TITLE: Proof of Lagrange theorem - Order of a subgroup divides order of the group
QUESTION [10 upvotes]: The Lagrange theorem states:

If $G$ is a finite group, and $H$ a subgroup of $G$, then the order of $H$ will divide the order of $G$. More precisely, $|G| = |H| \cdot (\text{number of left cosets of }H)$:
$$|G| = |H| \cdot (G:H)$$

The proof I have in my notes says:

$G$ consists of $\{G:H\}$ cosets, each of them consists of $|H|$ elements, the cosets are disjoint.

That's it. How does this prove the theorem?

REPLY [3 votes]: One simple approach is to show that the (left) cosets are distinct. Once we have that, the rest follows fairly obviously.
Step 1: We are given that $H$ is a group (a subgroup of $G$). In particular for every $h\in H$ we also have $h^{-1}\in H$. So for some element $j\in G, j  \not \in H$ we have that $jh \not \in H$, since otherwise $jhh^{-1}=j\in H$.
Step 2: Form the coset $jH$ from $\bigcup_{h\in H} jh$, and note $|jH| = |H|$. Of course since $e\in H$, $j\in jH$. Now consider the coset $kH$ with $k\in jH$. We must have $k=jh_0$ so also $j=kh_0^{-1}$. Then any $jh_i = kh_0^{-1}h_i$ and of course $h_0^{-1}h_i \in H$ so $kH = jH$.
Step 3: Thus cosets are distinct (since any common member implies that the cosets are identical), all are of size $|H|$, and every member of $G$ is in a coset, so $|H|$ divides $|G|$