Unit 4: Cosets, Lagrange's Theorem, Normal Subgroups & Quotient Groups

1. Definition of Left and Right Cosets

Let $G$ be a group and let $H$ be a subgroup of $G$ ($H \le G$). For any element $g \in G$:

Definition 4.1 (Left Coset): The left coset of $H$ in $G$ containing $g$ is the subset:

Definition 4.2 (Right Coset): The right coset of $H$ in $G$ containing $g$ is the subset:

The element $g$ is called a coset representative of $gH$ (or $Hg$). Notice that if $e \in G$ is the identity element, then $eH = H = He$. Therefore, the subgroup $H$ itself is always both a left coset and a right coset (represented by $e$ or any element $h \in H$).

2. Equivalence Relation Induced by a Subgroup

Cosets arise naturally from an equivalence relation on the elements of $G$.

Theorem 4.1 (Coset Equivalence Relation): Let $H \le G$. Define a binary relation $\sim_L$ on $G$ by:

Then $\sim_L$ is an equivalence relation on $G$, and the equivalence class of $a \in G$ is precisely the left coset $aH$.

Complete Proof:

1. Reflexivity:

For any $a \in G$, $a^{-1} a = e$. Since $H$ is a subgroup, $e \in H$. Thus $a^{-1} a \in H$, which implies $a \sim_L a$.

2. Symmetry:

Suppose $a \sim_L b$. Then $a^{-1} b \in H$. Since $H$ is closed under inversion:

Therefore, $b \sim_L a$.

3. Transitivity:

Suppose $a \sim_L b$ and $b \sim_L c$. Then $a^{-1} b \in H$ and $b^{-1} c \in H$. Since $H$ is closed under the group operation:

Therefore, $a \sim_L c$.

Thus $\sim_L$ is an equivalence relation. Now let us compute the equivalence class $[a]$:

If $a^{-1} x = h \in H$, multiplying on the left by $a$ yields $x = ah \in aH$. Conversely, if $x \in aH$, then $x = ah$ for some $h \in H$, so $a^{-1} x = h \in H$, meaning $x \in [a]$. Thus $[a] = aH$. $\blacksquare$

3. Fundamental Properties of Cosets

Lemma 4.1 (Coset Properties): Let $H \le G$ and $a, b \in G$.

1. $a \in aH$ (since $e \in H$, $a = ae \in aH$).
2. $aH = H \iff a \in H$.
3. $aH = bH \iff a^{-1} b \in H \iff b \in aH$.
4. Either $aH = bH$ or $aH \cap bH = \emptyset$ (left cosets partition $G$).
5. There exists a bijection between $H$ and any left coset $aH$, so $|aH| = |H|$.
6. There exists a bijection between any left coset $aH$ and the right coset $Ha^{-1}$, so the number of left cosets equals the number of right cosets.

Proof of Cardinality Equality ($|aH| = |H|$):

Define the mapping $\psi: H \to aH$ by $\psi(h) = ah$ for all $h \in H$.

• Surjectivity: By definition of $aH$, any element $x \in aH$ has the form $x = ah$ for some $h \in H$, so $\psi(h) = x$.
• Injectivity: Suppose $\psi(h_1) = \psi(h_2)$. Then $ah_1 = ah_2$.

Multiplying on the left by $a^{-1}$:

Thus $\psi$ is a bijection, and $|aH| = |H|$. In the case of finite groups, every left coset has identical cardinality $|H|$. $\blacksquare$

1. Lagrange's Theorem for Finite Groups

Lagrange's Theorem is widely regarded as the cornerstone of finite group theory.

Theorem 4.2 (Lagrange's Theorem): If $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$:

Moreover, the number of distinct left (or right) cosets of $H$ in $G$, denoted by $[G : H]$ (the index of $H$ in $G$), satisfies:

Complete Proof:

1. By Theorem 4.1, the relation $a \sim b \iff a^{-1} b \in H$ is an equivalence relation on $G$.
2. By the Fundamental Theorem of Equivalence Relations (Theorem 1.1), the equivalence classes—which are the distinct left cosets $a_1 H, a_2 H, \dots, a_k H$—form a partition of $G$.
3. Therefore:

where $a_i H \cap a_j H = \emptyset$ for all $i \ne j$.

