This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
When dealing with complex algebraic structures, one effective strategy is to examine how a given algebraic object acts on another, rather than analyzing the structure directly. We are particularly interested in group actions; but as always, we first consider the more general case of a monoid acting on a set.
Monoids Acting on Sets
Definition 1 Fix a monoidal category \((\mathcal{A},\otimes, I)\) and a monoid object \((A,\cdot, 1)\) in \(\mathcal{A}\). A morphism \(\rho: A\otimes E\rightarrow E\) is called a left action of \(A\) on an object \(E\in\obj(\mathcal{A})\) if the following two diagrams commute.
Here, \(I\otimes E \rightarrow E\) is the left unitor. We denote this situation by \(A\circlearrowright E\).
Similarly, a morphism \(\rho: E\otimes A\rightarrow E\) is called a right action of \(A\) on an object \(E\in\obj(\mathcal{A})\) if the following two diagrams commute.
Again, \(E\otimes I \rightarrow E\) is the right unitor. We denote this situation by \(E \circlearrowleft A\).
Fix a monoid object \((M,\cdot,1)\) in the monoidal category \((\Set,\times, I)\). Then we may consider a left action of \(M\) on an arbitrary set \(E\). By [Set Theory] §Product of Sets, ⁋Proposition 4, we have
\[\Hom_\Set(M\times E,E)\cong\Hom_\Set(M,\Hom_\Set(E,E))\cong\Hom_\Set(M, \End(E))\]so any left action defines a function \(M \rightarrow \End(E)\). The commutativity of the two diagrams in Definition 1 is equivalent to this function being a monoid homomorphism.
In other words, for \(M\) to act on \(E\) from the left means that, for any \(\alpha,\beta\in M\) and \(x\in E\), the equalities
\[(\alpha\beta)\cdot x=\alpha\cdot(\beta\cdot x),\qquad e\cdot x=x\]hold.
In general, we consider actions from the left as above, but sometimes acting from the right is more natural. The following definition shows that these are in fact equivalent.
Definition 2 For any magma \((M,\ast)\), the opposite magma \((M^\op,\ast^\op)\) is defined as follows:
- As a set, \(M^\op=M\).
- For any \(x,y\in A^\op\), we define \(x\ast^\op y\) to be \(y\ast x\).
We can verify that a right \(M\)-action is the same as a left \(M^\op\)-action. Rewriting this, we have
\[x\cdot(\beta\alpha)=(x\cdot\beta)\cdot\alpha,\qquad x\cdot e=x\]Thus left actions and right actions differ only in notation and are essentially the same concept. Henceforth, when developing general theory, we assume all actions are left actions.
Example 3 Let a monoid \(M\) and an \(M\)-set \(E\) be given. Then \(\mathcal{P}(E)\) naturally carries an \(M\)-set structure. For any \(\alpha\in M\) and \(A\in \mathcal{P}(E)\), define \(\alpha\cdot A\) by the formula
\[\alpha\cdot A=\{\alpha\cdot a\mid a\in A\}\]Then
\[(\alpha\beta)\cdot A=\{(\alpha\beta)\cdot a\mid a\in A\}=\{\alpha\cdot(\beta\cdot a)\mid a\in A\}=\alpha\cdot\{\beta\cdot a\mid a\in A\}=\alpha\cdot(\beta\cdot A)\]so this defines an \(M\)-action on \(\mathcal{P}(E)\).
For convenience, we make the following definition.
Definition 4 When a monoid \(M\) defines a left action on a set \(E\), we call \(E\) together with this action a (left) \(M\)-set.
\(M\)-set Homomorphisms
Definition 5 Fix a monoid \(M\), and let \(E,E'\) be \(M\)-sets. A function \(f:E\rightarrow E'\) is called an \(M\)-set homomorphism if for all \(x\in E\) and \(\alpha\in M\),
\[f(\alpha\cdot x)=\alpha\cdot f(x)\]holds.
