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Normal Subgroups and Quotient Groups

A subgroup NGN \leq G is normal (written NGN \trianglelefteq G) if gNg1=NgNg^{-1} = N for all gGg \in G I.e., gng1Ngng^{-1} \in N for all gGg \in G and all nNn \in N.

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

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

  1. NGN \trianglelefteq G.
  2. gN=NggN = Ng for all gGg \in G (left and right cosets coincide).
  3. The product of two left cosets is again a left coset: (aN)(bN)=(ab)N(aN)(bN) = (ab)N.

Proof of (1) \Rightarrow (3). Let a1n1aNa_1n_1 \in aN and b1n2bNb_1 n_2 \in bN. Then (a1n1)(b1n2)=a1b1(b11n1b1)n2(a_1 n_1)(b_1 n_2) = a_1 b_1 (b_1^{-1} n_1 b_1) n_2. Since NN is normal, b11n1b1Nb_1^{-1} n_1 b_1 \in N So (b11n1b1)n2N(b_1^{-1} n_1 b_1) n_2 \in NGiving (a1n1)(b1n2)a1b1N(a_1 n_1)(b_1 n_2) \in a_1 b_1 N. \blacksquare

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

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

Called the quotient group of GG by NN.

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

Example. S3/A3Z/2ZS_3 / A_3 \cong \mathbb{Z}/2\mathbb{Z}.

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

4.3 Worked Examples: Computing Quotient Groups

Section titled “4.3 Worked Examples: Computing Quotient Groups”

Problem. The quaternion group Q8={1,1,i,i,j,j,k,k}Q_8 = \{1, -1, i, -i, j, -j, k, -k\} has center Z(Q8)={1,1}Z(Q_8) = \{1, -1\}. Compute Q8/Z(Q8)Q_8 / Z(Q_8).

Solution

Solution. Since Q8=8|Q_8| = 8 and Z(Q8)=2|Z(Q_8)| = 2We have Q8/Z(Q8)=4|Q_8/Z(Q_8)| = 4. The cosets are:

Z(Q8)={1,1},iZ(Q8)={i,i},jZ(Q8)={j,j},kZ(Q8)={k,k}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)=i2Z=(1)Z=Z(iZ)(iZ) = i^2 Z = (-1)Z = Z (the identity coset). Similarly (jZ)(jZ)=Z(jZ)(jZ) = Z and (kZ)(kZ)=Z(kZ)(kZ) = Z. Also (iZ)(jZ)=ijZ=kZ(iZ)(jZ) = ijZ = kZ and (jZ)(iZ)=jiZ=(k)Z=kZ(jZ)(iZ) = jiZ = (-k)Z = kZ (since kkZ-k \in kZ). Every non-identity element has order 22 And the group is abelian. Therefore Q8/Z(Q8)V4Z/2Z×Z/2ZQ_8 / Z(Q_8) \cong V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}. \blacksquare

Problem. Let N=(1 2)(3 4),(1 3)(2 4)S4N = \langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle \leq S_4. Show that NS4N \trianglelefteq S_4 and identify S4/NS_4/N.

Solution

Solution. N={e,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}N = \{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\} is the Klein four-group V4V_4 With N=4|N| = 4. To verify NS4N \trianglelefteq S_4Note that conjugation preserves cycle type. Each non-identity element of NN is a product of two disjoint transpositions. Since S4S_4 acts Transitively on such elements (any pair of disjoint transpositions can be mapped to any other by relabeling), NN is closed under conjugation.

Thus S4/N=24/4=6|S_4/N| = 24/4 = 6. That S4/NS_4/N is non-abelian (e.g., the images of (1 2)(1\ 2) and (2 3)(2\ 3) do not commute), so S4/NS3S_4/N \cong S_3. \blacksquare

4.4 Worked Example: The First Isomorphism Theorem

Section titled “4.4 Worked Example: The First Isomorphism Theorem”

Problem. Define ϕ:GL2(R)R\phi : GL_2(\mathbb{R}) \to \mathbb{R}^* by ϕ(A)=det(A)\phi(A) = \det(A). Identify ker(ϕ)\ker(\phi) and GL2(R)/ker(ϕ)GL_2(\mathbb{R})/\ker(\phi).