1. Taking the cardinality of both sides:

1. By Lemma 4.1, every left coset has exactly $|H|$ elements: $|a_i H| = |H|$ for all $i = 1, \dots, k$.
2. Substituting this into the sum:

1. Here $k$ is by definition the number of distinct left cosets, denoted $[G : H]$.
2. Thus $|G| = [G : H] \cdot |H|$, which proves that $|H|$ divides $|G|$ and $[G : H] = |G| / |H|$. $\blacksquare$

2. Immediate Corollaries of Lagrange's Theorem

Lagrange's theorem yields profound structural consequences across group theory:

Corollary 4.2.1 (Element Order Divides Group Order): Let $G$ be a finite group. For any element $g \in G$, the order of $g$ divides the order of $G$:

Proof:

The cyclic subgroup generated by $g$, $\langle g \rangle = \{ g^k : k \in \mathbb{Z} \}$, has order $|\langle g \rangle| = |g|$. Since $\langle g \rangle \le G$, Lagrange's Theorem immediately implies that $|\langle g \rangle| \Big| |G|$, hence $|g| \Big| |G|$. $\blacksquare$

Corollary 4.2.2 (Group Power Exponent): Let $G$ be a finite group of order $n = |G|$. For every element $g \in G$:

Proof:

Let $m = |g|$. By Corollary 4.2.1, $m \mid n$, so there exists an integer $k \in \mathbb{Z}^+$ such that $n = m \cdot k$. Then:

Corollary 4.2.3 (Groups of Prime Order are Cyclic): Let $G$ be a group of prime order $p$. Then $G$ is cyclic, and any non-identity element $g \ne e$ is a generator of $G$:

Proof:

Since $|G| = p \ge 2$, choose any element $g \in G$ with $g \ne e$. Consider the subgroup $H = \langle g \rangle$. Since $g \ne e$, $|H| \ge 2$. By Lagrange's Theorem, $|H|$ must divide $|G| = p$. Because $p$ is a prime number, its only positive divisors are $1$ and $p$. Since $|H| \ne 1$, it must be that $|H| = p$. Therefore, $H = G$, meaning $G = \langle g \rangle$. Thus $G$ is cyclic and isomorphic to $\mathbb{Z}_p$. $\blacksquare$

1. The Group of Units $U(n)$

Recall from Unit 1 that for any integer $n \ge 2$, the set of invertible elements modulo $n$ under multiplication forms a group:

Definition 4.3 (Euler's Totient Function $\phi(n)$): The totient function $\phi(n)$ is defined as the number of integers in $\{1, 2, \dots, n\}$ that are coprime to $n$:

Formula for $\phi(n)$:

If $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ is the prime factorization of $n$, then:

2. Euler's Totient Theorem

We now obtain Euler's celebrated theorem in number theory as a direct 1-line corollary of Lagrange's Theorem!

Theorem 4.3 (Euler's Theorem): If $a$ and $n$ are coprime integers ($\gcd(a, n) = 1$ with $n \ge 1$), then:

Group-Theoretic Proof:

1. Consider the group of units $G = U(n) = (\mathbb{Z}_n)^\times$.
2. The order of this group is $|G| = \phi(n)$.
3. Since $\gcd(a, n) = 1$, the equivalence class $[a]$ is an element of $G$.
4. By Corollary 4.2.2 (Group Power Exponent), for every element $g \in G$, $g^{|G|} = e$.
5. Applying this to $g = [a]$ and $|G| = \phi(n)$:

1. In modular congruence notation, this translates directly to:

3. Fermat's Little Theorem

Corollary 4.3.1 (Fermat's Little Theorem): If $p$ is a prime number and $a \in \mathbb{Z}$ is not divisible by $p$ ($p \nmid a$), then:

Equivalent formulation for all $a \in \mathbb{Z}$:

Proof:

Since $p$ is prime, all integers $1, 2, \dots, p-1$ are coprime to $p$, so $\phi(p) = p - 1$. Applying Euler's Theorem directly with $n = p$:

Multiplying both sides by $a$ gives $a^p \equiv a \pmod p$. If $p \mid a$, then $a \equiv 0 \pmod p$, so $0^p \equiv 0 \pmod p$ trivially holds. Thus $a^p \equiv a \pmod p$ holds for all $a \in \mathbb{Z}$. $\blacksquare$

1. The Necessity of Normal Subgroups

When can the set of left cosets $G/H = \{ gH : g \in G \}$ be endowed with a natural group structure? Suppose we attempt to define coset multiplication by:

This operation is well-defined if and only if the product does not depend on the choice of representatives $a$ and $b$. That is, if $a_1 \in aH$ and $b_1 \in bH$, we must have $(a_1 b_1)H = (ab)H$. As we will prove, this holds if and only if $H$ is a normal subgroup.

Definition 4.4 (Normal Subgroup): A subgroup $N \le G$ is called a normal subgroup of $G$, denoted by $N \trianglelefteq G$ (or $N \triangleleft G$ if proper), if:

where $g N g^{-1} = \{ g n g^{-1} : n \in N \}$.

