## Group Actions

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 $g\left(hx\right)=\left(gh\right)x$,
2. (b) If $x\in X$ then $1x=x$.
• 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 ${G}_{x}=\left\{g\in G\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}gx=x\right\}$.
• The orbit of $x$ is the set $Gx=\left\{gx\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}g\in G\right\}$.

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

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

 Proof. To show: If ${h}_{1},{h}_{2}\in {G}_{x}$ then ${h}_{1}{h}_{2}\in {G}_{x}$. $1\in {G}_{x}$. If $h\in {G}_{x}$ then ${h}^{-1}\in {G}_{x}.$ (Proof) Assume ${h}_{1},{h}_{2}\in {G}_{x}$. Then $(h1h2)x =h1( h2x =h1x=x$. So ${h}_{1},{h}_{2}\in {G}_{x}$. (Proof) Since $1x=x$, $1\in {G}_{x}$. (Proof) Assume $h\in {G}_{x}$. Then $h-1x =h-1 (hx) =(h-1 h)x =1x=x$. So ${h}^{-1}\in {G}_{x}.$ So ${G}_{s}$ is a subgroup of $G$. To show: ${G}_{gx}\subseteq g{G}_{x}{g}^{-1}$. $g{G}_{x}{g}^{-1}\subseteq {G}_{gx}$. (Proof) Assume $h\in {G}_{gx}$. Then $hgs=hx$. So ${g}^{-1}hgx=x$ So ${g}^{-1}hg\in {G}_{x}$. Since $h=g\left({g}^{-1}hg{g}^{-1},h\in g{G}_{x}{g}^{-1}$. So ${G}_{gx}\subseteq g{G}_{s}{g}^{-1}$. (Proof) Assume $h\in g{G}_{x}{g}^{-1}$. So $h=ga{g}^{-1}$ for some $a\in {G}_{x}$. Then $hgx= (gag-1 ) gx=gax=gx$. So $h\in {G}_{gx}\subseteq g{G}_{x}{g}^{-1}$. So ${G}_{gx}=g{G}_{x}{g}^{-1}.$ $\square$

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 $x\in Gy$ for some $y\in X.$ If ${x}_{1},{x}_{2}\in S$ and $G{x}_{1}\cap G{x}_{2}\ne \varnothing$ then $G{x}_{1}=G{x}_{2}$. (Proof) Assume $x\in X$. Then, since $x=1x$, $x\in Gx$. (Proof) Assume ${x}_{1},{x}_{2}\in S$ and that $G{x}_{1}\cap G{x}_{2}\ne \varnothing$. 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 $x1= g1-1 g2x2 and x2 =g2-1g1x1.$ To show: $G{x}_{1}=G{x}_{2}$. To show: $G{x}_{1}\subseteq G{x}_{2}$. $G{x}_{2}\subseteq G{x}_{1}$. (Proof) Let ${y}_{1}\in G{x}_{1}.$ So ${y}_{1}={h}_{1}{x}_{1}$ for some ${h}_{1}\in G$. Then $y1= h1x1 =h1 g1-1 g2x2∈Gx2$. So $G{x}_{1}\subseteq G{x}_{2}$. (Proof) Let ${y}_{2}\in G{x}_{2}$. So ${y}_{2}={h}_{2}{x}_{2}$ for some ${h}_{2}\in G.$ Then $y2 =h2x2 =h2 g2-1 g1x1 ∈Gs1$. So $G{x}_{2}\subseteq G{x}_{1}$. So $G{x}_{1}=G{x}_{2}$. So the orbits partition $S$. $\square$

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 $Card(X) =∑distinct orbits Card(Gxi) .$

 Proof. By Proposition 1.2, $X$is a disjoint union of orbits. So $\mathrm{Card}\left(X\right)$ is the sum of the cardinalities of the orbits. $\square$

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 $Gs$ of an element $s\in S$ as an analogue of the image of a homomorphism. One might say

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 $Gx$ is the orbit containing $x$ and ${G}_{x}$ is the stabiliser of $x$ then $|G: Gx| =Card(Gx)$ where $|G:{G}_{x}|$ is the index of ${G}_{x}\in G.$

