Group Actions

Chentian Wu

2024-09-25

\newcommand{\R}{\mathbb R} \newcommand{\Z}{\mathbb Z} \newcommand{\Stab}{\text{Stab}} \newcommand{\Ker}{\text{ker}} \newcommand{\N}{\mathbb N} \newcommand{\Ord}{\text{ord}} \newcommand{\Op}{\text{op}} \newcommand{\lll}{\langle} \newcommand{\rrr}{\rangle} \newcommand{\Iso}{\cong} \newcommand{\P}{\mathcal P}

1. Group Actions

Definition. Group action, $$G$$ -set
Let $$G$$ be a group, $$X$$ be a set. A (left) $$G$$ -action on $$X$$ is a map $\cdot : G\times X \to X$ , such that

$\forall x\in X$ , $e\cdot x = x$ .
$\forall x\in X, a, b \in G, a\cdot (b\cdot x)=(ab)\cdot x$ .
The pair $(X, \cdot)$ is called a left $$G$$ -set.

when there is no ambiguity, we can omit this operation notation “ $\cdot$ ”

Example. several actions

Any group $$(G, *)$$ can be seen as a left $$G$$ -set by $$ga = g * a$$ . This is called the left action.
Any group $$(G, *)$$ can be seen as a left $$G$$ -set by $ga = g ∗ a ∗ g^{−1}$ . This is called the conjugate action.
It could be checked from definition that, if $$G$$ acts on $$X$$ and $$Y$$ , then $$G$$ acts on the direct product $X\times Y$ of $$X$$ and $$Y$$ , in a very intuitive way: $$g(x,y)=(gx,gy)$$ . This is good because we can construct left $$G$$ -sets from existing left $$G$$ -sets.
It could be checked from definition that, if $$G$$ acts on a nonempty set $$X$$ , then $$G$$ acts on power set $\P(X)$ of $$X$$ , in also a very intuitive way, $gA:=\{ga\mid a\in A\}$ . This action cannot be free or transitive.
For any nonempty set $$X$$ , $$S_X$$ acts on $$X$$ via the action $\sigma\cdot x=\sigma(x)$ . This utilizes the fact that we can construct a group simply a set itself.

Definition. Invariant subset
A subset $Y\subset X$ is called a $$G$$ -invariant subset if

\forall y\in Y. \forall g\in G. g\cdot y\in Y

A set is called an invariant subset if it’s closed under group action

Definition. Equivariant map
Let $(X, \cdot)$ , $(Z, \cdot')$ be two left $$G$$ -sets. A map $f : X\to Z$ is called $$G$$ -equivariant if $\forall g\in G, x\in X$ ,

f(g\cdot x) = g\cdot' f(x)

Example. vector space
$(G, ∗) = (\R\setminus \{0\}, ×)$ , $$X$$ is a $\R$ -vector space, then

The scalar multiplication is a left $$G$$ -action.
A subspace is an invariant subset and a linear transformation between vector spaces is an equivariant map.
$V \setminus\{0\}$ is a $$G$$ -equivariant subset but not a subspace.

2. Stabilizers

Theorem.
Let $$X$$ be a left $$G$$ -set. The followings are equivalent:

$Y\subset X$ is a $$G$$ invariant subset.
There is a left $$G$$ action on $$Y$$ such that the inclusion map is $$G$$ -equivariant.

Definition. Stabilizer
Let $$X$$ be a left $$G$$ -set, $x\in G$ , then $\{g\in G\mid gx = x\} \leq G$ . This subgroup is called the stabilizer of $$x$$ , denoted as $$G_x$$ or $\Stab_G(x)$ .

Definition. Free
If $$X$$ is a left $$G$$ -set, we say the left $$G$$ action on $$X$$ is free if $\forall x\in X, G_x = \{e\}$ where $$e$$ is the identity of $$G$$ .

It is obvious that $$e$$ should be in $$G_x$$ by definition of $$G$$ -action.

Example. Scalar multiplication
Let $G = (\R\setminus \{0\})$ , $$V$$ be an $\R$ -vector space, the left $$G$$ -action on $V \setminus\{0\}$ via scalar multiplication is free.

