Problem

Let \(H\) be a subgroup of the finite group \(G\) and let \(H\) act on \(G\) (here \(A = G\)) by left multiplication. Let \(x \in G\) and let \(O \) be the orbit of \(x\) under the action of \(H\). Prove that the map $$ H \rightarrow O \quad \text{defined by} \quad h \mapsto hx $$ is a bijection (hence all orbits have cardinality \(|H|\)). From this and the preceding exercise, deduce Lagrange’s Theorem:

$$ \text{If } G \text{ is a finite group and } H \text{ is a subgroup of } G \text{, then } |H| \text{ divides } |G|. $$

Solution

Let \(\varphi\) be the map in question. That is, \(\varphi(h) = hx\). Since \(x \in G\), there exists \(x^{-1}\). Therefore, if we define another map \(\psi\) to be \(\psi(k) = kx^{-1}\),

$$ \varphi\psi(k) = \varphi(kx^{-1}) = (kx^{-1})x = k $$ and $$ \psi\varphi(h) = \psi(hx) = (hx)x^{-1} = h. $$

That is, \(\varphi\) and \(\psi\) are an inverse of each other. Therefore, \(\varphi\) is a bijection. Consequently, since our choice of \(x\) was arbitrary, we can say that all orbits have cardinality \(|H|\).

$$\tag*{$\blacksquare$}$$

For all \(x \in G\), \(x\) belongs to the orbit of \(x\) under the action of \(H\) because \(1 \in H\) and \(x = 1x\).

Let \(O_g\) denote the orbit of \(g \in G\). For distinct \(x,y \in G\), suppose there is \(k \in G\) such that \(k \in O_x\) and \(k \in O_y\). Then for some \(h_1,h_2 \in H\), $$ h_1x = k = h_2y\\x = h_1^{-1}h_2y\\x \in O_y $$

This means that \(O_x = O_y\) since, as shown in the preceding exercise, an orbit is an equivalence class. Therefore, \(O_x \cap O_y \ne \emptyset\implies O_x = O_y\).

Hence, every element in \(G\) belongs to some orbit, and each pair of orbits are either identical or disjoint. That is, the orbits partition \(G\).

We also showed that all orbits have cardinality \(|H|\). Since \(G\) is partitioned by the orbits of the same size \(|H|\), there exists a positive integer \(n\) such that \(|G| = n|H|\). Therefore, \(|H|\) divides \(|G|\).

$$\tag*{$\blacksquare$}$$