Normal Subgroups and Quotient Groups

⚡ Key Takeaways

One-line statement: A subgroup $N \leq G$ is normal (written $N \trianglelefteq G$ ) when it is invariant under conjugation, and exactly these subgroups allow the set of cosets $G/N$ to inherit a group structure called the quotient group. Key result in words: Normal subgroups are precisely the kernels of group homomorphisms, and quotienting by a kernel recovers the homomorphic image (First Isomorphism Theorem). Where it applies: Throughout group theory, Galois theory, topology (fundamental groups), and the classification of finite simple groups. Who/when: Évariste Galois (c. 1831) implicitly; formalized by Camille Jordan, Otto Hölder, and others in the late 19th century.

💡 The Core Insight

Coset multiplication "wants" to be well-defined by the rule $(aN)(bN) = abN$ , but this requires that the answer not depend on which representatives $a, b$ we chose. That well-definedness is logically equivalent to normality. So normality is not an arbitrary technical hypothesis — it is the condition that lets a partition of $G$ into cosets become a quotient group. Normality $\Leftrightarrow$ "the cosets form a group" $\Leftrightarrow$ " $N$ is a kernel."

DefinitionNormal Subgroup

Let $G$ be a group and $N \leq G$ a subgroup. Then $N$ is normal in $G$ , written $N \trianglelefteq G$ , if any of the following equivalent conditions hold:

1

\quad \forall g \in G:\quad gNg^{-1} = N

Derivation hint: This is the definition by conjugation invariance of the whole set; $gNg^{-1} = \{gng^{-1} : n \in N\}$ .

2

\quad \forall g \in G:\quad gN = Ng

Derivation hint: Multiply $gNg^{-1}=N$ on the right by $g$ ; left and right cosets coincide.

3

\quad \forall g \in G,\ \forall n \in N:\quad gng^{-1} \in N

Derivation hint: Condition [1] reduces to the inclusion $gNg^{-1} \subseteq N$ for all $g$ ; applying it to $g^{-1}$ gives the reverse inclusion, so $\subseteq$ for all $g$ already forces equality. Here $G$ is the ambient group, $N$ the candidate subgroup, $g$ ranges over all of $G$ , and $n$ over $N$ . The element $gng^{-1}$ is the conjugate of $n$ by $g$ .

⚠️ Common mistake

Condition [1], $gNg^{-1} = N$ , is a statement about the set $N$ , not about individual elements. It does not say $gng^{-1} = n$ (that would be centrality). It says the conjugation map permutes $N$ among itself. Conflating these is the single most common error.

DefinitionQuotient Group

If $N \trianglelefteq G$ , the quotient group $G/N$ is the set of left cosets

4

\quad G/N = \{\, gN : g \in G \,\}

Derivation hint: A left coset is $gN = \{gn : n \in N\}$ ; distinct cosets partition $G$ . equipped with the operation

5

\quad (aN)(bN) = (ab)N

Derivation hint: Define multiplication of cosets via representatives; well-definedness requires normality (proved below). The identity is $eN = N$ , and the inverse of $aN$ is $a^{-1}N$ . The order of the quotient (when finite) is the index:

6

\quad |G/N| = [G : N] = \frac{|G|}{|N|}

Derivation hint: Lagrange's theorem — cosets all have size $|N|$ and partition $G$ , so the number of cosets is $|G|/|N|$ .

TheoremQuotient Group Theorem

Let $N \leq G$ . The operation $(aN)(bN) = abN$ on the set of cosets is well-defined if and only if $N \trianglelefteq G$ . When it is well-defined, $G/N$ is a group.

TheoremFirst Isomorphism Theorem

Let $\varphi : G \to H$ be a group homomorphism. Then:

7

\quad \ker\varphi = \{ g \in G : \varphi(g) = e_H \} \ \trianglelefteq\ G

Derivation hint: For $k \in \ker\varphi$ and $g \in G$ , $\varphi(gkg^{-1}) = \varphi(g)\varphi(k)\varphi(g)^{-1} = \varphi(g)e_H\varphi(g)^{-1} = e_H$ . and the induced map is an isomorphism:

8

\quad G/\ker\varphi \ \cong\ \operatorname{im}\varphi

Derivation hint: Define $\bar\varphi(g\ker\varphi) = \varphi(g)$ ; check well-defined, injective (trivial kernel), and surjective onto the image.

