lbwei space

Group Action

#Abstract Algebra, Group
2023/02/13

Table of Content

Definition (Group action)

Let \(X\) be a set and \(G\) a group. An action of \(G\) on \(X\) is a map \(*: G\times X\to X\) such that

  1. \(*(e, x) = x\) for all \(x\in X\), where \(e\) is the identity element of \(G\).
  2. \(*(g_1, *(g_2, x)) = *(g_1g_2, x)\) for all \(x\in X\) and all \(g_1, g_2 \in G\).

We write \(*(g, x)\) as \(gx\) for simplicity. Thus the above two conditions can be rephrased as:

  1. \(ex = x\) for all \(x\in X\).
  2. \(g_1(g_2x) = (g_1g_2)x\) for all \(x\in X\) and all \(g_1, g_2 \in G\).

Under these conditions, \(X\) is referred to as a \(G\)-set.

Remark

  • We say \(G\) acts on \(X\).
  • The second condition has nothing to do with associativity, since \(g_i\) and \(x\) belong to different sets.

Example (prove to be a \(G\)-set)

See this.

Proposition (Group actions as permutations)

Suppose that a group \(G\) acts on a set \(X\). Then for a given \(g\in G\), the function \(X\to X\) represented by \(g\) is one-to-one and onto, that is:

  1. If \(gx_1 = gx_2\), then \(x_1 = x_2\).
  2. For all \(y\in X\), there exists \(x\in X\) such that \(gx = y\).

Proof (Ignored.)

Q: Do we must have \(gx \in X\) for any \(g\in G, x\in X\)?

A: Certainly! Since the range of the mapping \(*\) is \(X\).

Remark

回想起來，permutation 是一 one-to-one 且 onto 的 mapping，與此處的 group action 不謀而和。也就是說，以上的定理可以如下表示：

Let \(S_X\) be the set of permutations on \(X\). A group action \(*\) of \(G\) on \(X\) can be regarded as a group homomorphism \(\rho: G\to S_X\) such that

\[*(g, x) = gx = \rho(g)x,\]

where \(\rho(g) \in S_X\) and hence \(\rho(g): X\to X\).

於是

\[\text{An action of } G \text{ on } X \iff \text{a group homomorphism from } G \text{ to } S_X.\]

注意，我們討論的是 \(X\) 的 permutation，是否存在 \(\rho(g_i) = \rho(g_j), i\not =j\) 並不重要！

相當於問 \(g_ix\) 是否等於 \(g_jx\)。

底下證明 \(\rho\) 是 group homomorphism：

Proof

For all \(g_1, g_2\in G\),

\[\begin{align*} *(g_1, *(g_2, x)) &= \rho(g_1)\big(\rho(g_2)x\big) \\ &= \big(\rho(g_1)\rho(g_2)\big)x \\ &= *(g_1g_2, x) = \rho(g_1g_2)x. \tag*{$\blacksquare$} \\ \end{align*}\]

Examples

Below \(G\)s act on \(X\)s, repsectively.

  • Let \(G\) be a group of symmetries of an object \(X\).
  • Let \(G=S_n\), \(X=\{1, 2,\cdots, n\}\).
    • \(S_n\) acts on \(X\) by \(\sigma x = \sigma(x)\).
  • Let \(G=\text{GL}(n, \Bbb{R}), X=\Bbb{R}^n\).
    • \(\text{GL}(n, \Bbb{R})\) acts on \(X\) by \((M, v) \mapsto Mv\).
  • Let \(X=G\); \((g, x)\mapsto gx\).
  • Let \(X=G/H\); \((g, aH) \mapsto (ga)H\).
  • Let \(X\) be a non-empty set and \(n\) be a positive integer. Let \(G\) be any subgroup of \(S_n\).
    • \(S_n\) acts on \(X^n\) by \((\sigma, (x_1, \cdots, x_n)) \mapsto (x_{\sigma(1)}, \cdots, x_{\sigma(n)})\).
  • The map \(*: G\times G\to G\) given by \(*(g, x) = gxg^{-1}\) is a group action, since \(\rho: G \to \text{Aut}(G)\) is a group homomorphism.