Definition. Permutation representation.
Let $$X$$ be a left $$G$$ -set, then

$\forall g\in G$ , $x \mapsto gx$ is a bijection from $$X$$ to $$X$$ . (this even holds for infinite cardinalities)
The map $g \mapsto (x \mapsto gx)$ is a group homomorphism from $$G$$ to $$S_X$$ . This homomorphism is called the permutation representation

We can think of a permutation representation as an analogue of the matrix representation of linear transformations. We know that if $$A$$ is a finite set of $$n$$ numbers then $S_A\Iso S_n$ , where using the same idea fixing the basis of a vector space allows us to view a linear transformation as a matrix.

Definition. Effective
Let $$X$$ be a left $$G$$ -set. The kernel of the corresponding permutation representation is called the kernel of the left $$G$$ action, and if this kernel equals $\{e\}$ we say the action is effective.

Remark. kernel of action and stabilizer
Let $$f$$ be the group action from $$G$$ to $$X$$ , then by definition

\Ker(f)=\{g\in G\mid \forall x\in X.gx=x\}=\bigcap_{x\in X}\Stab_{G}(x)

Comparison between effective and free actions: If an action is free, it should be effective

free: $\exists x.gx=x\implies g=e$
effective: $\forall x.gx=x\implies g=e$
what free actions require is weaker but the results are as strong

Definition. Center
The kernel of the conjugate action is a subgroup $\{g\in G\mid \forall h\in G, hg = gh\}$ , called the center, denoted as $$C(G)$$ .

Definition. Transitive
We say a left $$G$$ action on a non-empty set $$X$$ is transitive if

\forall x, y\in X.\exists g\in G.y = gx

Theorem. Cayley’s Theorem.
Any group is isomorphic to a subgroup of a permutation group
Pf.

3. Orbit decomposition

Definition. Orbit
Let $$X$$ be a left $$G$$ set, $x\in X$ . A $$G$$ -orbit containing $$x$$ (denoted by $$Gx$$ ) is defined as

Gx:=\{y\in X\mid \exists g\in G.gx=y\}

The set of all $$G$$ -orbits can then be written as $\{Gx\mid x\in X\}$

Definition. Orbit decomposition
The decomposition of $$X$$ into disjoint unions of $$G$$ -orbits is called the orbit decomposition.

The following two lemmas show that orbits indeed disjoint, thus the saying of decomposition makes sense

Lemma. Orbits are equivalent classes
Let $$X$$ be a left $$G$$ set. The relation $\sim:= \{(x, y)\in X\times X\mid \exists g\in G, y = gx\}$ is a equivalence relation, and

(X/\sim)=\{Gx\mid x\in X\}

Pf. First we need to show this constructed relation is an equivalence relation

reflexivity. $$x=ex$$
symmetry. $y=gx\implies x=g^{-1}y$
transitivity. $y=gx\land z=hy\implies z=(hg)x$
By definition, $[x]=[y]\iff \exists g\in G.gx=y\iff Gx=Gy$

Lemma. Orbits are transitive invariant subsets

By definition, $[x] = \{gx\mid g\in G\}$ . Hence

\forall g'\in G, gx\in [x], g'(gx) = (g'g)x ∈ [x],

By definition $$[x]$$ is $$G$$ -invariant.

\forall y\in [x].\exists g' ∈ G. y = g'x

hence the $$G$$ action on $$[x]$$ is transitive.

Lemma. The set of left cosets are disjoint

Let $$G$$ be a group, $$X$$ is a left $$G$$ set. $H\leq G$ . Then

\{gH\mid g\in G\}

forms a partition of $$G$$ .
Pf. Consider the $H^{\Op}$ action defined on $$G$$ .

h\cdot g\mapsto gh

Pick any $g\in G$ , $\exists h.g'=h\cdot g\iff \exists h.g^{-1}g'=h$ , so

g'\in Hg\iff gH=g'H

This means that every left coset of $$H$$ is a $H^\Op$ -orbit. By definition orbits are partitions.

This is a really elegant proof.

Theorem. Lagrange’s Theorem / Orbit-Stablizer Theorem.