Solution

Solution. ϕ\phi is a homomorphism since det(AB)=det(A)det(B)\det(AB) = \det(A)\det(B). It is surjective: For any rRr \in \mathbb{R}^*The matrix (r001)\begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix} has determinant rr.

The kernel is ker(ϕ)={AGL2(R):det(A)=1}=SL2(R)\ker(\phi) = \{A \in GL_2(\mathbb{R}) : \det(A) = 1\} = SL_2(\mathbb{R}).

By the first isomorphism theorem (Theorem 5.3), GL2(R)/SL2(R)RGL_2(\mathbb{R})/SL_2(\mathbb{R}) \cong \mathbb{R}^*. \blacksquare

Problem. Show that C/S1R+\mathbb{C}^* / S^1 \cong \mathbb{R}^+Where S1={zC:z=1}S^1 = \{z \in \mathbb{C}^* : |z| = 1\}.

Solution

Solution. Define ϕ:CR+\phi : \mathbb{C}^* \to \mathbb{R}^+ by ϕ(z)=z\phi(z) = |z|. This is a homomorphism since zw=zw|zw| = |z||w|. It is surjective since for any r>0r > 0 ϕ(r)=r\phi(r) = r. The kernel is ker(ϕ)={zC:z=1}=S1\ker(\phi) = \{z \in \mathbb{C}^* : |z| = 1\} = S^1The unit circle. By the first isomorphism theorem, C/S1R+\mathbb{C}^* / S^1 \cong \mathbb{R}^+. \blacksquare

4.5 Intuition: What Are Normal Subgroups and Quotients?

Section titled “4.5 Intuition: What Are Normal Subgroups and Quotients?”

A normal subgroup is a subgroup that is invariant under conjugation: gNg1=NgNg^{-1} = N for all gGg \in G. This means the subgroup “looks the same” from every perspective in the group. Normality is the algebraic condition that makes quotient groups possible: when NN is normal, the cosets G/NG/N inherit a group structure because the product of two cosets is well-defined.

The quotient group G/NG/N collapses all elements of NN to the identity, creating a simpler group that captures the “large-scale” structure of GG while ignoring the internal structure of NN. The first isomorphism theorem says that G/ker(ϕ)im(ϕ)G/\ker(\phi) \cong \mathrm{im}(\phi): every homomorphism factors through its quotient. This means quotient groups are the natural objects that arise from homomorphisms. For example, Z/nZ\mathbb{Z}/n\mathbb{Z} is the quotient that collapses all multiples of nn to zero, creating a finite cyclic group. The cosets of a normal subgroup partition the group into equal-sized pieces, and the quotient group describes how these pieces fit together.

  • Forgetting to check all cosets. When verifying normality via gNg1NgNg^{-1} \subseteq N, you must check every gGg \in G, not just generators.
  • Confusing G/NG/N with N/GN/G. The quotient G/NG/N is defined only when NGN \trianglelefteq G; the notation is not symmetric.
  • Assuming all subgroups are normal. In non-abelian groups, most subgroups are not normal. For instance, (1 2)\langle (1\ 2) \rangle is not normal in S3S_3.
  • Miscalculating coset products. Always reduce representatives: (aN)(bN)=(ab)N(aN)(bN) = (ab)N, but (ab)N(ab)N may simplify further if abab can be rewritten.
ConceptConditionConsequence
NGN \trianglelefteq GgNg1=NgNg^{-1} = N for all ggG/NG/N is a group
[G:N]=2[G:N] = 2Index 2 subgroupNN is automatically normal
First Isomorphism Theoremϕ:GH\phi : G \to H homomorphismG/ker(ϕ)im(ϕ)G/\ker(\phi) \cong \operatorname{im}(\phi)
Center Z(G)Z(G)Commutes with everythingAlways a normal subgroup
Commutator subgroup [G,G][G,G]Generated by g1g2g11g21g_1 g_2 g_1^{-1} g_2^{-1}Always normal; G/[G,G]G/[G,G] is abelian
flowchart TD
A[4_Normal Subgroups And Quotient Groups] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Ensure you have mastered the prerequisite material before attempting this advanced content.