Group Actions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 4 August 2012

Group Actions

Let $G$ be a group.

An action of G on a set X is a function α: G×X → X (g,x) ↦ gx such that
1. (a) If $g, h \in G$ and $x \in X$ then $g (h x) = (g h) x$ ,
2. (b) If $x \in X$ then $1 x = x$ .
A $G$ -set is a set $G$ with an action of $G$ on $X$ .

Examples of group actions are given below in this section and in the Exercises.

Let $G$ be a group and let $X$ be a $G$ -set. Let $x \in X$ .

The stabiliser of $x$ is the set $G_{x} = {g \in G | g x = x}$ .
The orbit of $x$ is the set $G x = {g x | g \in G}$ .

Suppose $G$ is a group acting on a set $X$ and let $x \in X$ and $g \in G$ . Then

(a) $G_{x}$ is a subgroup of $G$ .
(b) $G_{g x} = g G_{x} g^{- 1}$ .

Proof.

To show:
1. If $h_{1}, h_{2} \in G_{x}$ then $h_{1} h_{2} \in G_{x}$ .
2. $1 \in G_{x}$ .
3. If $h \in G_{x}$ then $h^{- 1} \in G_{x} .$
4. (Proof) Assume $h_{1}, h_{2} \in G_{x}$ . Then $(h_{1} h_{2}) x = h_{1} (h_{2} x = h_{1} x = x$ . So $h_{1}, h_{2} \in G_{x}$ .
5. (Proof) Since $1 x = x$ , $1 \in G_{x}$ .
6. (Proof) Assume $h \in G_{x}$ . Then $h^{- 1} x = h^{- 1} (h x) = (h - 1 h) x = 1 x = x$ . So $h^{- 1} \in G_{x} .$
So Gs is a subgroup of G.
To show:
1. $G_{g x} \subseteq g G_{x} g^{- 1}$ .
2. $g G_{x} g^{- 1} \subseteq G_{g x}$ .
3. (Proof) Assume $h \in G_{g x}$ .
4. Then $h g s = h x$ .
5. So $g^{- 1} h g x = x$
6. So $g^{- 1} h g \in G_{x}$ .
7. Since $h = g (g^{- 1} h g g^{- 1}, h \in g G_{x} g^{- 1}$ .
8. So $G_{g x} \subseteq g G_{s} g^{- 1}$ .
9. (Proof) Assume $h \in g G_{x} g^{- 1}$ .
10. So $h = g a g^{- 1}$ for some $a \in G_{x}$ .
11. Then $h g x = (g a g^{- 1}) g x = g a x = g x$ .
12. So $h \in G_{g x} \subseteq g G_{x} g^{- 1}$ .
So Ggx =gGx g-1.

$□$

The following is an analogue of Proposition 1.1.3.

Let $G$ be a group which acts on a set $X$ . Then the orbits partition the set $X$ .

Proof.

To show:

If $x \in X$ then $x \in G y$ for some $y \in X .$
If $x_{1}, x_{2} \in S$ and $G x_{1} \cap G x_{2} \neq \emptyset$ then $G x_{1} = G x_{2}$ .
(Proof) Assume $x \in X$ .
Then, since $x = 1 x$ , $x \in G x$ .
(Proof) Assume $x_{1}, x_{2} \in S$ and that $G x_{1} \cap G x_{2} \neq \emptyset$ .
Then let $t \in G x_{1} \cap G x_{2} .$
So $y = g_{1} x_{1}$ and $y = g_{2} x_{2}$ for some elements $g_{1}, g_{2} \in G$ .
So $x_{1} = {g_{1}}^{- 1} g_{2} x_{2} and x_{2} = {g_{2}}^{- 1} g_{1} x_{1} .$
To show: Gx1 =Gx2.
1. To show:
2. $G x_{1} \subseteq G x_{2}$ .
3. $G x_{2} \subseteq G x_{1}$ .
4. (Proof) Let $y_{1} \in G x_{1} .$
5. So $y_{1} = h_{1} x_{1}$ for some $h_{1} \in G$ .
6. Then $y_{1} = h_{1} x_{1} = h_{1} {g_{1}}^{- 1} g_{2} x_{2} \in G x_{2}$ . So $G x_{1} \subseteq G x_{2}$ .
7. (Proof) Let $y_{2} \in G x_{2}$ .
8. So $y_{2} = h_{2} x_{2}$ for some $h_{2} \in G .$
9. Then $y_{2} = h_{2} x_{2} = h_{2} {g_{2}}^{- 1} g_{1} x_{1} \in G s_{1}$ . So $G x_{2} \subseteq G x_{1}$ .
So Gx1 =Gx2.

