3.14 \(G\)-sets and the orbit-stabilizer theorem
In this section \(G\) is a group with identity element \(1\).
A \(G\)-set is an ordered pair \((X,\rho )\), where \(X\) is a set and \(\rho :G\to A(X)\) is an action of \(G\) on \(X\).
We have a few examples of \(G\)-sets already:
\(S_n\) acting on \([n]\).
\(\operatorname{\mathbf{GL}}_2(\mathbb {R})\) acting on \(\mathbb {R}^2\).
\(G\) acting on itself on the left.
\(G\) acting on itself by inner autorphisms.
One of the nice things about \(G\)-sets is that, once we have one \(G\)-set \(X\), we can often get many others. Here’s one example.
Suppose \(X\) is a \(G\)-set, and write \(\mathcal{P}(X)\) for the power set of \(X\). I.e., \(\mathcal{P}(X)\) is the set of all subsets of \(X\). For each \(g\in G\) and \(S\subset X\), set \(gS = \{ gs: s\in S\} \). Then the map \(a:G\times \mathcal{P}(X)\to \mathcal{P}(X)\) given by \(a(g,S) = gS\) defines an action of \(G\) on \(\mathcal{P}(X)\).
Exercise (maybe obvious).
If \(X\) is a \(G\)-set and \(Y\subseteq X\), then we say that \(Y\) is a sub \(G\)-set if \(gy\in Y\) for all \(y\in Y\).
Clearly, a sub \(G\)-set of a \(G\)-set \(X\) is itself a \(G\)-set. Also, clearly orbits are sub \(G\)-sets. In fact, the orbits of \(G\) acting on \(X\) are exactly the minimal nonemtpy sub \(G\)-sets
Suppose \(H\leq G\). The \(G/H\) is a sub \(G\)-set of \(\mathcal{P}(G)\).
Suppose \(g_1,g_2\in G\). Then \(g_1(g_2H) = (g_1g_2)H\in G/H\).
Suppose \(X\) and \(Y\) are \(G\)-sets. A \(G\)-equivariant map from \(X\) to \(Y\) or a morphism of \(G\)-sets is a map \(f:X\to Y\) such that, for all \(g\in G\) and \(x\in X\), \(f(gx) = gf(x)\). We say a \(G\)-equivariant map \(f:X\to Y\) is an isomorphisms of \(G\)-sets if it is a bijection.
There are several things you can say about morphisms of \(G\)-sets, which are easy to prove and left as exercises. I’ll list some of them in the next proposition.
Suppose \(S:X\to Y\) and \(T:Y\to Z\) are \(G\)-equivariant maps of \(G\)-sets.
The image \(S(X)\) is a sub \(G\)-set of \(Y\).
The composition \(T\circ S\) is \(G\)-eqivariant.
The idenity map \(\operatorname{\mathrm{id}}_X\) is \(G\)-equivariant.
If \(S\) is a bijection, then its inverse map \(S^{-1}:Y\to X\) is a \(G\)-equivariant map.
We say that a \(G\)-set \(X\) is transitive if \(G\backslash X\) is a singleton. In other words, \(X\) is transitive if it has exactly one orbit.
Suppose \(H\leq G\). Then
\(G/H\) is a transitive \(G\)-set.
\(\operatorname{\mathrm{Stab}}_G H = H\).
Suppose \(X\) is a \(G\)-set, \(x\in X\) and \(g\in G\). Then \(\operatorname{\mathrm{Stab}}_G(gx) = g\operatorname{\mathrm{Stab}}_G(x) g^{-1}\).
Exercise.
Suppose \(H\leq G\) and \(g\in G\). Then \(\operatorname{\mathrm{Stab}}_G(gH) = gHg^{-1}\).
Suppose \(X\) is a \(G\)-set and \(x\in X\). Then the map \(m_x:G\to X\) given by \(m_x(g) = gx\) is a map of \(G\)-sets. Conversely, if \(f:G\to X\) is any map of \(G\)-sets, then \(f = m_{f(1)}\).
We have \(m_x(g_1g_2) = (g_1g_2)(x) = g_1(g_2 x) = g_1 m_x(g_2)\). So \(m_x\) is a map of \(G\)-sets.
On the other hand, if \(f:G\to X\) is any map of \(G\)-sets and \(x = f(1)\), then \(f(g) = f(g1) = gf(1) = gx = m_x(g)\).
