Exercise 1.7.18 (Orbits under the action of a group $H$)

Proof. For all $a,b\in A$,

Let $a,b,c$ be any elements in $A$. Let $e=1_H$ denote the identity element of $H$.

R.

Since $e\in H$ and $a=e\cdot a$, then $a\sim a$. The relation is reflexive.

S.

If $a\sim b$, then $a=h\cdot b$ for some $h\in H$, thus $h^{-1}a=h^{-1}\cdot (h\cdot a)=(h^{-1}h)\cdot a=e\cdot a=a$, where $h^{-1}\in H$, so $b\sim a$. The relation is symmetric.

T.

If $a\sim b$ and $b\sim c$, then $a=h\cdot b$ and $b=k\cdot c$ for some elements $h,k\in H$. Therefore $a=h\cdot (k\cdot c)=(\mathit{hk})\cdot c$, where $\mathit{hk}\in H$, so $a\sim c$. The relation is reflexive.

Hence the equivalence classes partition the set $G$: if we write ${𝒪}_x$ the orbit of $x\in A$ (the equivalence class of $x$ for the relation $\sim$), then

(i)

$x\in {𝒪}_x$ for every $x\in A$, so $${𝒪}_x\ne \varnothing .$$

(ii)

Distinct orbits are disjoint: for all $x,y\in A$, $${𝒪}_x\ne {𝒪}_y\Rightarrow {𝒪}_x\cap {𝒪}_y=\varnothing .$$

(iii)

The union of all orbits is $A$: $$A=\underset{x\in A}{\bigcup }{𝒪}_x.$$

To obtain a disjoint union in (iii), we choose a unique element $c$ in every orbit. The ensemble $C$ of these chosen elements is called a complete system of representatives of the orbits. It satisfies $C\cap {𝒪}_c=\{c\}$ for every $c\in C$. Then

where the union is disjoint:

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