 Proof. Recall that $|G:{G}_{x}|=\mathrm{Card}\left(G/{G}_{x}\right)$. To show: There is a bijective map $φ:G/Gx →Gx.$ Define $φ: G/ Gx → Gx gGx ↦ gx.$ To show: $\phi$ is well defined. $\phi$ is bijective. (Proof) To show $\phi \left(g{G}_{x}\right)\in Gx$ for every $g\in G$. If ${g}_{1}{G}_{x}={g}_{2}{G}_{x}$ then $\phi \left({g}_{1}{G}_{s}\right)=\phi \left({g}_{2}{G}_{s}\right)$. (Proof) Is clear from the definition of $\phi$ that $\phi \left(g{G}_{x}\right)=gx\in Gx$. (Proof) Assume ${g}_{1},{g}_{2}\in G$ and ${g}_{1}{G}_{x}={g}_{2}{G}_{x}$. Then ${g}_{1}={g}_{2}h$ for some $h\in {G}_{s}$. To show: ${g}_{1}s={g}_{2}s$. Then $g1x =g2hx =g2x,$ since $h\in {G}_{x}$. So $\phi \left({g}_{1}{G}_{x}\right)=\phi \left({g}_{2}{G}_{x}\right)$. So $\phi$ is well defined. To show: $\phi$ is injective, i.e. if $\phi \left({g}_{1}{G}_{x}\right)=\phi \left({g}_{2}{G}_{2}\right)$ then ${g}_{1}{G}_{x}={g}_{2}{G}_{x}$. $\phi$ is surjective, i.e. if $gx\in {G}_{x}$ then there exists $h{G}_{x}\in G/{G}_{x}$ such that $\phi \left(h{G}_{s}\right)=gs$. (Proof) Assume $\phi \left({g}_{1}{G}_{s}\right)=\phi \left({g}_{2}{G}_{s}\right)$. Then ${g}_{1}x={g}_{2}x$. So $s={{g}_{1}}^{-1}{g}_{2}x$ and ${{g}_{2}}^{-1}{g}_{1}x=x$. So ${{g}_{1}}^{-1}{g}_{2}\in {G}_{x}$ and ${{g}_{2}}^{-1}{g}_{1}\in {G}_{x}$. To show: $\phi$ is injective. To show: ${g}_{1}{G}_{x}={g}_{2}{G}_{x}$. To show: baa) ${g}_{1}{G}_{x}\subseteq {g}_{2}{G}_{x}$. bab) ${g}_{2}{G}_{x}\subseteq {g}_{1}{G}_{x}$. baa) (Proof) Let ${k}_{1}\in {g}_{1}{G}_{x}$. So ${k}_{1}={g}_{1}{h}_{1}$ for some ${h}_{1}\in {G}_{x}$. Then $k1 =g1h1 =g1 g1-1 g2 g2-1 g1h1 =g2( g2-1 g1h1) ∈g2Gx.$ So ${g}_{1}{G}_{x}\subseteq {g}_{1}{G}_{x}$. bab) (Proof) Let ${k}_{2}\in {g}_{2}{G}_{x}$. So ${k}_{2}={g}_{2}{h}_{2}$ for some ${h}_{2}\in {G}_{x}$. Then $k2 =g2h2 =g2 g2-1 g1 g1-1 g2h2 =g2( g2-1 g1h1) ∈g1Gx.$ So ${g}_{2}{G}_{x}\subseteq {g}_{1}{G}_{x}$. So ${g}_{1}{G}_{x}={g}_{2}{G}_{x}$. So $\phi$ is injective. To show: $\phi$ is surjective. Assume $y\in {G}_{x}$. Then $y=gx$ for some $g\in G$. Thus $φ(gGx) =gx=y.$ So $\phi$ is surjective. So $\phi$ is bijective. $\square$

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

 Proof. Multiply both sides of the identity in Proposition 1.4 by $\mathrm{Card}\left({G}_{x}\right)$ and use Corollary 2.3 from 'Groups, Basic Definitions and Cosets'. $\square$

## Conjugation

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

• The normalizer of $X$ in $G$ is the set $NX ={g∈G | gXg-1 =X},$ where $gX{g}^{-1}=\left\{gx{g}^{-1}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}x\in X\right\}$.