We easily verify that the composition of \(M\)-set homomorphisms is again an \(M\)-set homomorphism, and that the identity function is an \(M\)-set homomorphism. Thus the collection of (left) \(M\)-sets forms a category, which we denote by \(\lset{M}\).
Fix a monoid homomorphism \(\phi:M \rightarrow M'\). Then for any \(M'\)-set \(E\), the composition
\[M\overset{\phi}{\longrightarrow}M'\overset{\rho}{\longrightarrow}\End(E)\]allows us to regard \(E\) as an \(M\)-set. We denote this action by \(\phi^\ast\rho\). Explicitly, \(\phi^\ast\rho\) is the action defined for any \(\alpha\in M\) and \(x\in E\) by
\[(\phi^\ast\rho)(\alpha)(x)=\rho(\phi(\alpha))(x)\]Now suppose that two \(M'\)-actions \(\rho:M' \rightarrow \End(E)\) and \(\rho':M' \rightarrow \End(E)\) are given, along with an \(M'\)-homomorphism \(f:E \rightarrow E'\) between them. Then for any \(\alpha\in M\) and \(x\in E\),
\[f((\phi^\ast\rho)(\alpha)(x))=f(\rho(\phi(\alpha))(x))=\rho'(\phi(\alpha))(f(x))=(\phi^\ast\rho')(f(x))\]holds. Thus any monoid homomorphism \(\phi:M \rightarrow M'\) defines a functor from \(\lset{M'}\) to \(\lset{M}\). In particular, if \(\iota\) is the inclusion of a submonoid, this is precisely the restriction of the monoid action.
On the other hand, if \((E_i)\) is a collection of \(M\)-sets, their product \(\prod E_i\) becomes an \(M\)-set under the action defined by
\[\alpha\cdot(x_i)_{i\in I}=(\alpha\cdot x_i)_{i\in I}\]Similarly, a subset \(F\) of an \(M\)-set \(E\) is called an \(M\)-subset if it satisfies
\[x\in F\implies \alpha\cdot x\text{ for all $\alpha\in F$}\]Also, if an equivalence relation \(\sim\) on an \(M\)-set is compatible with the \(M\)-action, that is, if
\[x\sim y\implies\alpha\cdot x\sim\alpha\cdot y\]always holds, then \(E/\mathnormal{\sim}\) naturally carries an \(M\)-set structure.
Stabilizer, Fixer
Definition 6 Let \(A\) be a subset of an \(M\)-set \(E\).
- The stabilizer of \(A\) is the set of all \(\alpha\) such that \(\alpha A\subseteq A\), denoted \(\stab (A)\).
- The strict stabilizer of \(A\) is the set of all \(\alpha\) such that \(\alpha A=A\), denoted \(\Stab(A)\).
- The fixer of \(A\) is the set of all \(\alpha\) such that \(\alpha a=a\) for all \(a\in A\), denoted \(\Fix(A)\).
For any subset \(A\), we have \(\Fix(A)\subseteq \Stab(A)\subseteq \stab(A)\). Also, \(e\in\Fix(A)\) is immediate.
Proposition 7 For an \(M\)-set \(E\) and its subset \(A\), the sets \(\stab(A)\), \(\Stab (A)\), and \(\Fix(A)\) are submonoids of \(M\).
Proof
It suffices to show that these sets are closed under multiplication. If \(\alpha,\beta\in\stab(A)\), then
\[(\alpha\beta)A=\alpha(\beta A)\subseteq \alpha A\subseteq A\]so \(\alpha\beta\in \stab(A)\). Similarly, if \(\alpha,\beta\in\Stab(A)\), then
\[(\alpha\beta)A=\alpha(\beta A)=\alpha A=A\]so \(\alpha\beta\in \Stab(A)\). Finally, if \(\alpha,\beta\in\Fix(A)\), then for any \(a\in A\),
\[(\alpha\beta)a=\alpha(\beta a)=\alpha a=a\]so \(\alpha\beta\in \Fix(A)\).