ProofWell-definedness

\Leftrightarrow

Normality

Goal: Show $(aN)(bN)=abN$ is well-defined exactly when $N \trianglelefteq G$ .

Step 1 (Setup). Suppose $aN = a'N$ and $bN = b'N$ . Then $a' = an_1$ and $b' = bn_2$ for some $n_1, n_2 \in N$ . Well-definedness means $a'b'N = abN$ , i.e. $(ab)^{-1}(a'b') \in N$ .

Step 2 (Compute the discrepancy).

(ab)^{-1}(a'b') = b^{-1}a^{-1}\,(an_1)(bn_2) = b^{-1}a^{-1}a n_1 b n_2 = b^{-1} n_1 b\, n_2.

For this to lie in $N$ for all choices, since $n_2 \in N$ already, we need $b^{-1}n_1 b \in N$ .

Step 3 (Normality is necessary). Taking $a=a'=e$ (so $n_1$ arbitrary in $N$ ) and $b$ arbitrary, well-definedness forces $b^{-1}n_1 b \in N$ for all $b \in G$ , $n_1 \in N$ . By [3] (applied to $g = b^{-1}$ ), this is exactly $N \trianglelefteq G$ .

Step 4 (Normality is sufficient). Conversely, if $N \trianglelefteq G$ then $b^{-1}n_1 b \in N$ by [3], so $b^{-1}n_1 b\, n_2 \in N$ , giving $a'b'N = abN$ . The operation is well-defined.

Step 5 (Group axioms). Associativity: $((aN)(bN))(cN) = (ab)N\,(cN) = abcN = (aN)((bN)(cN))$ , inherited from $G$ . Identity: $(aN)(eN) = aN$ . Inverse: $(aN)(a^{-1}N) = aa^{-1}N = N$ . $\blacksquare$

∎

ProofFirst Isomorphism Theorem

Step 1 (Kernel is normal). Shown in the derivation hint for [7].

Step 2 (Define the map). Let $K = \ker\varphi$ . Define $\bar\varphi : G/K \to H$ by $\bar\varphi(gK) = \varphi(g)$ .

Step 3 (Well-defined). If $gK = g'K$ then $g^{-1}g' \in K$ , so $\varphi(g^{-1}g') = e_H$ , hence $\varphi(g)=\varphi(g')$ . The value does not depend on the representative.

Step 4 (Homomorphism). $\bar\varphi(gK \cdot g'K) = \bar\varphi(gg'K) = \varphi(gg') = \varphi(g)\varphi(g') = \bar\varphi(gK)\bar\varphi(g'K)$ .

Step 5 (Injective). If $\bar\varphi(gK) = e_H$ then $\varphi(g)=e_H$ , so $g \in K$ , i.e. $gK = K$ , the identity coset. Trivial kernel $\Rightarrow$ injective.

Step 6 (Surjective onto image). By construction $\operatorname{im}\bar\varphi = \operatorname{im}\varphi$ . Hence $\bar\varphi : G/K \to \operatorname{im}\varphi$ is a bijective homomorphism, i.e. an isomorphism. $\blacksquare$

∎

mathematician · Évariste Galois | 1811–1832

Galois, who died in a duel at twenty, introduced the conceptual seed of normality while studying the solvability of polynomial equations by radicals. In his memoir he distinguished subgroups whose cosets behaved "regularly" — what we now call normal subgroups — because these correspond to field extensions that can be peeled off cleanly. His notion of a normal series with simple quotients underlies the modern statement that a polynomial is solvable by radicals iff its Galois group is solvable. The word "normal" and the systematic theory came later (Jordan, Hölder), but the strategic insight — that the invariant subgroups are the ones you can quotient by — is Galois's.