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

1. $H$ is a normal subgroup of ${N}_{H}$.
2. 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: $k\in {N}_{H}$. $kh{k}^{-1}\in H$ for all $h\in H$. $\phantom{\rule{2em}{0ex}}$ This is true since $H$is normal in $K$. So $K\subseteq {N}_{H}$. This is the special case of b) when $K=H$. $\square$

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

1. $G$ acts on $𝒮$ by $G×𝒮 → 𝒮 (gS) → gSg-1$ where $gS{g}^{-1}=\left\{gs{g}^{-1}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}s\in S\right\}$. We say that $G$ acts on $𝒮$ by conjugation.
2. 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: $\alpha$ is well defined. $\alpha \left(1S\right)=S$, for all $S\in 𝒮$. $\alpha \left(g\alpha \left(hS\right)\right)=\alpha \left(gh\right)S\right)$, for all $g,h\in G$ and $S\in 𝒮$. (Proof) To show: aaa) $gS{g}^{-1}\in 𝒮$. aab) If $S=T$ and $g=h$ then $gS{g}^{-1}=hT{h}^{-1}$. Both of these are clear from the definitions. (Proof) Let $S\in 𝒮$. Then $α(1S) =1S 1-1 =S.$ (Proof) Let $g,h\in G$ and $S\in 𝒮$. Then $α(g,α( h,S) = α(g, hSh-1) =g( hSh-1 ) g-1 = (gh)S( h-1 g-1 ) =(gh)S (gh)-1 =α(gh,S) .$ This follows immediately from the definitions of ${N}_{S}$ and of stabiliser. $\square$

• 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 $Zg={ x∈G | xgx-1 =g }.$

Let $G$ be a group. Then

1. $G$ acts on $G$ by $G×G → G (gs) ↦ gsg-1.$ We say that $G$ acts on itself by conjugation.
2. 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.
3. The conjugacy class ${𝒞}_{g}$ of $g\in G$ is the orbit of $g$ under the action of $G$ on itself by conjugation.
4. 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. $\square$

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

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

1. For each subset $S\subseteq G$, $ZS= ⋂s∈S Gs.$
2. $Z\left(G\right)={Z}_{G}$, where $Z\left(G\right)$ denotes the center of $G$.
3. $s\in Z\left(G\right)$ if and only if ${Z}_{s}=G$.
4. $s\in Z\left(G\right)$ if and only if ${𝒞}_{s}=\left\{s\right\}$.

Proof.
1. Assume $s\in {Z}_{s}$.
2. Then $sx{s}^{-1}=s$, 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 $xs{x}^{-1}=s$, for all $s\in S$.
8. So $x\in {Z}_{s}$.
9. So ${\cap }_{s\in S}{G}_{s}$.
1. This is clear from the definitions of ${Z}_{G}$ and $Z\left(G\right)$.
2.  $⇒:$ Let $s\in Z\left(G\right)$. To show: ${Z}_{S}=G$. By definition ${Z}_{S}\subseteq G$. To show: $G\subseteq {Z}_{S}$. Let $g\in G$. Then $gs{g}^{-1}=s$ since $s\in Z\left(G\right)$. So $g\in {Z}_{S}$. So $G\subseteq {Z}_{S}$. So ${Z}_{S}=G$. $⇐$ Assume ${Z}_{S}=G$. Then $gs{g}^{-1}=s$, for all $g\in G$. So $sg=gs$, for all $g\in G$. So $s\in Z\left(G\right)$.
3.  $⇒:$ Assume $s\in Z\left(G\right)$. Then $gs{g}^{-1}=s$, for all $s\in G$. So ${𝒞}_{s}=\left\{gs{g}^{-1}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}g\in G\right\}=\left\{s\right\}$. $⇐:$ Assume ${𝒞}_{s}=\left\{s\right\}$. Then $gs{g}^{-1}=s$, for all $g\in G$. So $s\in Z\left(g\right)$.

$\square$

(The Class Equation) Let ${𝒞}_{{g}_{i}}$ denote the conjugacy classes in a group $G$ and let $|{𝒞}_{{g}_{i}}|$ denote $\mathrm{Card}\left({𝒞}_{{g}_{i}}\right).$ Then $|G| =| Z(G)| +∑| Z(G)| >1 Card(𝒞gi) .$

 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 $|G| = ∑𝒞gi Card(𝒞gi ) .$ By Lemma 2.4 (d) we know that $Z(G) = ⋃ |𝒞gi | =1 𝒞gi.$ So $|G|= ∑ | 𝒞gi | =1 Card(𝒞gi ) + ∑ |𝒞gi | >1 Card(𝒞gi ) =Card ( Z(G) ) + ∑ | 𝒞gi | >1 Card(𝒞gi ) .$ $\square$

## 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.