Corollary 8 Let \(G\) be a group. For a \(G\)-set \(E\) and a subset \(A\subseteq E\), the sets \(\Stab (A)\) and \(\Fix(A)\) are subgroups of \(G\), and in particular \(\Fix(A)\) is a normal subgroup of \(\Stab(A)\).
Proof
For the first claim, it suffices to show that these sets are closed under taking inverses. For any \(\alpha\in\Stab(A)\), we have
\[A=(\alpha^{-1}\alpha)A=\alpha^{-1}(\alpha A)=\alpha^{-1}A\]and for any \(\alpha\in\Fix(A)\) and \(a\in A\),
\[a=(\alpha^{-1}\alpha)a=\alpha^{-1}(\alpha a)=\alpha^{-1}a\]which establishes the first claim. For the second claim, let \(\alpha\in\Fix(A)\) and \(\beta\in\Stab(A)\). Then for any \(a\in A\), we compute
\[(\beta\alpha\beta^{-1})a=\beta(\alpha(\beta^{-1}a))=\beta\beta^{-1}a=a\]so \(\beta\alpha\beta^{-1}\in\Fix(A)\), as desired.
From the proof of the above corollary, we see that when a group \(G\) acts on a set \(E\), each \(\rho_g\) is necessarily bijective. That is, \(\im\rho\subseteq \Aut(E)\) always holds.
Inner Automorphisms
We now consider the case where the set \(E\) carries additional structure. For instance, suppose \(E\) itself has a monoid structure and a monoid \(M\) acts on \(E\). Then the \(M\)-action is given by a monoid homomorphism \(M \rightarrow\End(E)=\End_\Mon(E)\).
In particular, consider the case where a group \(G\) acts on itself. That is, suppose \(\rho:G\rightarrow\End(G)=\End_\Grp(G)\) is given. From the proof of Corollary 8, we know that the image of \(\rho\) consists entirely of bijections. Since a bijective group homomorphism is always a group isomorphism (§Algebraic Structures, ⁋Definition 6), if \(G\) acts on itself then this action must be of the form of a group homomorphism \(G \rightarrow \Aut(G)\).
Among group actions on itself, the following is particularly worth remembering.
Proposition 9 For any element \(g\) in a group \(G\), define \(\rho_g\in\Aut(G)\) by the formula
\[\rho_g(x)=gxg^{-1}\]Then the map \(\rho:g\mapsto \rho_g\) is a group homomorphism.
Proof
For any \(x,y\in G\),
\[\rho_g(xy)=g(xy)g^{-1}=(gxg^{-1})(gyg^{-1})=\rho_g(x)\rho_g(y)\]so \(\rho_g\) is a group homomorphism, and hence \(\im\rho\subseteq\Aut(G)\).
Moreover, for any \(g,h\in G\) and \(x\in G\),
\[\rho_{gh}(x)=(gh)x(gh)^{-1}=g(hxh^{-1})g^{-1}=(\rho_g\circ\rho_h)(x)\]so \(\rho_{gh}=\rho_g\circ\rho_h\). Thus the map \(\rho:g\mapsto \rho_g\) is a group homomorphism from \(G\) to \(\Aut(G)\).
Definition 10 Let \(G\) be a group. The automorphism \(\rho_g\) from Proposition 9 is called the inner automorphism defined by \(g\), and the set of all such automorphisms is denoted by \(\Inn(G)\).
Proposition 11 For a group \(G\), the set \(\Inn(G)\) of inner automorphisms is a normal subgroup of \(\Aut(G)\).
Proof
Since \(\Inn(G)\) is the image of the group homomorphism \(\rho:G\rightarrow\Aut(G)\), it is clearly a subgroup of \(\Aut(G)\). Thus it suffices to show that \(\Inn(G)\) is a normal subgroup.
Fix any \(f\in\Aut(G)\) and \(g\in G\). We must show that \(f\circ\rho_g\circ f^{-1}\in \Inn(G)\). For any \(x\in G\),
\[(f\circ\rho_g\circ f^{-1})(x)=f(gf^{-1}(x)g^{-1})=f(g)xf(g^{-1})=\rho_{f(g)}(x)\]which is clear.