Definition (faithful and transitive)

A group action \(*\) by \(G\) on \(X\) is said to be faithful if \(e\in G\) (identity) is the only element that leaves every element of \(X\) unchanged. That is to say, \(*\) is faithful iff \(\text{ker}(\rho) = \{e\}\), where \(\rho: G\to S_X\), by this remark.

Then, \(*\) is called transitive if for all \(x_1,x_2\in X\), there exists \(g\in G\) such that \(gx_1 = x_2\).

Remark

If \(*\) is transitive, the induced homomorphism must be non-trivial.

不 transitive 也可以 non-trivial！存在 \(gx_1=x_2\) 就代表 \(\rho(g)\) 不是 identity map。

Orbit and Stabilizer

Definition (orbit)

Let \(X\) be a \(G\)-set. The \(G\)-orbit of \(x\) is defined as

\[G_x := \{gx\mid g\in G\}.\]

See Symmetry Groups on N-gons. Note that \(G\)-orbit itself is a \(G\)-set.

令 \(H\) 為 \(G\) 的 subgroup。若 \(H\) 作用在 \(G\) 上，則對於任意 \(g\in G\)，\(H\)-orbit of \(g\) 就是 \(H\) 的 right coset！也就是說，\(H_g = Hg\)。

Lemma (transitive and orbit)

If an action of \(G\) on \(X\) is transitive, then for any \(x\in X\),

\[G_x = X.\]

Proof

By the definition of transitivity.

Definition (stabilizer)

Let \(X\) be a \(G\)-set. The set \(\text{Stab}_G(x) = \{g\in G\mid gx = x\}\) for all \(x \in X\), is called the stabilizer of \(x\) in \(G\).

\(\text{Stab}_G(x)\) is a subgroup of \(G\).

Theorem (eq. relation)

Let \(X\) be a \(G\)-set. Define a relation \(\sim\) on \(X\) by \(x_1\sim x_2\) if there exists \(g\in G\) such that \(gx_1 = x_2\). Then \(\sim\) is an equivalence relation.

See Relation.

Remark

The equivalence relation defined above defines a partition of \(X\), and the cell of \(x\in X\) is exaclty the \(G\)-orbit of \(x\). Hence, \(X\) can be written as a disjoint union of \(G\)-orbits.

為什麼是 \(G\)-orbit？因為 \(G\)-orbit 收集 \(gx\)，而如果 \(gx = x'\)，那麼 \(x\sim x'\)！

Corollary (disjoint union of orbits)

\(X\) can be written as a disjoint union of \(G\)-orbits.

Theorem (the orbit-stabilizer theorem)

For all \(x\in X\), where \(X\) is a \(G\)-set, we have \(\vert G_x\vert = [G:\text{Stab}_G(x)]\). If \(\vert G\vert\) is finite, we have

\[\vert G\vert = \vert G_x\vert \vert\text{Stab}_G(x) \vert.\]

See Symmetry Groups on N-gons.

Proof

We are to show that there is a bijection between \(G_x\) anf the left cosets \(G/H\), where \(H=\text{Stab}_G(x)\).

Let \(\phi: gH\to G_x\) be the function given by \(\phi(gH) = gx\). First, we should establish the well-defiinedness of it.

well-definedness

Suppose \(g_1H = g_2H\), and thus \(g_1 = g_2h\) for some \(h\in H\). Then we have \(\phi(g_1H) = g_1x = g_2hx = g_2x = \phi(g_2H)\), since \(h\in \text{Stab}_G(x)\) and \(hx = x\). Therefore \(\phi\) is well-defined.

Then show \(\phi\) is bijective.

one-to-one

Suppose that \(\phi(g_1H) = \phi(g_2H)\). This means that \(g_1x = g_2x \implies g_1^{-1}g_2x = x\). Hence, \(g_1^{-1}g_2 \in H\), which implies that \(g_1H=g_2H\).

onto

Let \(y\in G_x\) Then \(y=gx\) for some \(g\in G\). It follows that \(\phi(gH) = gx = y\). ◼