2. Equivalent Characterizations of Normality

Theorem 4.4 (Characterization of Normal Subgroups): Let $N \le G$. The following statements are mutually equivalent:

1. $N \trianglelefteq G$ (i.e., $g N g^{-1} = N$ for all $g \in G$).
2. $g N g^{-1} \subseteq N$ for all $g \in G$.
3. $g N = N g$ for all $g \in G$ (every left coset is a right coset).
4. $(aN)(bN) = (ab)N$ is a well-defined binary operation on the set of cosets $G/N$.

Proof of Equivalence ($2 \implies 1$ and $1 \iff 3$):

• $2 \implies 1$: Suppose $g N g^{-1} \subseteq N$ for all $g \in G$.

Replacing $g$ with $g^{-1} \in G$:

Now multiply on the left by $g$ and on the right by $g^{-1}$:

Combining $g N g^{-1} \subseteq N$ and $N \subseteq g N g^{-1}$ yields $g N g^{-1} = N$.

• $1 \iff 3$: Suppose $g N g^{-1} = N$. Multiplying both sides on the right by $g$:

Conversely, if $g N = N g$, multiplying on the right by $g^{-1}$ gives $g N g^{-1} = N$. $\blacksquare$

Proposition 4.1 (Subgroups of Index 2 are Normal): If $H \le G$ with $[G : H] = 2$, then $H \trianglelefteq G$.

Proof:

Since $[G : H] = 2$, there are exactly two left cosets of $H$ in $G$. One is $eH = H$. The other must be the complement of $H$ in $G$:

Similarly, there are exactly two right cosets of $H$ in $G$: $He = H$ and $Hg = G \setminus H$ for all $g \notin H$. Thus, for all $g \in G$:

• If $g \in H$, $gH = H = Hg$.
• If $g \notin H$, $gH = G \setminus H = Hg$.

In both cases $gH = Hg$. Therefore $H \trianglelefteq G$. $\blacksquare$

3. Conjugacy Classes and the Class Equation

Recall the action of $G$ on itself by conjugation: $g \cdot x = g x g^{-1}$. The orbit of an element $x \in G$ under this action is its conjugacy class:

The stabilizer of $x$ is the centralizer $C_G(x) = \{ g \in G : gx = xg \}$. By the Orbit-Stabilizer Theorem (Theorem 3.4):

An element $x$ has $|\text{Cl}(x)| = 1 \iff g x g^{-1} = x$ for all $g \in G \iff gx = xg$ for all $g \in G \iff x \in Z(G)$, where $Z(G)$ is the center of $G$.

Theorem 4.5 (The Class Equation): Let $G$ be a finite group. Let $x_1, x_2, \dots, x_k$ be representatives of the distinct conjugacy classes of size $> 1$. Then:

Proof:

Since conjugacy is an equivalence relation, $G$ is partitioned into disjoint conjugacy classes:

The classes of size 1 are precisely the elements in $Z(G)$, of which there are $|Z(G)|$. Summing the cardinalities of all disjoint classes yields:

1. Construction of the Quotient Group (Factor Group)

Let $N \trianglelefteq G$. The collection of all cosets of $N$ in $G$ is denoted:

Read as "$G$ modulo $N$" or "$G \bmod N$".

Theorem 4.6 (Quotient Group Axioms): Let $N$ be a normal subgroup of $G$. The set $G/N$ forms a group under coset multiplication:

Complete Verification of Group Axioms:

1. Well-Definedness:

Suppose $a_1 N = a_2 N$ and $b_1 N = b_2 N$. Then $a_2 = a_1 n_1$ and $b_2 = b_1 n_2$ for some $n_1, n_2 \in N$. We must show $(a_2 b_2)N = (a_1 b_1)N$, which is equivalent to showing $(a_1 b_1)^{-1} (a_2 b_2) \in N$:

Since $N \trianglelefteq G$, $b_1^{-1} n_1 (b_1^{-1})^{-1} = b_1^{-1} n_1 b_1 \in N$. Let $n_3 = b_1^{-1} n_1 b_1 \in N$. Then $(a_1 b_1)^{-1} (a_2 b_2) = n_3 n_2 \in N$ (by closure of $N$). Thus $(a_2 b_2)N = (a_1 b_1)N$. The operation is well-defined!

2. Associativity:

For any $a, b, c \in G$:

Inherited directly from associativity of $G$.

3. Identity Element:

The coset $eN = N$ serves as the identity element:

4. Inverses:

For any $aN \in G/N$, the inverse is $(aN)^{-1} = a^{-1}N$:

Thus $(G/N, \cdot)$ is a group! $\blacksquare$

2. The Canonical (Natural) Projection

Definition 4.5 (Canonical Projection): The map $\pi: G \to G/N$ defined by $\pi(g) = gN$ is a surjective group homomorphism called the canonical projection (or natural projection). Its kernel is precisely $N$:

This establishes the profound duality: every normal subgroup is the kernel of a homomorphism, and the kernel of every homomorphism is a normal subgroup.