(x)\), then

\[\alpha_{h^{-1}g}(x) = \alpha_{h^{-1}}(\alpha_{g}(x)) = \alpha_{h^{-1}}(\alpha_{h}(x)) = \alpha_{1_{G}}(x) = x,\]

so that \(h^{-1}g \in G_{x}\) and hence \(g \cdot G_{x} = h \cdot G_{x}\). \(\Phi\) is injective since if \(y,z \in \mathcal{O}_{x}\) are such that \(\Phi(y) = \Phi(z)\), there exist \(g,h \in G\) such that \(\alpha_{g}(x) = y\) and \(\alpha_{h}(x) = z\). Thus, we have that

\[g G_{x} = h G_{x} \implies h^{-1} g \in G_{x} \implies \alpha_{h^{-1}g}(x)=x.\]

Thus, \(\alpha_{h^{-1}}(\alpha_{g}(x)) = \alpha_{h}^{-1}(\alpha_{g}(x)) = x \implies \alpha_{g}(x)=\alpha_{h}(x)\). Finally \(\Phi\) is surjective, since if \(g \cdot G_{x} \in G / G_{x}\), take \(y = \alpha_{g}(x)\). Note that \(\Phi(y) = g \cdot G_{x}\). Thus, \(\Phi\) is a bijection and hence the given sets have the same cardinality.

Corollary (The orbit divides the order of the group)

Let \(G\) be a finite group and \(X\) a set such that \(G\) acts on \(X\). Then for every \(x \in X\), \(\lvert \mathcal{O}_{x} \rvert \Big\lvert \lvert G \rvert\).

Proof: We have that \(\lvert \mathcal{O}_{x} \rvert = [G:G_{x}] =\frac{ \lvert G \rvert}{\lvert G_{x} \rvert}\), from which we conclude the result

Corollary (Size of the conjugacy class)

Let \(G\) be a finite group and \(x \in G\), Then \(\lvert x^{G} \rvert = [G:C_{G}(x)]\).

Theorem (Cauchy)

Let \(G\) be a finite group and \(p\) a prime such that \(p \Big\lvert \lvert G \rvert\). Then there exists \(g \in G\) such that \(\lvert g \rvert = p\).

Proof: [Proof 1, for abelian groups.] Suppose that \(G\) is abelian. We proceed by strong induction on \(\lvert G \rvert=n\). The base case (\(n=1\)) is trivially true, vacuously. Suppose as inductive hypothesis that the result is true for every group \(H\) such that \(\lvert H \rvert < n\). Let \(a \in G\) with \(a \neq 1_{G}\) and \(k = \lvert a \rvert\). Consider the following cases: Case 1: Suppose that \(p \mid k\). Then, there exists \(\ell \in \mathbb{N}\) such that \(p \ell = k\). Hence \(\lvert a^{\ell} \rvert = p\). Case 2: Suppose that \(p \not\mid k\). Let \(H = \langle a \rangle\). Since \(G\) is abelian, \(H \triangleleft G\), so that \(G/H\) is a group and \(\lvert H \rvert = k\). Note that

\[\lvert G / H \rvert = \frac{\lvert G \rvert }{\lvert H \rvert } = \frac{n}{k} < n.\]

Since \(p \mid n\) and \(p \not\mid k\), \(p \mid \frac{n}{k}\). By the inductive hypothesis, there exists \(sH \in G /H\) such that \(\lvert sH \rvert = p\). Let \(m = \lvert s \rvert\). Note that \((sH)^{m} = s^{m}H = H\), so that \(p \mid m\). Finally, apply case 1 to \(s\). We conclude the result

Proof: [Proof 2, general.] Let \(x \in G\). Recall that \(x^{G} = \{ g x g^{-1}: g \in G \}\) and that \(\lvert x^{G} \rvert = [G:C_{G}(x)]\). We proceed by strong induction on \(\lvert G \rvert=n\). Consider the following cases: Case 1: If \(x \in Z(G)\), then \(x^{G}=\{ x \} \implies G = C_{G}(x)\). Case 2: If \(x \not\in Z(G)\), then \(\lvert x^{G} \rvert > 1\). Then \(C_{G}(x) < G \implies \lvert C_{G}(x) \rvert < \lvert G \rvert\). If \(p \Big\lvert \lvert C_{G}(x) \rvert\), by the inductive hypothesis, there exists \(g \in C_{G}(x) \subseteq G\) such that \(\lvert g \rvert=p\), from which we conclude the result. If \(p \not\Big\lvert \lvert C_{G}(x) \rvert\), we have that

\[[G:C_{G}(x)] = \frac{\lvert G \rvert }{\lvert C_{G}(x) \rvert } \implies [G:C_{G}(x)] \cdot \lvert C_{G}(x) \rvert = \lvert G \rvert \implies p \mid [G:C_{G}(x)].\]

Let \(x_{1},\dots,x_{k}\) be the distinct conjugacy classes obtained through the relation of being conjugate and which are not in \(Z(G)\). Then,

\[G = Z(G) \dot{\cup} \bigcup_{i=1}^{k} x_{i}^{G} \implies \lvert G \rvert = \lvert Z(G) \rvert + \sum_{i=1}^{k} \lvert x_{i}^{G} \rvert \implies \lvert G \rvert - \sum_{i=1}^{k} \lvert x_{i}^{G} \rvert = \lvert Z(G) \rvert,\]

from which we conclude that \(p \Big\lvert \lvert Z(G) \rvert\). Thus, \(Z(G)\) is a proper subgroup of \(G\), and we conclude the result by applying the inductive hypothesis to \(Z(G)\).