Moreover, \(\rho:G\rightarrow\Inn(G)\) is surjective, so by the first isomorphism theorem,
\[G/\ker\rho\cong\Inn(G)\]holds. The kernel \(\ker\rho\) also has a special name.
Definition 12 Let \(G\) be a group. For the group homomorphism \(\rho:G\rightarrow\Inn(G)\) defined in Proposition 9, the kernel \(\ker\rho\) is called the center of \(G\) and is denoted by \(C(G)\).
By definition,
\[g\in\ker\rho\iff\rho_g=\id_G\iff gxg^{-1}=x\quad\text{for all $x\in G$}\]so the fixer \(\Fix(G)\) in the situation where \(G\) acts on itself by inner automorphisms is precisely \(C(G)\). More generally, for any subset \(A\subseteq G\) we define the centralizer \(C_G(A)\) of \(A\) to be its fixer \(\Fix(A)\). Similarly, we define the normalizer \(N_G(A)\) of \(A\) to be \(\Stab(A)\).
Orbit-Stabilizer Theorem
We now return to group actions on general sets. First, we make the following definition.
Definition 13 Let a group \(G\) act on a set \(E\). The orbit of an element \(x\in E\) is the set
\[G\cdot x=\{g\cdot x\mid g\in G\}\]Then the relation on \(E\) defined by
\[x\sim y\iff G\cdot x=G\cdot y\tag{$\ast$}\]is an equivalence relation, so the quotient set \(E/{\sim}\) is defined and consists of orbits.
Theorem 14 (Orbit-stabilizer theorem) Let a group \(G\) act on a set \(E\). Then
\[\lvert G\cdot x\rvert=[G:\Stab(x)]\]holds.
Proof
Define a function \(p:G \rightarrow G\cdot x\) by \(g\mapsto g\cdot x\). By definition of \(G\cdot x\), this function is surjective. Moreover, \(p(g_1)=p(g_2)\iff g_1^{-1}g_2\in \Stab(x)\), so the desired result follows from the canonical decomposition discussed after [Set Theory] §Examples of Equivalence Relations, ⁋Proposition 7.
Therefore, if \(G\) is finite, then by §Quotient Groups, ⁋Proposition 5 we obtain
\[\lvert G\cdot x\rvert=\frac{\lvert G\rvert}{\lvert\Stab(x)\rvert}\tag{$\ast\ast$}\]Similarly, suppose \(G\) is finite and acts on a finite set \(E\). Define \(E^g\) to be the set of elements fixed by \(g\):
\[E^g=\{x\in E\mid g\cdot x=x\}\]Then
\[\sum_{g\in G}\lvert E^g\rvert=\# \{(g, x)\in G\times E: g\cdot x=x\}=\sum_{x\in X}\lvert \Stab(x)\rvert\]holds. From \((\ast\ast)\) we have
\[\sum_{x\in X}\lvert \Stab(x)\rvert=\sum_{x\in X}\frac{\lvert G\rvert}{\lvert G\cdot x\rvert}\]On the other hand, considering the quotient set \(E/{\sim}\) defined by \((\ast)\), the above sum becomes
\[\sum_{x\in X}\frac{\lvert G\rvert}{\lvert G\cdot x\rvert}=\lvert G\rvert\sum_{O\in E/{\sim}}\sum_{x\in O}\frac{1}{\lvert O\rvert}=\lvert G\rvert\sum_{O\in E/{\sim}} 1=\lvert G\rvert\lvert E/{\sim}\rvert\]From this we obtain the following lemma.
Lemma 15 Let a finite group \(G\) act on a finite set \(E\), and let \(E/{\sim}\) be the quotient set of \(E\) consisting of orbits. Then
\[\lvert E/{\sim}\rvert=\frac{1}{\lvert G\rvert}\sum_{g\in G}\lvert E^g\rvert\]holds.
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
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