So the orbits partition

S

$□$

If $G$ is a group acting on a set $X$ and $G_{x_{i}}$ denote the orbits of the action of $G$ on $X$ then $Card (X) = \sum_{distinct orbits} Card (G_{x_{i}}) .$

Proof.

By Proposition 1.2, $X$ is a disjoint union of orbits.
So $Card (X)$ is the sum of the cardinalities of the orbits.

$□$

It is possible to view the stabiliser $G_{s}$ of an element $s \in S$ as an analogue of the kernel of a homomorphism and the orbit $G s$ of an element $s \in S$ as an analogue of the image of a homomorphism. One might say $\begin{array}{lcl} group actions α : G \times X \to X & are to & group homomorphisms f : G \to H & as \\ stabilisers G_{x} & are to & kernels \ker f & as \\ orbits G x & are to & images im f . \end{array}$

From this point of view the following corollary is an analogue of Corollary 1.1.5.

Let $G$ be a group acting on a set $X$ and let $x \in X$ . If $G x$ is the orbit containing $x$ and $G_{x}$ is the stabiliser of $x$ then $| G : G_{x} | = Card (G x)$ where $| G : G_{x} |$ is the index of $G_{x} \in G .$

Proof.

Recall that $| G : G_{x} | = Card (G / G_{x})$ .
To show: There is a bijective map $φ : G / G_{x} \to G x .$ Define $\begin{array}{rcl} φ : & G / G_{x} & \to & G x \\ g G_{x} & \mapsto & g x . \end{array}$ To show:

$φ$ is well defined.
$φ$ is bijective.
(Proof) To show
1. $φ (g G_{x}) \in G x$ for every $g \in G$ .
2. If $g_{1} G_{x} = g_{2} G_{x}$ then $φ (g_{1} G_{s}) = φ (g_{2} G_{s})$ .
3. (Proof) Is clear from the definition of $φ$ that $φ (g G_{x}) = g x \in G x$ .
4. (Proof) Assume $g_{1}, g_{2} \in G$ and $g_{1} G_{x} = g_{2} G_{x}$ .
5. Then $g_{1} = g_{2} h$ for some $h \in G_{s}$ .
6. To show: g1s =g2s.
  1. Then $g_{1} x = g_{2} h x = g_{2} x,$ since $h \in G_{x}$ .
  So φ(g1Gx ) =φ(g2Gx ) .
So φ is well defined.
To show:
1. $φ$ is injective, i.e. if $φ (g_{1} G_{x}) = φ (g_{2} G_{2})$ then $g_{1} G_{x} = g_{2} G_{x}$ .
2. $φ$ is surjective, i.e. if $g x \in G_{x}$ then there exists $h G_{x} \in G / G_{x}$ such that $φ (h G_{s}) = g s$ .
3. (Proof) Assume $φ (g_{1} G_{s}) = φ (g_{2} G_{s})$ .
4. Then $g_{1} x = g_{2} x$ .
5. So $s = {g_{1}}^{- 1} g_{2} x$ and ${g_{2}}^{- 1} g_{1} x = x$ .
6. So ${g_{1}}^{- 1} g_{2} \in G_{x}$ and ${g_{2}}^{- 1} g_{1} \in G_{x}$ .
7. To show: φ is injective.
  1. To show: g1Gx =g2Gx.
    1. To show:
    2. baa) $g_{1} G_{x} \subseteq g_{2} G_{x}$ .
    3. bab) $g_{2} G_{x} \subseteq g_{1} G_{x}$ .
    4. baa) (Proof) Let $k_{1} \in g_{1} G_{x}$ .
    5. So $k_{1} = g_{1} h_{1}$ for some $h_{1} \in G_{x}$ .
    6. Then $k_{1} = g_{1} h_{1} = g_{1} {g_{1}}^{- 1} g_{2} {g_{2}}^{- 1} g_{1} h_{1} = g_{2} ({g_{2}}^{- 1} g_{1} h_{1}) \in g_{2} G_{x} .$ So $g_{1} G_{x} \subseteq g_{1} G_{x}$ .
    7. bab) (Proof) Let $k_{2} \in g_{2} G_{x}$ .
    8. So $k_{2} = g_{2} h_{2}$ for some $h_{2} \in G_{x}$ .
    9. Then $k_{2} = g_{2} h_{2} = g_{2} {g_{2}}^{- 1} g_{1} {g_{1}}^{- 1} g_{2} h_{2} = g_{2} ({g_{2}}^{- 1} g_{1} h_{1}) \in g_{1} G_{x} .$ So $g_{2} G_{x} \subseteq g_{1} G_{x}$ .
    So g1Gx =g2Gx.
  So φ is injective.
8. To show: φ is surjective.
  1. Assume $y \in G_{x}$ .
  2. Then $y = g x$ for some $g \in G$ .
  3. Thus $φ (g G_{x}) = g x = y .$
  So φ is surjective.
9. So $φ$ is bijective.

