Group actions

Section 1.7 Group actions

Definition 1.7.1. Group action.

A (left) group action is a triple \((G,A,\cdot)\text{,}\) where \(G\) is a group, \(A\) is a nonempty set, and \(\cdot\) is a binary operation

\begin{align*} G\times A \amp \rightarrow A\\ (g,a) \amp \mapsto g\cdot a \end{align*}

satisfying the following axioms.

Associativity.
\(g\cdot(h\cdot a)=(gh)\cdot a\) for \(g,h\in G\) and \(a\in A\text{.}\)
🔗

🔗
Identity action.
\(e\cdot a=a\) for all \(a\in A\text{.}\)
🔗

🔗

🔗

Example 1.7.2. Group actions.

The group \(\Isom(\R^n)\) acts on \(\R^n\) via evaluation: i.e., given \(g\in \Isom(\R^n)\) and \(v\in \R^n\text{,}\) we define \(g\cdot v=g(v)\text{.}\)
🔗

🔗
The group \(D_n\) acts on the regular polygon \(P_n\) via evaluation: i.e., given \(g\in D_n\) and \(Q\in P_n\text{,}\) we define \(g\cdot Q=g(Q)\text{.}\)
🔗

Furthermore, \(D_n\) acts on the set of vertices \(V\) of \(P_n\) and the set \(E\) of its edges.
🔗

🔗
The group \(S_n\) acts on \(\{1,2,\dots, n\}\) via evaluation: i.e., given \(\sigma\in S_n\) and \(k\in \{1,2,\dots, n\}\text{,}\) we define \(\sigma\cdot k=\sigma(k)\text{.}\)
🔗

🔗
Any group \(G\) acts on itself (treated as a set) via multiplication on the left:

\begin{align*} G\times G \amp \rightarrow G\\ (g,h) \amp \mapsto g\star h=gh\text{.} \end{align*}

Note that multiplication on the right

\begin{align*} (g,h) \amp \mapsto g\star h=hg \end{align*}

is not in a general a group action since by definition we have

\begin{align*} g\star (h\star k) \amp= (h\star k)g \\ \amp = (kh)g\\ \amp = k(hg) \end{align*}

and

\begin{align*} (gh)\star k \amp = k(gh)\text{.} \end{align*}

🔗

🔗
Any group \(G\) acts on itself (treated as a set) via conjugation:

\begin{align*} G\times G \amp \rightarrow G\\ (g,h) \amp \mapsto g\star h=ghg^{-1} \end{align*}

🔗

🔗
Given \(R\in \{\Z, \Q, \R, \C\}\) or \(R=\Z/m\Z\) the group of units \(R^*\) acts on \(R\) via multiplication:

\begin{equation*} (u,r)\mapsto ur\text{.} \end{equation*}

🔗

🔗

🔗

Theorem 1.7.3. Group actions and homomorphisms.

Assume \(G\) acts on the the nonempty set \(A\) via the binary operation

\begin{equation*} (g,a)\mapsto g\cdot a\text{.} \end{equation*}

Given \(g\in G\text{,}\) the map \(\sigma_g\colon A\rightarrow A\) defined as

\begin{equation*} \sigma_g(a)=g\cdot a\text{.} \end{equation*}

is a permutation of \(A\text{:}\) i.e., \(\sigma_g\in S_A\text{.}\)

🔗

🔗
The map \(\phi\colon G\rightarrow S_A\) defined as

\begin{equation*} \phi(g)=\sigma_g \end{equation*}

is a group homomorphism.

🔗

🔗
There is a bijection between the set of all group actions

\begin{equation*} G\times A\xrightarrow{\star} A \end{equation*}

and the set of all homomorphisms

\begin{equation*} \phi\colon G\rightarrow S_A \end{equation*}

given by

\begin{align*} \{\text{group actions } G\times A\xrightarrow{\star} A\} \amp \longrightarrow \{\phi\colon G\rightarrow S_A\mid \phi \text{ a homomorphism}\} \\ \star\amp \longmapsto \phi_\star\colon G\rightarrow S_A \end{align*}

where given \(g\in G\text{,}\) \(\phi_\star(g)\) is the permutation \(\sigma_g\in S_A\) defined as

\begin{equation*} \sigma_g(a)=g\star a\text{.} \end{equation*}

🔗

🔗

🔗

Proof.

By definition \(g\cdot a\in A\) for all \(a\in A\text{,}\) so the recipe for \(\sigma_g\) does indeed yield a well-defined function \(\sigma_g\colon A\rightarrow A\text{.}\) To prove that \(\sigma_g\) is a permutation, we will show that it has an inverse: in fact, we claim \(\sigma_g^{-1}=\sigma_{g^{-1}}\text{.}\) We must show that composition of these two functions, in any order, yields the identity map on \(A\text{.}\) This follows from the group action axioms:

\begin{align*} \sigma_g(\sigma_{g^{-1}}(a)) \amp =g\cdot (g^{-1}\cdot a)\\ \amp =(gg^{-1})\cdot a\\ \amp =e\cdot a\\ \amp =a\\ \sigma_{g^-1}(\sigma_{g}(a)) \amp =g^{-1}\cdot (g\cdot a)\\ \amp =(g^{-1}g)\cdot a\\ \amp =e\cdot a\\ \amp =a\text{.} \end{align*}

🔗

🔗
🔗

🔗

🔗

Definition 1.7.4. Permutation representation.

Given a group \(G\text{,}\) a homomorphism \(\phi\colon G\rightarrow S_A\) for some nonempty set \(A\) is called a permutation representation.

🔗

Example 1.7.5. Isomorphisms.

Prove: \(D_3\) is isomorphic to \(S_3\text{.}\) Use a group action to produce your homomorphism.

🔗

Solution.

Recall the definition of \(D_3\) as the set of rigid motions of \(\R^2\) fixing the equilateral triangle \(T\) centered at the origin with one vertex at \((1,0)\text{.}\) Let \(V=\{Q_1,Q_2,Q_3\}\) be the set of vertices of \(T\text{.}\) Since \(D_3\) acts on \(V\) (as described in Example 1.7.2), by Theorem 1.7.3 we get a group homomorphism \(\phi \colon D_3\rightarrow S_V\) that maps a function \(f\in D_3\) to the permutation it defines on \(V\) via evaluation. This means that \(\phi(f)=f\vert_V\text{,}\) the function restriction of \(f\) to the set \(V\) of vertices of \(T\text{.}\) We claim \(\phi\) is in fact an isomorphism. Since \(\abs{D_3}=\abs{S_V}=6\text{,}\) it suffices to show that \(\phi\) is injective. To this end, we note that given \(f,g\in D_3\text{,}\) we have

\begin{align*} \phi(f)=\phi(g) \amp \implies f\vert_V=g\vert_V \\ \amp \implies f(Q_i)=g(Q_i) \text{ for all } Q_i\in V\\ \amp \implies f=g\text{,} \end{align*}

where the last implication follows from the fact that two rigid motions of \(\R^2\) are uniquely determined by their values at three non-colinear points.

🔗

We have shown that \(\phi\) is an isomorphism and thus that \(D_3\cong S_V\text{.}\) Lastly, by Example 1.6.7, we have that \(S_V\cong S_3\) and thus that \(D_3\cong S_3\text{.}\) (The isomorphism relation between groups is easily seen to be an equivalence relation since compositions of isomorphisms are isomorphisms.)

🔗

Prev Top Next