Suppose \(H\leq G\). Then the map \(\pi _H:G\to G/H\) given by \(g\mapsto gH\) is a map of \(G\)-sets.
\(\pi _H(g) = g(1H)\).
Suppose \(X\) is a \(G\)-set and \(x\in X\). Set \(H = \operatorname{\mathrm{Stab}}_G(x)\), and set \(m = m_x:G\to X\). Then there exists a unique \(G\)-equivariant map \(j:G/H\to X\) such that \(m = j\circ \pi _H\). Moreover, \(j\) is an isomorphism of \(G\)-sets of from \(G/H\) to \(\operatorname{\mathrm{Orb}}_G(x)\).
Suppose \(g_1,g_2\in G\) and \(g_1H = g_2H\). Then \(g_1^{-1} g_2 \in H\). So \( g_1x = g_1 (g_1^{-1} g_2) x = g_2 x\). Therefore, we get a well-defined map \(j:G/H\to X\) by setting \(j(gH) = gx\). For \(g\in G\), we have \(j\circ \pi _H(g) = j(gH) = gx = m(g)\). So \(j\circ \pi _H = m\).
That proves the first part of the theorem. For the second part, since \(m = j\circ \pi _H\) and \(m(G) = \operatorname{\mathrm{Orb}}_G(x)\), we must have \(j(G/H) = \operatorname{\mathrm{Orb}}_G(x)\). On the other hand, suppose \(j(g_1H) = j(g_2H)\). Then \(g_1 x = g_2 x\). So \(x = g_1^{-1} g_2 x\), and, therefore, \(g_1^{-1}g_2\in \operatorname{\mathrm{Stab}}_G(x) = H\). But then \(g_1H = g_2 H\). It follows that \(j:G/H\to X\) is one-one with image equal \(\operatorname{\mathrm{Orb}}_G(x)\).
Suppose \(X\) is a \(G\)-set and \(x\in X\). Then we have \(|\operatorname{\mathrm{Orb}}_G(x)| = [G:H]\). In particular, if \(|G| {\lt}\infty \), \(|\operatorname{\mathrm{Orb}}_G(x)| \bigm | |G|\).
Clear.
Recall that the conjugacy class of an element \(h\in G\) is that set \([h] = \{ ghg^{-1}: g\in G\} \), and two elements in the same conjugacy class are said to be conjugate in \(G\). G is a disjoint union of its conjugacy classes, and we write \(h\sim h'\) if \(h\) and \(h'\) are conjugate. The conjugacy classes are just the orbits of \(G\) acting on itself by inner automorphisms, and, with that inner action, we have \(\operatorname{\mathrm{Stab}}_G(h) = \{ g\in G: ghg^{-1} = h\} = C_G(h)\). In other words, the stabilizer of \(h\) is just the centralizer of \(h\).
Suppose \(g\in G\). Then \([h]\) is isomorphic as a \(G\)-set to \(G/C_G(h)\). Consequently, if \(G\) is finite, \(|[h]| = |G|/|C_G(h)|\). In particular, if \(G\) is fnite, \(|[h]| \bigm | |G|\).
If \(H\leq G\) and \(g\in G\), then \(gHg^{-1}\leq G\). So we get an action of \(G\) on the set \(\operatorname{\mathrm{Sub}}G\) of subgroups of \(G\). Another way of saying this is that \(\operatorname{\mathrm{Sub}}G\) is a sub-\(G\)-set of the \(G\)-set \(\mathcal{P}(G)\) where \(G\) acts on subsets of \(\mathcal{P}(G)\) via the inner action. The stabilizer of a subgroup \(H\leq G\) under this action is \(\{ g\in G: gHg^{-1} = H\} = N_G(H)\). In other words, the stabilizer is just the normalizer of \(H\). Two subgroups \(H_1\) and \(H_2\) are said to be conjugate and we write \(H_1\sim H_2\) if \(H_2 = gH_1g^{-1}\) for some \(g\in G\). In other words, two subgroups are conjugate if and only if they are in the same orbit of \(G\) acting on \(\operatorname{\mathrm{Sub}}G\) via the inner action.
Suppose \(G\) is finite and \(H\leq G\). Then the number of conjugate subgroups of \(H\) is \(|G|/|N_G(H)|\).
Let \(G = S_3\).