Lagrange's Theorem (Group Theory)

Theorem
Let $$H$$ be a subgroup of a finite group $$G$$. Then $$|H|$$ divides $$|G|$$.

Proof
First, we need the following lemma:

Lemma: Let $$G$$ be a finite group, $$H \leq G$$. Let $$\sim$$ be the relation on $$G$$ such that $$a\sim b \leftrightarrow a\in Hb$$

Then $$\sim$$ is an equivalence relation, with the cosets of $$H$$ as its equivalence classes.

Proof of Lemma
Reflexivity:As $$H$$ is a subgroup, $$e\in H$$, so $$a = ea\in Ha$$. Thus $$a\sim a$$.

Symmetry: $$ a\sim b\rightarrow b\in Ha \rightarrow b=ha,~h\in H \rightarrow h^{-1}b = a\rightarrow a\in Hb~ \mbox{as}~h^{-1}\in H\rightarrow b\sim a$$

Transitivity: $$ a\sim b~\mbox{and}~b\sim c\mbox{, then}~a=h_1b,~b=h_2c;~h_1,h_2\in H$$

$$\rightarrow a = h_1b = h_1(h_2c) = (h_1h_2)c \rightarrow a\in Hc \rightarrow a\sim c$$

For the equivalence classes:

$$ [a] = \{b\in G~|~b\sim a\} = \{b\in G~|~b\in Ha\} = Ha$$. Q.E.D.

Proof of Lagrange's Theorem
By the above lemma, the cosets of $$H$$ partition $$G$$. Note that $$ |H| = |Hg|~\forall~g\in G$$ since multiplication by group elements induces an injective map. That is, $$gh_1=gh_2\rightarrow h_1=h_2$$. Thus, presuming there are $$k$$ distinct cosets of $$H$$, we have

$$|G| = |H|\cdot k$$

Thus $$ |H|$$ divides $$|G|$$.

Q.E.D.