$□$

Let $G$ be a group acting on a set $X$ . Let $G_{x}$ denote the stabiliser of $x$ and let $G x$ denote the orbit of $x$ . Then $Card (G) = Card (G x) Card (G_{x}) .$

		Proof.
	Multiply both sides of the identity in Proposition 1.4 by $Card (G_{x})$ and use Corollary 2.3 from 'Groups, Basic Definitions and Cosets'. $□$

Conjugation

Let $X$ be a subset of a group $G$ .

The normalizer of $X$ in $G$ is the set $N_{X} = {g \in G | g X g^{- 1} = X},$ where $g X g^{- 1} = {g x g^{- 1} | x \in X}$ .

Let $H$ be a subgroup of $G$ and let $N_{H}$ be the normaliser of $H$ in $G .$ Then

$H$ is a normal subgroup of $N_{H}$ .
If $K$ is a subgroup of $G$ such that $H \subseteq K \subseteq G$ and $H$ is a normal subgroup of $K$ then $K \subseteq N_{H}$ .

Proof.

Let $k \in K$ .
To show: k∈NH.
1. $k h k^{- 1} \in H$ for all $h \in H$ .
2. This is true since $H$ is normal in $K$ .
So $K \subseteq N_{H}$ .
This is the special case of b) when $K = H$ .

$□$

This proposition says that $N_{H}$ is the largest subgroup of $G$ such that $H$ is normal in this subgroup.

Let $G$ be a group and let $𝒮$ be the set of subsets of $G$ . Then

$G$ acts on $𝒮$ by $\begin{array}{rcl} G \times 𝒮 & \to & 𝒮 \\ (g S) & \to & g S g^{- 1} \end{array}$ where $g S g^{- 1} = {g s g^{- 1} | s \in S}$ . We say that $G$ acts on $𝒮$ by conjugation.
If $S$ is a subset of $G$ then $N_{S}$ is the stabiliser of $S$ under the action of $G$ on $𝒮$ by conjugation.

Proof.

To show:
1. $α$ is well defined.
2. $α (1 S) = S$ , for all $S \in 𝒮$ .
3. $α (g α (h S)) = α (g h) S)$ , for all $g, h \in G$ and $S \in 𝒮$ .
4. (Proof) To show:
  1. aaa) $g S g^{- 1} \in 𝒮$ .
  2. aab) If $S = T$ and $g = h$ then $g S g^{- 1} = h T h^{- 1}$ .
  Both of these are clear from the definitions.
5. (Proof) Let $S \in 𝒮$ .
6. Then $α (1 S) = 1 S 1^{- 1} = S .$
7. (Proof) Let $g, h \in G$ and $S \in 𝒮$ .
8. Then $\begin{matrix} α (g, α (h, S) & = & α (g, h S h^{- 1}) = g (h S h^{- 1}) g^{- 1} \\ = & (g h) S (h^{- 1} g^{- 1}) = (g h) S {(g h)}^{- 1} = α (g h, S) . \end{matrix}$
This follows immediately from the definitions of $N_{S}$ and of stabiliser.

