Group Actions

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 20 April 2010

Group Actions

An action of a group $G$ on a set $S$ is a mapping $α : G \times S \to S$ (the convention is to write $g s$ for $α (g, s)$ ) such that

$g (h s) = (g h) s$ for all $g, h \in G, s \in S .$
$1 s = s$ for all $s \in S .$

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

Suppose a group $G$ , a set $S$ and an action of $G$ on $S$ are given.

The stabiliser of an element $s \in S$ under the action of $G$ is the set $G_{s} = \{g \in G | g s = s\} .$

The orbit of an element $s \in S$ under the action of $G$ is the set $G s = \{s' \in S | g s = s' for some g \in G\} .$

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

$G_{s}$ is a subgroup of $G .$
$G_{g s} = g G_{s} g^{- 1}$

Proof.

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

$□$

The following is an analogue of Proposition 1.1.3

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

Proof.

To show:

If $s \in S$ then $s \in G t$ for some $t \in S .$
If $s_{1}, s_{2} \in S$ and $G s_{1} \cap G s_{2} \neq \emptyset$ then $G s_{1} = G s_{2} .$
(Proof) Assume $s \in S .$
Then, since $s = 1 s, s \in G s .$
(Proof) Assume $s_{1}, s_{2} \in S$ and that $G s_{1} \cap G s_{2} \neq \emptyset .$
Then let $t \in G s_{1} \cap G s_{2} .$
So $t = g_{1} s_{1}$ and $t = g_{2} s_{2}$ for some elements $g_{1}, g_{2} \in G .$
So $s_{1} = {g_{1}}^{- 1} g_{2} s_{2} and s_{2} = {g_{2}}^{- 1} g_{1} s_{1} .$
To show: G s1 =G s2 .
1. To show:
2. $G s_{1} \subseteq G s_{2} .$
3. $G s_{2} \subseteq G s_{1} .$
4. (Proof) Let $t_{1} \in G s_{1} .$
5. So $t_{1} = h_{1} s_{1}$ for some $h_{1} \in G .$
6. Then $t_{1} = h_{1} s_{1} = h_{1} {g_{1}}^{- 1} g_{2} s_{2} \in G s_{2} .$ So $G s_{1} \subseteq G s_{2} .$
7. (Proof) Let $t_{2} \in G s_{2} .$
8. So $t_{2} = h_{2} s_{2}$ for some $h_{2} \in G .$
9. Then $t_{2} = h_{2} s_{2} = h_{2} {g_{2}}^{- 1} g_{1} s_{1} \in G s_{1}$ So $G s_{2} \subseteq G s_{1} .$
So G s 1 =G s2 .

So the orbits partition

S .

$□$

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

Proof.

By Proposition 1.2, $S$ is a disjoint union of orbits.
So $Card (S)$ 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 $\begin{array}{lcl} group actions α : G \times S \to S & are to & group homomorphisms f : G \to H & as \\ stabilisers G_{s} & are to & kernels \ker f & as \\ orbits G s & are to & images im f . \end{array}$

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 $S$ and let $s \in S .$ If $G s$ is the orbit containing $s$ and $G_{s}$ is the stabiliser of $s$ then $|G : G_{s}| = Card (G s)$ where $|G : G_{s}|$ is the index of $G_{s} \in G .$

Proof.

