Normal Subgroups and Quotient Groups

The concept of normal subgroups was introduced by Évariste Galois in his 1830 letter to Augustin-Louis Cauchy, written the night before his fatal duel. Galois showed that a polynomial equation is solvable by radicals if and only if its Galois group has a composition series whose factors are all abelian — a result that connected group theory to the ancient problem of solving polynomial equations. The quotient group $G/N$ was formalised later by Camille Jordan (1870) in his Traité des substitutions. Emmy Noether’s work in the 1920s-30s showed that normal subgroups correspond to congruence relations, connecting group theory to lattice theory and universal algebra. Today, normal subgroups are fundamental to the classification of finite simple groups — one of the great achievements of 20th-century mathematics.

4.1 Normal Subgroups

A subgroup $N \leq G$ is normal (written $N \trianglelefteq G$ ) if $gNg^{-1} = N$ for all $g \in G$ I.e., $gng^{-1} \in N$ for all $g \in G$ and all $n \in N$ .

Proposition 4.1. Every subgroup of an abelian group is normal.

Proposition 4.2. The following are equivalent for $N \leq G$ :

$N \trianglelefteq G$ .
$gN = Ng$ for all $g \in G$ (left and right cosets coincide).
The product of two left cosets is again a left coset: $(aN)(bN) = (ab)N$ .

Proof of (1) $\Rightarrow$ (3). Let $a_1n_1 \in aN$ and $b_1 n_2 \in bN$ . Then $(a_1 n_1)(b_1 n_2) = a_1 b_1 (b_1^{-1} n_1 b_1) n_2$ . Since $N$ is normal, $b_1^{-1} n_1 b_1 \in N$ So $(b_1^{-1} n_1 b_1) n_2 \in N$ Giving $(a_1 n_1)(b_1 n_2) \in a_1 b_1 N$ . $\blacksquare$

4.2 The Quotient Group

When $N \trianglelefteq G$ The set $G/N = \{gN : g \in G\}$ of cosets forms a group under

$(aN)(bN) = (ab)N$

Called the quotient group of $G$ by $N$ .

Theorem 4.3. If $G$ is finite and $N \trianglelefteq G$ Then $|G/N| = [G : N] = |G|/|N|$ .

Example. $S_3 / A_3 \cong \mathbb{Z}/2\mathbb{Z}$ .

Example. $\mathbb{Z}/n\mathbb{Z}$ is the quotient of $\mathbb{Z}$ by $n\mathbb{Z}$ .

4.3 Worked Examples: Computing Quotient Groups

Problem. The quaternion group $Q_8 = \{1, -1, i, -i, j, -j, k, -k\}$ has center $Z(Q_8) = \{1, -1\}$ . Compute $Q_8 / Z(Q_8)$ .

Solution

Solution. Since $|Q_8| = 8$ and $|Z(Q_8)| = 2$ We have $|Q_8/Z(Q_8)| = 4$ . The cosets are:

$Z(Q_8) = \{1, -1\}, \quad iZ(Q_8) = \{i, -i\}, \quad jZ(Q_8) = \{j, -j\}, \quad kZ(Q_8) = \{k, -k\}$

Multiplication in the quotient: $(iZ)(iZ) = i^2 Z = (-1)Z = Z$ (the identity coset). Similarly $(jZ)(jZ) = Z$ and $(kZ)(kZ) = Z$ . Also $(iZ)(jZ) = ijZ = kZ$ and $(jZ)(iZ) = jiZ = (-k)Z = kZ$ (since $-k \in kZ$ ). Every non-identity element has order $2$ And the group is abelian. Therefore $Q_8 / Z(Q_8) \cong V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ . $\blacksquare$

Problem. Let $N = \langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle \leq S_4$ . Show that $N \trianglelefteq S_4$ and identify $S_4/N$ .

Solution

Solution. $N = \{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$ is the Klein four-group $V_4$ With $|N| = 4$ . To verify $N \trianglelefteq S_4$ Note that conjugation preserves cycle type. Each non-identity element of $N$ is a product of two disjoint transpositions. Since $S_4$ acts Transitively on such elements (any pair of disjoint transpositions can be mapped to any other by relabeling), $N$ is closed under conjugation.

Thus $|S_4/N| = 24/4 = 6$ . That $S_4/N$ is non-abelian (e.g., the images of $(1\ 2)$ and $(2\ 3)$ do not commute), so $S_4/N \cong S_3$ . $\blacksquare$

4.4 Worked Example: The First Isomorphism Theorem

Problem. Define $\phi : GL_2(\mathbb{R}) \to \mathbb{R}^*$ by $\phi(A) = \det(A)$ . Identify $\ker(\phi)$ and $GL_2(\mathbb{R})/\ker(\phi)$ .

Solution

Solution. $\phi$ is a homomorphism since $\det(AB) = \det(A)\det(B)$ . It is surjective: For any $r \in \mathbb{R}^*$ The matrix $\begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$ has determinant $r$ .

The kernel is $\ker(\phi) = \{A \in GL_2(\mathbb{R}) : \det(A) = 1\} = SL_2(\mathbb{R})$ .

By the first isomorphism theorem (Theorem 5.3), $GL_2(\mathbb{R})/SL_2(\mathbb{R}) \cong \mathbb{R}^*$ . $\blacksquare$

Problem. Show that $\mathbb{C}^* / S^1 \cong \mathbb{R}^+$ Where $S^1 = \{z \in \mathbb{C}^* : |z| = 1\}$ .

Solution

Solution. Define $\phi : \mathbb{C}^* \to \mathbb{R}^+$ by $\phi(z) = |z|$ . This is a homomorphism since $|zw| = |z||w|$ . It is surjective since for any $r > 0$ $\phi(r) = r$ . The kernel is $\ker(\phi) = \{z \in \mathbb{C}^* : |z| = 1\} = S^1$ The unit circle. By the first isomorphism theorem, $\mathbb{C}^* / S^1 \cong \mathbb{R}^+$ . $\blacksquare$

4.5 Common Pitfalls

Forgetting to check all cosets. When verifying normality via $gNg^{-1} \subseteq N$ , you must check every $g \in G$ , not just generators.
Confusing $G/N$ with $N/G$ . The quotient $G/N$ is defined only when $N \trianglelefteq G$ ; the notation is not symmetric.
Assuming all subgroups are normal. In non-abelian groups, most subgroups are not normal. For instance, $\langle (1\ 2) \rangle$ is not normal in $S_3$ .
Miscalculating coset products. Always reduce representatives: $(aN)(bN) = (ab)N$ , but $(ab)N$ may simplify further if $ab$ can be rewritten.

4.6 Key Relationships

Concept	Condition	Consequence
$N \trianglelefteq G$	$gNg^{-1} = N$ for all $g$	$G/N$ is a group
$[G:N] = 2$	Index 2 subgroup	$N$ is automatically normal
First Isomorphism Theorem	$\phi : G \to H$ homomorphism	$G/\ker(\phi) \cong \operatorname{im}(\phi)$
Center $Z(G)$	Commutes with everything	Always a normal subgroup
Commutator subgroup $[G,G]$	Generated by $g_1 g_2 g_1^{-1} g_2^{-1}$	Always normal; $G/[G,G]$ is abelian

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