Note (The primality hypothesis is necessary)

The theorem is false if \(p\) is not prime. Note that \(8 \mid 120 \implies 8 \Big\lvert \lvert S_{5} \rvert\). But we have seen that there are no subgroups of order 8 in \(S_{5}\).

Definition (\(p\)-group)

Let \(p\) be prime. A group \(G\) is a \(p\)-group if there exists \(k \in \mathbb{N}\) such that \(\lvert G \rvert=p^{k}\).

Exercise (Centre of a \(p\)-group)

If \(G\) is a \(p\)-group, \(Z(G) \neq \{ 1_{G} \}\).

Corollary (Groups of order \(p^{2}\))

If \(p\) is prime and \(\lvert G \rvert = p^{2}\), then \(G\) is abelian.

Proof: Suppose that \(G\) is not abelian. Then \(Z(G)\) is a proper subgroup of \(G\). By Lagrange, \(\lvert Z(G) \rvert \in \{ 1,p \}\). By the previous exercise, \(\lvert Z(G) \rvert = p\). We know that \(Z(G) \triangleleft G\). Thus, \(\lvert G / Z(G) \rvert = \frac{\lvert G \rvert}{\lvert Z(G) \rvert} = p\), from which we conclude that \(G / Z(G)\) is cyclic and hence \(G\) is abelian.

Theorem (Subgroups of a finite abelian group)

Let \(G\) be a finite abelian group. Then \(G\) has a subgroup of order \(d\) for every \(d\) such that \(d \Big\lvert \lvert G \rvert\).

Proof: Let \(n = \lvert G \rvert\). Proceed by strong induction on \(d\), with \(d \mid n\). The base case is trivial, since if \(H = \{ 1_{G} \}\implies \lvert H \rvert=1\). Let \(d>1\). Suppose as inductive hypothesis that for every \(e < d\) and every group \(\tilde{G}\) with \(e \Big\lvert \lvert \tilde{G} \rvert\) there exists \(\tilde{H} < \tilde{G}\) such that \(\lvert \tilde{H} \rvert = e\). Since \(d \mid n\), there exists \(k \in \mathbb{N}\) such that \(kd=n\). Since \(d>1\), there exists a prime \(p\) such that \(p \mid d\), i.e., there exists \(\ell \in \mathbb{N}\) such that \(\ell p = d\). By Cauchy’s theorem, there exists \(g \in G\) such that \(\lvert g \rvert=p\). Let \(H = \langle g \rangle\). Note that \(H\) is cyclic, and hence abelian and normal. Hence, \(G / H\) is a group. Thus,

\[\lvert G / H \rvert = \frac{\lvert G \rvert }{\lvert H \rvert } = \frac{n}{p} = \frac{kd}{p} = \frac{k \ell p}{p} = k \ell.\]

Thus, \(\ell \Big\lvert \lvert G / H \rvert\) and \(\lvert G/H \rvert < n\). Since \(\ell < d\), by the inductive hypothesis, there exists \(S^{\ast} < G / H\) such that \(\lvert S^{\ast} \rvert = \ell\). By the correspondence theorem, there exists \(S < G\) such that \(H \leq S \leq G\) and \(S^{\ast} = S/H\). We conclude by noting that

\[\ell = \lvert S^{\ast} \rvert = \frac{\lvert S \rvert}{\lvert H \rvert } = \frac{\lvert S \rvert}{p} \implies \lvert S \rvert = \ell p =d,\]

from which the result follows.

Theorem (Subgroups of a \(p\)-group)

Let \(G\) be a \(p-grupo\) such that \(\lvert G \rvert = p^{e}\). Then for every \(k \leq e\), there exists \(H \leq G\) such that \(\lvert H \rvert = p^{k}\).

Proof: We proceed by induction on \(e\). For \(e=0\), we have that \(G=\{ 1_{G} \}.\) Take \(H = G\). Let \(e > 0\) and let \(k \leq e\). Suppose as inductive hypothesis that the result is true for \(f<e\). Since \(\lvert G \rvert = p^{e}\) and \(e>0\), then \(Z(G)\) is non-trivial. Consider the following cases.

If \(G = Z(G)\), then \(G\) is abelian. By the previous theorem, we conclude the result.
Now, if \(G \neq Z(G)\), there exists \(c < e\) such that \(\lvert Z(G) \rvert = p^{c}\) with \(c<e\). If \(k \leq c\), then by the inductive hypothesis applied to \(Z(G)\), there exists \(H \triangleleft Z(G)\) such that \(\lvert H \rvert=p^{k}\). On the other hand, if \(k>c\), we know that \(G / Z(G)\) is a group (since \(Z(G) \triangleleft G\)). Now, note that
\[\lvert G / Z(G) \rvert = \frac{\lvert G \rvert }{\lvert Z(G) \rvert } = p^{e-c}.\]

Applying the inductive hypothesis to \(G / Z(G)\), there exists \(S^{\ast} \leq G/Z(G)\) such that \(\lvert S^{\ast} \rvert = p^{k-c}\). By the correspondence theorem, there exists \(S \leq G\) such that \(Z(G) \leq S \leq G\) and \(S^{\ast} = S / Z(G)\). Thus,

\[\lvert S^{\ast} \rvert \cdot \lvert Z(G) \rvert = \lvert S \rvert \implies p^{k-c} \cdot p^{c} = \lvert S \rvert,\]

from which we conclude the result.