🧠 Intuition

Think of cosets as the "fibers" of a measurement. A homomorphism $\varphi: G \to H$ groups together all elements with the same image; those clumps are exactly the cosets of $\ker\varphi$ . Quotienting is deliberate forgetting: we declare two elements equal if they differ by something in $N$ , then ask whether the group law survives this forgetting. It survives precisely when $N$ is "symmetric enough" — invariant under conjugation — so that the act of forgetting commutes with multiplication. A second picture: $\mathbb{Z}/n\mathbb{Z}$ is the prototype. Here $G = \mathbb{Z}$ , $N = n\mathbb{Z}$ , and we forget multiples of $n$ — that's modular arithmetic. Every quotient group generalizes "clock arithmetic" to a setting where the underlying group need not be abelian. In an abelian group every subgroup is normal (conjugation does nothing), so abelian quotients are unobstructed; the subtlety only appears in nonabelian groups.

Figure

Cosets of a normal subgroup partition G; the quotient collapses each coset to a point

ExampleThe integers modulo 6

Given: $G = (\mathbb{Z}, +)$ and $N = 6\mathbb{Z} = \{\dots,-6,0,6,12,\dots\}$ . Since $\mathbb{Z}$ is abelian, $N$ is automatically normal.

Step 1 — List cosets. A coset is $k + 6\mathbb{Z}$ . There are exactly $6$ distinct cosets:

0+6\mathbb{Z},\ 1+6\mathbb{Z},\ 2+6\mathbb{Z},\ 3+6\mathbb{Z},\ 4+6\mathbb{Z},\ 5+6\mathbb{Z}.

Step 2 — Order. By [6], $|G/N| = [\mathbb{Z}:6\mathbb{Z}] = 6$ .

Step 3 — Operation. $(4+6\mathbb{Z}) + (5+6\mathbb{Z}) = 9 + 6\mathbb{Z} = 3 + 6\mathbb{Z}$ , because $9 \equiv 3 \pmod 6$ .

Step 4 — Identify. This is precisely the additive group $\mathbb{Z}/6\mathbb{Z}$ , the integers mod 6.

\boxed{\ \mathbb{Z}/6\mathbb{Z} \cong (\{0,1,2,3,4,5\},\ +\!\!\!\mod 6),\quad |\mathbb{Z}/6\mathbb{Z}| = 6\ }

ExampleQuotient of S₃ by A₃

Given: $G = S_3$ (symmetric group on 3 letters, $|G|=6$ ) and $N = A_3 = \{e, (123), (132)\}$ , the alternating group, $|N| = 3$ .

Step 1 — Normality. $A_3$ has index $[S_3:A_3] = 6/3 = 2$ . Any subgroup of index 2 is normal (its two left cosets are $N$ and $G\setminus N$ , identical to the two right cosets). So $A_3 \trianglelefteq S_3$ .

Step 2 — Cosets. Two cosets: $N = A_3$ (the even permutations) and $tN$ where $t$ is any transposition, e.g. $(12)$ , giving the odd permutations.

Step 3 — Operation (sign multiplication).

(tN)(tN) = t^2 N = eN = N,\qquad (N)(tN) = tN.

This matches the rule even $\times$ even $=$ even, odd $\times$ odd $=$ even.

Step 4 — Order and identification. $|S_3/A_3| = 2$ . The unique group of order 2 is $\mathbb{Z}/2\mathbb{Z}$ ; here it records the sign (parity) of a permutation, i.e. the sign homomorphism $\operatorname{sgn}: S_3 \to \{\pm 1\}$ with kernel $A_3$ .

\boxed{\ S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z},\quad |S_3/A_3| = 2\ }

By the First Isomorphism Theorem, this is $S_3/\ker(\operatorname{sgn}) \cong \operatorname{im}(\operatorname{sgn}) = \{\pm 1\}$ .

⚠️ Common mistake

"Normal means elements commute with everything." False. Normality is $gNg^{-1} = N$ as sets, not $gn = ng$ for each $n$ . The condition that every element commutes defines the center $Z(G)$ , a much stronger (and always-normal) special case.

"Every subgroup is normal." True only in abelian groups (and a few others, e.g. quaternion-like Hamiltonian groups). In $S_3$ , the subgroup $\langle (12)\rangle = \{e,(12)\}$ is not normal.

" $G/N$ is a subgroup of $G$ ." Almost never. $G/N$ is a group whose elements are cosets, not elements of $G$ . Confusing the quotient with a complementary subgroup leads to false claims like $G \cong N \times (G/N)$ , which can fail (see below).

