CHAPTER 11 · MA-0561

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Group Actions

Elementary definitions and properties

Definition (Group Action)

Let \((G, \cdot)\) be a group and let \(X\) be a set. We say that \(G\) acts on \(X\) (or that \(X\) is a \(G\)-set) if there exists a function \(\alpha: G \times X \to X\) such that

  1. \(\forall g,h \in G \quad\forall x \in X \quad \Big( \alpha\big(g, \alpha(h,x)\big) = \alpha(g \cdot h, x)\Big)\),
  2. \(\forall x \in X \quad\Big( \alpha(1_{G}, x) = x \Big)\).

Notation (Action of an element)

If \(G\) acts on \(X\), given \(g \in G\), we denote by \(\alpha_{g}:X \to X\) the map such that \(x \mapsto \alpha(g,x)\). If we know the specific action, it is common to write \(g \cdot x\).

Note (Immediate properties)

  1. For all \(g,h \in G\) and for every \(x \in X\), we have that \(\alpha_{g}(\alpha_{h}(x)) = \alpha_{g \cdot h}(x)\).
  2. \[\alpha_{1_{G}} = \mathrm{id}_{X}\]

Examples (Standard actions)

  1. Let \(n \in \mathbb{N}\). Consider \(G = S_{n}\) and \(X = \{ 1,\dots,n \}\). Define \(\alpha: S_{n} \times X \to X\) such that \((\sigma, i) \mapsto \sigma(i)\). Note that if \(\gamma,\sigma \in S_{n}\), given \(i \in X\),

    \[\alpha_{\sigma}(\alpha_{\gamma}(i)) = \alpha_{\sigma}(\gamma(i)) = (\sigma \circ \gamma)(i) = \alpha_{\sigma \circ \gamma} (i).\]

Moreover, \(\alpha_{\mathrm{id}}(i) = \mathrm{id}(i) = i\). Therefore, \(\alpha\) is a group action and \(S_{n}\) acts on \(X\).

  1. Let \(G = (\mathbb{Z}, +)\) and \(X = \mathbb{R}\). Define \(\tau: G \times X \to X\) such that \((n,x) \mapsto x+ n\). Note that, given \(m,n \in \mathbb{Z}\) and \(x \in \mathbb{R}\),

    \[\tau_{n}(\tau_{m(x)}) = \tau_{n}(x+m) = (x+m)+n = x + (m+n) = \tau_{m+n}(x).\]

Moreover, \(\tau_{0}(x) = x+0 = x\). Therefore, \(\mathbb{Z}\) acts on \(\mathbb{R}\).

  1. Let \(G = \mathrm{GL}_{n}(\mathbb{R})\) and \(X = \mathbb{R}^{n}\). Define \(\alpha: G \times X \to X\) such that \((A,v) \mapsto A v\). Showing that \(\alpha\) is a group action is left as an exercise.

  2. Given a group \(G\), define \(\mu:G \times G \to G\) such that \((g,h) \mapsto g \cdot h\). It is easy to see that \(\mu\) is a group action. Thus, \(G\) acts on itself.

Lemma (The action is bijective on \(X\))

Let \(G\) be a group and \(X\) a set. Suppose that \(\alpha:G \times X \to X\) is a group action. Then, if \(g \in G\), the map \(\alpha_{g}: X \to X\) such that \(x \mapsto \alpha(g,x)\) is a bijection, i.e., \(\alpha_{g} \in S_{x}\).

Proof: Consider \(\alpha_{g^{-1}}:X \to X\) such that \(x \mapsto \alpha(g^{-1}, x)\). Note that

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

In the same way, \((\alpha_{g^{-1}} \circ \alpha_{g}) = \mathrm{id}_{X}\), from which we conclude that the map is bijective.

Theorem (Actions and homomorphisms are in bijection)

Let \(G\) be a group and \(X\) a set. Define

\[\begin{aligned} \Omega:&= \{ \alpha: G \times X \to X: \alpha \text{ is an action of } G \text{ on }X\},\\ \Sigma:&= \{ \phi: G \to S_{X} : \phi \text{ is a group homomorphism} \}. \end{aligned}\]

Then \(\Omega\) and \(\Sigma\) are in bijection

Proof: Define \(\Phi: \Omega \to \Sigma\) as follows. Given an action \(\alpha: G \times X \to X\), define \(\phi:G \to S_{X}\) such that \(\phi(g) = \alpha_{g}\) where \(\alpha_{g}:X \to X\) is such that \(x \mapsto\alpha(g,x)\). It is easy to see that \(\phi\) is a homomorphism and that \(\alpha_{g} \in S_{X}\). Take \(\Phi(\alpha) = \phi\). Now, define \(\Psi: \Sigma \to \Omega\) as follows. Given a homomorphism \(\phi:G\to S_{X}\), let \(\alpha:G \times X \to X\) be such that \((g,x) \mapsto \phi(g)(x) \in X\). Note that \(\alpha\) is an action, since if \(g,h \in G\) and \(x \in X\), we have that

\[\begin{aligned} \alpha(g \cdot h,x) = \phi(gh) (x) &= (\phi(g) \circ \phi(h))(x) \\ &= \phi(g) \big(\phi(h)(x)\big) \\ &=\phi(g) (\alpha(h, x)) \\ &=\alpha(g, \alpha(h,x)), \end{aligned}\]

and moreover, given \(x \in X\), \(\alpha(1_{G}, x) = \phi(1_{G})(x) = \mathrm{id}_{X}(x) = x\). Take \(\Psi(\phi) = \alpha\). Verifying that \(\Phi \circ \Psi = \mathrm{id}_{\Sigma}\) and that \(\Psi \circ \Phi = \mathrm{id}_{\Omega}\) is left as an exercise.

Theorem (Cayley)

Given a group \(G\), \(G\) is isomorphic to a subgroup of \(S_{G}\).

Proof: Consider \(\mu_{g}:G \to G\) such that \(h \mapsto g\cdot h\). Let \(\phi: G \to S_{G}\) be such that \(g \mapsto \mu_{g}\). We have already seen that \(\phi\) is a homomorphism. Let us see that \(\phi\) is injective:

\[g \in \operatorname{Ker}(\phi) \iff \phi(g) = \mu_{g}= \mathrm{id}_{G} \iff g \cdot x = x \quad \forall x \in G \iff g =1_{G},\]

so that \(\operatorname{Ker}(\phi) = \{ 1_{G} \}\). Now, using the first isomorphism theorem, we obtain that

\[G / \operatorname{Ker}(\phi) = G/\{ 1_{G} \} = G\cong \operatorname{Im}(\phi) \leq S_{G},\]

from which we conclude the result.