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

$φ$ is well defined.
$φ$ is bijective.
(Proof) To show
1. $φ (g G_{s}) \in G s$ for every $g \in G .$
2. If $g_{1} G_{s} = g_{2} G_{s}$ then $φ (g_{1} G_{s}) = φ (g_{2} G_{s}) .$
3. (Proof) Is clear from the definition of $φ, φ (g G_{s}) = g s \in G s .$
4. (Proof) Assume $g_{1}, g_{2} \in G$ and $g_{1} G_{s} = g_{2} G_{s} .$
5. Then $g_{1} = g_{2} h$ for some $h \in G_{s} .$
6. To show: g1 s= g2 s.
  1. Then $g_{1} s = g_{2} h s = g_{2} s,$ since $h \in G_{s} .$
  So φ g1 Gs =φ g2 Gs .
So φ is well defined.
To show:
1. $φ$ is injective, ie if $φ (g_{1} G_{s}) = φ (g_{2} G_{2})$ then $g_{1} G_{s} = g_{2} G_{s} .$
2. $φ$ is surjective, ie if $g s \in G_{s}$ then there exists $h G_{s} \in G / G_{s}$ such that $φ (h G_{s}) = g s .$
3. (Proof) Assume $φ (g_{1} G_{s}) = φ (g_{2} G_{s}) .$
4. Then $g_{1} s = g_{2} s .$
5. So $s = {g_{1}}^{- 1} g_{2} s$ and ${g_{2}}^{- 1} g_{1} s = s .$
6. So ${g_{1}}^{- 1} g_{2} \in G_{s}$ and ${g_{2}}^{- 1} g_{1} \in G_{s} .$
7. To show: φ is injective.
  1. To show: g1 Gs = g2 Gs
    1. To show:
    2. baa) $g_{1} G_{s} \subseteq g_{2} G_{s} .$
    3. bab) $g_{2} G_{s} \subseteq g_{1} G_{s} .$
    4. baa) (Proof) Let $k_{1} \in g_{1} G_{s} .$
    5. So $k_{1} = g_{1} h_{1}$ for some $h_{1} \in G_{s} .$
    6. Then $k_{1} = g_{1} h_{1} = g_{1} {g_{1}}^{- 1} g_{2} {g_{2}}^{- 1} g_{1} h_{1} = g_{2} ({g_{2}}^{- 1} g_{1} h_{1}) \in g_{2} G_{s} .$ So $g_{1} G_{s} \subseteq g_{1} G_{s} .$
    7. bab) (Proof) Let $k_{2} \in g_{2} G_{s} .$
    8. So $k_{2} = g_{2} h_{2}$ for some $h_{2} \in G_{s} .$
    9. Then $k_{2} = g_{2} h_{2} = g_{2} {g_{2}}^{- 1} g_{1} {g_{1}}^{- 1} g_{2} h_{2} = g_{2} ({g_{2}}^{- 1} g_{1} h_{1}) \in g_{1} G_{s} .$ So $g_{2} G_{s} \subseteq g_{1} G_{s} .$
    So g1 Gs = g2 Gs .
  So φ is injective.
8. To show: φ is surjective.
  1. Assume $t \in G_{s} .$
  2. Then $t = g s$ for some $g \in G .$
  3. Thus $φ (g G_{s}) = g s = t .$
  So φ is surjective.
9. So $φ$ is bijective.

$□$

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

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

Conjugation

Let $S$ be a subset of a group $G .$ The normaliser of $S$ in $G$ is the set $N_{S} = \{x \in G | x S x^{- 1} = S\},$ where $x S x^{- 1} = \{x s x^{- 1} | s \in S\} .$

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 said to be 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 centraliser 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 iff they are in the same orbit under the action of $G$ on itself via conjugation.
The conjugacy class $𝒞_{g}$ of $g \in G$ is the orbit of $g$ under the action of $G$ on itself via conjugation.
The centraliser $Z_{g}$ of $g \in G$ is the stabiliser of $g \in G$ under the action of $G$ on itself via 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 easily from the definitions.

$□$

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

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

For each subset $S \subseteq G,$ $Z_{S} = \underset{s \in S}{\cap} G_{s} .$
$Z (G) = Z_{G},$ where $Z (G)$ denotes the center of $G .$
$s \in Z (G)$ iff $Z_{s} = G .$
$s \in Z (G)$ iff $𝒞_{s} = \{s\} .$

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 $Z (G) = \underset{\|𝒞_{g_{i}}\| = 1}{\cup} 𝒞_{g_{i}} .$ So $\|G\| = \sum_{\|𝒞_{g_{i}}\| = 1} Card (𝒞_{g_{i}}) + \sum_{\|𝒞_{g_{i}}\| > 1} Card (𝒞_{g_{i}}) = Card (Z (G)) + \sum_{\|𝒞_{g_{i}}\| > 1} Card (𝒞_{g_{i}}) .$ $□$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

page history