$□$

Two elements $g_{1}, g_{2} \in G$ are conjugate if $g_{1} = h g_{2} h^{- 1}$ , for some $h \in G$ .
Let $G$ be a group and let $g \in G$ . The conjugacy class $𝒞_{g}$ of $g$ is the set of all conjugates of $g .$
Let $g$ be an element of a group $G$ . The centralizer or normaliser of $g$ is the set $Z_{g} = {x \in G | x g x^{- 1} = g} .$

Let $G$ be a group. Then

$G$ acts on $G$ by $\begin{array}{rcl} G \times G & \to & G \\ (g s) & \mapsto & g s g^{- 1} . \end{array}$ We say that $G$ acts on itself by conjugation.
Two elements $g_{1}, g_{2} \in G$ are conjugate if and only if they are in the same orbit under the action of $G$ on itself by conjugation.
The conjugacy class $𝒞_{g}$ of $g \in G$ is the orbit of $g$ under the action of $G$ on itself by conjugation.
The centraliser $Z_{g}$ of $g \in G$ is the stabiliser of $g \in G$ under the action of $G$ on itself by conjugation.

Proof.

The proof is exactly the same as in the proof of (a) in Proposition 2.2. One simply replaces all the capital $S$ 's by lower case $s$ 's.
and (c) and (d) follow from the definitions.

$□$

Let $S$ be a subset of a group $G$ . The centraliser of $S$ in $G$ is the set $Z_{S} = {x \in G | x s x^{- 1} = s for all s \in S} .$

Let $G_{s}$ be the stabiliser of $s \in G$ under the action of $G$ on itself by conjugation. Then

For each subset $S \subseteq G$ , $Z_{S} = ⋂_{s \in S} G_{s} .$
$Z (G) = Z_{G}$ , where $Z (G)$ denotes the center of $G$ .
$s \in Z (G)$ if and only if $Z_{s} = G$ .
$s \in Z (G)$ if and only if $𝒞_{s} = {s}$ .

Proof.

1. Assume $s \in Z_{s}$ .
2. Then $s x s^{- 1} = s$ , for all $s \in S$ .
3. So $x \in G_{s}$ for all $s \in S$ .
4. So $x \in \cap_{s \in S} G_{s}$ .
5. So $Z_{s} \subseteq \cap_{s \in S} G_{s}$ .
6. Assume $x \in \cap_{s \in S} G_{s}$ .
7. Then $x s x^{- 1} = s$ , for all $s \in S$ .
8. So $x \in Z_{s}$ .
9. So $\cap_{s \in S} G_{s}$ .
This is clear from the definitions of $Z_{G}$ and $Z (G)$ .

$\Rightarrow :$	Let $s \in Z (G)$ .
	To show: $Z_{S} = G$ . By definition $Z_{S} \subseteq G$ . To show: $G \subseteq Z_{S}$ . Let $g \in G$ . Then $g s g^{- 1} = s$ since $s \in Z (G)$ . So $g \in Z_{S}$ . So $G \subseteq Z_{S}$ . So $Z_{S} = G$ .
$\Leftarrow$	Assume $Z_{S} = G$ .
	Then $g s g^{- 1} = s$ , for all $g \in G$ .
	So $s g = g s$ , for all $g \in G$ .
	So $s \in Z (G)$ .

$\Rightarrow :$ Assume $s \in Z (G)$ .

Then $g s g^{- 1} = s$ , for all $s \in G$ .

So $𝒞_{s} = {g s g^{- 1} | g \in G} = {s}$ .

$\Leftarrow :$ Assume $𝒞_{s} = {s}$ .
Then $g s g^{- 1} = s$ , for all $g \in G$ .

So $s \in Z (g)$ .

$□$

(The Class Equation) Let $𝒞_{g_{i}}$ denote the conjugacy classes in a group $G$ and let $| 𝒞_{g_{i}} |$ denote $Card (𝒞_{g_{i}}) .$ Then $| G | = | Z (G) | + \sum_{| Z (G) | > 1} Card (𝒞_{g_{i}}) .$

		Proof.
	By Corollary 1.3 and the fact that $𝒞_{g_{i}}$ are the orbits of $G$ acting on itself by conjugation we know that $\| G \| = \sum_{𝒞_{g_{i}}} Card (𝒞_{g_{i}}) .$ By Lemma 2.4 (d) we know that $Z (G) = ⋃_{\| 𝒞_{g_{i}} \| = 1} 𝒞_{g_{i}} .$ So $\| G \| = \sum_{\| 𝒞_{g_{i}} \| = 1} Card (𝒞_{g_{i}}) + \sum_{\| 𝒞_{g_{i}} \| > 1} Card (𝒞_{g_{i}}) = Card (Z (G)) + \sum_{\| 𝒞_{g_{i}} \| > 1} Card (𝒞_{g_{i}}) .$ $□$

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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