Groups and Subgroups - Key Proof

We present the complete proof of Lagrange's theorem, one of the foundational results in finite group theory.

TheoremLagrange's Theorem (Restatement)

Let $G$ be a finite group and $H$ a subgroup of $G$ . Then: $|G| = |H| \cdot [G:H]$

where $[G:H]$ is the number of distinct left cosets of $H$ in $G$ .

ProofProof of Lagrange's Theorem

We prove this in three steps.

Step 1: Cosets partition $G$

Define an equivalence relation on $G$ by $a \sim b$ if $aH = bH$ , which holds if and only if $a^{-1}b \in H$ . This is easily verified:

Reflexive: $a^{-1}a = e \in H$
Symmetric: If $a^{-1}b \in H$ , then $(a^{-1}b)^{-1} = b^{-1}a \in H$
Transitive: If $a^{-1}b \in H$ and $b^{-1}c \in H$ , then $a^{-1}c = (a^{-1}b)(b^{-1}c) \in H$

The equivalence classes are precisely the left cosets $gH$ , so they partition $G$ : $G = g_1H \cup g_2H \cup \cdots \cup g_kH$

where the cosets are pairwise disjoint and $k = [G:H]$ .

Step 2: All cosets have the same size

We show $|gH| = |H|$ for any $g \in G$ . Define $\phi: H \to gH$ by $\phi(h) = gh$ . This map is:

Well-defined: $gh \in gH$ by definition
Injective: If $gh_1 = gh_2$ , then $h_1 = h_2$ by left cancellation
Surjective: Every element of $gH$ has the form $gh$ for some $h \in H$

Thus $\phi$ is a bijection, so $|gH| = |H|$ .

Step 3: Count elements

Since $G$ is partitioned into $k$ disjoint cosets, each of size $|H|$ : $|G| = k \cdot |H| = [G:H] \cdot |H|$

This completes the proof.

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Remark

The key insight is that cosets are translates of $H$ that tile $G$ without overlap. The left multiplication map $h \mapsto gh$ preserves cardinality, making each coset an identical copy of $H$ shifted by $g$ .

ExampleCorollary: Element Orders

For any $g \in G$ , the cyclic subgroup $\langle g \rangle$ has order $|g|$ . By Lagrange's theorem, $|g|$ divides $|G|$ . Therefore: $g^{|G|} = (g^{|g|})^{|G|/|g|} = e^{|G|/|g|} = e$

This generalizes Fermat's little theorem: for $\gcd(a,p) = 1$ , we have $a^{p-1} \equiv 1 \pmod{p}$ since $a$ has order dividing $p-1$ in $(\mathbb{Z}/p\mathbb{Z})^*$ .

The proof technique—partitioning by equivalence classes and counting—recurs throughout group theory. Similar arguments appear in the orbit-counting theorem, Burnside's lemma, and the class equation.