Lagrange’s theorem

Lagrange's Theorem

For any finite group $G$ and any Subgroup $H\le G$,

$$\boxed{|H|\mid |G|.}$$

The order of $H$ divides the order of $G$. The quotient $|G|/|H|$ is the index $[G:H]$ - the number of distinct left (or right) cosets.

Proof Outline

The left Coset of $H$ partition $G$, all have the same size $|H|$, and there are $[G:H]$ of them. Hence $|G|=|H|\cdot [G:H]$, so $|H|\mid |G|$. $■$

Consequences

1. Order of an element divides the group order. For $g\in G$, $⟨g⟩\le G$ and $|⟨g⟩|=o(g)$, so $o(g)\mid |G|$.
2. Groups of prime order are cyclic. If $|G|=p$ prime, then any non-identity $g\in G$ has $o(g)\mid p$, so $o(g)=p$, hence $⟨g⟩=G\cong C_p$.
3. Fermat’s little theorem. Apply Lagrange in $(\mathbb{Z}/p{)}^{\ast }$ of order $p-1$: $a^{p-1}\equiv 1(\mathrm{mod}p)$.
4. No subgroup of forbidden size. If $|G|=12$, no subgroup of order $5$ exists, since $5\nmid 12$.

What Lagrange Doesn’t Say

The converse fails: knowing $d\mid |G|$ does NOT guarantee a subgroup of order $d$. The classic example is $A_4$ (order $12$), which has no subgroup of order $6$.

The partial converse via prime powers is Sylow’s theorems.

Index Counting

A useful reformulation:

$$|G|=|H|\cdot [G:H].$$

Combined with the Orbit-Stabiliser theorem $|Gx|=[G:{\mathrm{Stab}}_G(x)]$, this often pins down orbit sizes via divisibility arguments.