Let $$G$$ be a finite group, $H\leq G$ , then

|G/H| = \frac{|G|}{|H|}

Pf. We’ve shown a lemma in the section of cosets, that, any two left cosets are equinumerous. Since $$G$$ is finite, then all the cosets of $$H$$ have the same cardinality. We’ve shown in the above lemma that any two cosets are disjoint. Then the proof is done.

Having proved Lagrange’s theorem, we’ll then prove two augmented theorems (Cauchy’s theorem and Sylow’s theorem) in the following section

Definition. Index
If $$G$$ is a group (possibly infinite) and $H\leq G$ , the number of left cosets of $$H$$ in $$G$$ is called the index of $$H$$ in $$G$$ , and is denoted by $$[G:H]$$

If $|G|<\infty$ , by Lagrange’s theorem $[G:H]:=\frac{|G|}{|H|}$
Otherwise, $$[G:H]$$ can be $\infty$

Definition. Order (指数) of group elements
Let $$G$$ be a finite group, $g\in G$ . The order of $$g$$ (denoted by $\Ord(g)$ ) is defined as follows

\Ord(g):=\{n\in \N\mid g^n=e\}

The subgroup generated by $$g$$ is $⟨g⟩ = \{e, g, \ldots , g^{\Ord(g)−1}\}$ , and is a group of $\Ord(g)$ elements. Using Lagrange’s theorem, it’s obvious that

|G|<\infty\implies \forall g\in G.\Ord(g)\mid |G|

Theorem. Fermat’s Little Theorem.
If $$p$$ is a prime, then

\forall a\in \Z.p\mid (a^p-a)

Pf. We’ll firstly show that $A=(\Z/p\Z\setminus \{[0]\}, [a][b]\mapsto[ab])$ is a well-defined group.

4. Applications

There are tons of applications of group theory in modern number theory.

Lemma. If $$|G|=p$$ where $$p$$ is a prime, then $$G$$ is cyclic and $G\Iso \Z/p\Z$
Pf. Let $g\in G$ where $g\neq e$ , then $|\lll g\rrr|>1$ , and by definition

\lll g\rrr\leq G

So by Lagrange’s theorem we have $\lll g\rrr=G$ , which means $$G$$ is cyclic. Then $G\Iso \Z/p\Z$

Theorem. Cauchy’s Theorem
If $$G$$ is a finite group and $$p$$ is a prime dividing $$G$$ , then $$G$$ has a element of order $$p$$ .
Pf. Let $$p$$ be a prime number that divides $$|G|$$ , consider the set $$X$$ of $$p$$ -tuples where

X:=\{(x_1, x_2, \ldots, x_p)\in G^p\mid x_1x_2\ldots x_p=e\}

By definition, every $$p$$ -tuple in $$X$$ is uniquely determined by all its components except the last one, and those $$p-1$$ elements can be chosen freely, so $$X$$ has $|G|^{p−1}$ elements, which is divisible by $$p$$ . Now from the fact that in a group $ab = e\implies ba = e$ , it follows that any cyclic permutation of the components of an element of $$X$$ again gives an element of $$X$$ . Therefore one can define an action of the cyclic group $$C_p$$ of order $$p$$ on $$X$$ by cyclic permutations of components, in other words in which a chosen generator of $$C_p$$ sends

{\displaystyle (x_{1},x_{2},\ldots ,x_{p})\mapsto (x_{2},\ldots ,x_{p},x_{1})}.

As remarked, orbits in $$X$$ under this action either have

size $$1$$ , precisely when $(x,x,\ldots ,x)$ and $$x^p=e$$
size $$p$$ . Counting the elements of $$X$$ by orbits, and dividing by p, one sees that the number of elements satisfying xp=e $x^{p}=e$ is divisible by p. But x = e is one such element, so there must be at least p − 1 other solutions for x, and these solutions are elements of order p.
This completes the proof.

This is a really elegant construction proposed by James McKay, 1959.

Theorem. Sylow’s Theorem
If $$G$$ is a finite group of order $p^{\alpha}m$ , where $$p$$ is a prime and $p\not\mid m$ , then $$G$$ has a subgroup of order $p^\alpha$