CounterexampleA non-normal subgroup, and a non-splitting quotient

Non-normal subgroup. In $S_3$ , let $H = \{e, (12)\}$ . Conjugate by $(13)$ :

(13)(12)(13)^{-1} = (13)(12)(13) = (23) \notin H.

So $H$ is not normal, and indeed $(12)H \neq H(12)$ : the left coset $\{(12), e\}\cdot$ ... — concretely $aH \ne Ha$ for $a=(13)$ . The coset product $(aH)(bH)$ is genuinely ill-defined here.

Quotient does not "split." In $G = \mathbb{Z}/4\mathbb{Z}$ take $N = 2\mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$ . Then $G/N \cong \mathbb{Z}/2\mathbb{Z}$ , but

\mathbb{Z}/4\mathbb{Z} \;\not\cong\; (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}).

Both sides have order 4, yet the left has an element of order 4 and the right does not. Knowing $N$ and $G/N$ does not determine $G$ — that ambiguity is the subject of the Extension Problem.

connection

Normal subgroups and quotients are the central machinery linking several pillars of algebra. They are inseparable from the Group Homomorphisms picture via the kernel/image correspondence and the First Isomorphism Theorem; the Second and Third Isomorphism Theorems and the Correspondence Theorem all describe how subgroups of $G/N$ mirror subgroups of $G$ containing $N$ . Repeated quotienting builds composition series, whose simple quotients are governed by the Jordan–Hölder Theorem and lead to the Classification of Finite Simple Groups — the "atoms" left when no further nontrivial normal subgroup exists. In Galois Theory, normal subgroups of the Galois group correspond exactly to normal (Galois) intermediate field extensions, and solvable groups (chains of abelian quotients) characterize solvability by radicals. In topology, the quotient of a Fundamental Group by a normal subgroup corresponds to a covering space.

🔬 Where this appears

⏳ History

- c. 1831 — Galois: Implicitly identifies "invariant" (normal) subgroups while studying permutation groups of polynomial roots; introduces the strategic idea of solvability via chains of such subgroups.

- 1846 — Liouville: Publishes Galois's manuscripts posthumously, bringing the ideas to a wider audience.

- 1854 — Cayley: Gives an abstract definition of a (finite) group, setting the stage for intrinsic, axiomatic treatment.

- 1870 — Camille Jordan: Traité des substitutions systematizes normal subgroups ("sous-groupes invariants") and proves a version of what becomes the composition-series uniqueness theorem.

- 1889 — Otto Hölder: Completes and refines the Jordan–Hölder Theorem, establishing uniqueness of composition factors and crystallizing the role of simple quotients.

- 1893 — Heinrich Weber: Provides a fully abstract axiomatic definition of groups, normal subgroups, and quotient groups in modern style.

- 1895–1896 — Frobenius & others: Develop character theory and deepen the structural study of normal subgroups in finite groups.

- 1920s — Emmy Noether: Elevates the isomorphism theorems to general algebraic structures, framing quotients as a universal construction; her formulation is the one taught today.

- 1955–2004 — Classification of Finite Simple Groups: A vast collaborative effort culminates in listing all groups with no nontrivial normal subgroups — the endpoint of repeated quotienting.

open-problem

Normality is not transitive. If $K \trianglelefteq H \trianglelefteq G$ it need not follow that $K \trianglelefteq G$ . (Example: in $G = A_4$ or in dihedral groups, one can find such a "subnormal but not normal" chain.) Subgroups reachable by a chain of normal inclusions are called subnormal, and the study of subnormality is a genuine subfield.

The extension problem. Given $N \trianglelefteq G$ with quotient $Q = G/N$ , the pair $(N, Q)$ does not determine $G$ . Classifying all groups $G$ fitting into $1 \to N \to G \to Q \to 1$ is the Extension Problem, controlled by group cohomology $H^2(Q, Z(N))$ ; it remains computationally formidable in general.

Detecting simplicity. Determining whether a given group has any nontrivial normal subgroup (i.e. is simple) was historically extremely hard; the full classification of finite simple groups runs to thousands of pages and its complete write-up ("the second-generation proof") is still being consolidated.

Infinite groups. For infinite (especially topological/Lie) groups one must require closed normal subgroups for the quotient to be well-behaved as a topological group — set-theoretic normality alone is insufficient.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also