Lagrange's Theorem | Mathematics

Let $H \leq G$. For $a \in G$The left coset of $H$ containing $a$ is

$aH = \{ah : h \in H\}$

The right coset is $Ha = \{ha : h \in H\}$.

Proposition 3.1. The cosets of $H$ in $G$ partition $G$.

Proof. Define $a \sim b$ if $a^{-1}b \in H$. This is an equivalence relation: reflexive ($a^{-1}a = e \in H$), symmetric ($a^{-1}b \in H \Rightarrow b^{-1}a = (a^{-1}b)^{-1} \in H$), Transitive ($a^{-1}b, b^{-1}c \in H \Rightarrow a^{-1}c = (a^{-1}b)(b^{-1}c) \in H$). The equivalence class of $a$ is exactly $aH$. $\blacksquare$

Proposition 3.2. $|aH| = |H|$ for all $a \in G$.

Proof. The map $\phi : H \to aH$ given by $\phi(h) = ah$ is a bijection. $\blacksquare$

Theorem 3.3 (Lagrange’s Theorem). If $H$ is a subgroup of a finite group $G$ Then $|H|$ divides $|G|$.

Proof. The cosets of $H$ partition $G$ into disjoint sets, each of size $|H|$. If there are $k$ cosets, Then $|G| = k \cdot |H|$ So $|H|$ divides $|G|$. $\blacksquare$

The number of cosets is called the index of $H$ in $G$Denoted $[G : H]$.

Corollary 3.4. The order of any element of $G$ divides $|G|$.

Proof. $|g| = |\langle g \rangle|$ And $\langle g \rangle \leq G$ So $|g|$ divides $|G|$ by Lagrange. $\blacksquare$

Corollary 3.5 (Fermat’s Little Theorem). If $p$ is prime and $\gcd(a, p) = 1$ Then $a^{p-1} \equiv 1 \pmod{p}$.

Proof. $\mathbb{Z}/p\mathbb{Z}$ has $p$ elements. The multiplicative group $(\mathbb{Z}/p\mathbb{Z})^*$ has order $p - 1$. The order of $[a]$ divides $p - 1$ So $[a]^{p-1} = [1]$. $\blacksquare$

Corollary 3.6 (Euler’s Theorem). If $\gcd(a, n) = 1$ Then $a^{\phi(n)} \equiv 1 \pmod{n}$Where $\phi$ Is Euler’s totient function.

Problem. Show that every group of prime order $p$ is cyclic.

Solution. Let $G$ be a group of order $p$ and $g \in G$ with $g \neq e$. By Corollary 3.4, $|g|$ divides $p$. Since $g \neq e$, $|g| \neq 1$. Since $p$ is prime, $|g| = p$. Thus $\langle g \rangle = G$ And $G$ is cyclic. $\blacksquare$

Problem. Let $H = \langle (1\ 2\ 3) \rangle \leq S_3$. Find all left cosets of $H$ in $S_3$.

Problem. Let $H = \langle 4 \rangle \leq \mathbb{Z}/12\mathbb{Z}$. Find all cosets of $H$.

Corollary 3.7. If $[G : H] = 2$ Then $H \trianglelefteq G$.

Proof. There are exactly two left cosets $H$ and $aH$ And exactly two right cosets $H$ and $Ha$. Since the cosets partition $G$We have $aH = G \setminus H = Ha$. Thus $gH = Hg$ for all $g \in G$ So $H$ is normal. $\blacksquare$

Corollary 3.8 (Product Formula). If $H, K \leq G$ are finite subgroups, then

$|HK| = \frac{|H||K|}{|H \cap K|}$

Proof. The map $H \times K \to HK$ given by $(h, k) \mapsto hk$ is surjective. For any $x = hk \in HK$ The fiber is $\{(hc^{-1}, ck) : c \in H \cap K\}$Which has size $|H \cap K|$. Thus $|H||K| = |HK| \cdot |H \cap K|$. $\blacksquare$

• Confusing index with order. The index $[G:H]$ is the number of cosets, not the order of $H$.
• Assuming Lagrange’s converse. If $d \mid |G|$, there need not exist a subgroup of order $d$. The converse holds for cyclic groups but fails for $A_4$ (order 12, no subgroup of order 6).
• Forgetting that cosets partition $G$. Each element of $G$ belongs to exactly one left coset and exactly one right coset of $H$.
• Misapplying Fermat’s Little Theorem. It requires $p$ prime and $\gcd(a,p) = 1$; omitting the coprimality condition gives incorrect results.

Lagrange’s theorem says that the order of any subgroup divides the order of the group. The proof is beautifully simple: the left cosets of $H$ partition $G$ into disjoint sets, each of size $|H|$. If there are $k$ cosets, then $|G| = k \cdot |H|$, so $|H|$ divides $|G|$.

Think of it as tiling: if you have a floor of area $|G|$ and tiles of area $|H|$, you need exactly $k = |G|/|H|$ tiles, and this must be a whole number. The cosets are the tiles.

Why the converse fails. Lagrange’s theorem says the size of any subgroup must divide $|G|$. But the converse --- that every divisor of $|G|$ gives rise to a subgroup --- is false. The obstruction is that the “tiling” might not be achievable by a subgroup. For $A_4$ (order 12), there is no subgroup of order 6, even though $6 \mid 12$. The reason is that a subgroup of order 6 would have index 2, hence would be normal. But $A_4$ has no normal subgroup of order 6 (its only proper normal subgroup is the Klein four-group $V_4$ of order 4).

Connection to number theory. Lagrange’s theorem applied to $(\mathbb{Z}/p\mathbb{Z})^*$ gives Fermat’s Little Theorem: the order of any element divides $p - 1$, so $a^{p-1} \equiv 1 \pmod{p}$. This is the foundation of RSA encryption and primality testing.

Problem. List all subgroups of $D_4$ (the dihedral group of the square, order 8) and verify Lagrange’s theorem for each.

• Number theory: Fermat’s Little Theorem and Euler’s Theorem underpin RSA encryption and primality testing.
• Coding theory: The structure of cosets of subgroups in finite groups is used in linear codes and syndrome decoding.
• Computational group theory: Lagrange’s Theorem bounds the search space when testing subgroup membership; the index determines the number of coset representatives needed.

flowchart TD
    A[3_Lagrange S Theorem] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Group Actions — The orbit-stabilizer theorem generalises Lagrange’s theorem to group actions, linking subgroup indices to orbit sizes.

• The Sylow Theorems — Sylow’s theorems refine Lagrange’s theorem by guaranteeing subgroups of prime-power order and constraining their count.

• Classification of Groups of Small Order — Lagrange’s theorem limits the possible subgroup structure used in classifying small-order groups.

• Common Pitfalls — The common pitfalls section warns against assuming the converse of Lagrange’s theorem and confusing index with order.

• Classical Mechanics

• Electromagnetism