Exercise 1.7.19 (Lagrange's Theorem)

Proof. Let $x\in G$ and let $𝒪={𝒪}_x$ be the orbit of $x$ under the action of $H$. Consider the map

• $\phi$ is injective:

  If $h,k\in H$ satisfy $\phi (h)=\phi (k)$, then $\mathit{hx}=\mathit{kx}$, thus $\mathit{hx}{x}^{-1}=\mathit{kx}{x}^{-1}$, so $h=k$.

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->h\in H,\forall <!--\; FUNCTION\; APPLICATION\; -->k\in H,\phi (h)=\phi (k)\Rightarrow h=k.$$

• $\phi$ is surjective:

  Let $y$ be any element in ${𝒪}_x$. By definition of ${𝒪}_x$, $y\sim x$, so there is some $h\in H$ such that $y=\mathit{hx}=\phi (x)$:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->y\in {𝒪}_x,\exists <!--\; FUNCTION\; APPLICATION\; -->h\in H,y=\phi (x).$$

  (This shows also that ${𝒪}_x=\mathit{Hx}=\{\mathit{hx}\mid h\in H\}$.)

This proves that $\phi$ is bijective, therefore for any $x\in G$,

(Note that $H={𝒪}_e$ is one of the orbits.)

If $C$ is a complete system of representatives of the orbits (see the solution of Exercise 18), then

Therefore

For all $c\in C$, $|{𝒪}_c|=|H|$, thus

(where $|C|$ is the number of orbits). This shows that $|H|$ divides $|G|$.

If $G$ is a finite group and $H$ is a subgroup of $G$ then $|H|$ divides $|G|$. □