---
title: "Normal Subgroups and Quotient Groups"
description: "A subgroup is normal exactly when multiplying its cosets is independent of the chosen representatives. The first isomorphism theorem identifies a homomorphism's image with its domain modulo its kernel"
canonical: https://lightmysky.com/learn/mathematics/normal-subgroups-and-quotient-groups-mt_RU7A-t7wQ5
source: https://lightmysky.com/learn/mathematics/normal-subgroups-and-quotient-groups-mt_RU7A-t7wQ5.md
retrieved: 2026-11-12
---

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# Normal Subgroups and Quotient Groups

A subgroup is normal exactly when multiplying its cosets is independent of the chosen representatives. The first isomorphism theorem identifies a homomorphism's image with its domain modulo its kernel.

Subject: Mathematics · Area: Abstract Algebra · Ages 22 to 23
Page: https://lightmysky.com/learn/mathematics/normal-subgroups-and-quotient-groups-mt_RU7A-t7wQ5

## Ready when they can

- Decide whether a subgroup is normal, by conjugation or by its index
- Write the multiplication table of a small quotient group
- Use the first isomorphism theorem to identify the image of a homomorphism

## Lesson: When cosets form a group

Recall that cosets partition a group. Write e for its identity and g^-1 for an inverse. A subgroup N is normal in G if gng^-1 belongs to N for every g in G and n in N. In an abelian group this conjugate equals n. An index-two subgroup is normal too: its left and right cosets are itself and its complement. Index counts left cosets; another index alone need not decide normality.

**Example.** S4 is the group of permutations of four letters. The notation (34) means swap 3 and 4; products act rightmost first. Let H = {e, (34)}. Conjugating by (14) gives (14)(34)(14) = (13). Conjugation relabels the moved points: 3 stays 3 and 4 becomes 1. Since (13) is outside H, this subgroup is not normal. One escaped conjugate suffices to disprove normality.

*(drawing: One conjugate outside H disproves normality.)*

For normal N, define (aN)(bN) = abN. Replacing a and b by an and bm gives ab(b^-1nb)m. Normality puts b^-1nb in N, so the product coset remains abN. Conversely, independence of representatives gives (nN)(gN) = (eN)(gN), hence ngN = gN and g^-1ng belongs to N. Thus this rule works exactly when N is normal. The quotient identity is N; the inverse of aN is a^-1N.

**Example.** Z_n denotes residues modulo n under addition. In Z8, H = {0, 4} is normal because the group is abelian. Name its cosets H, A = 1 + H, B = 2 + H and C = 3 + H. Add representatives to calculate: A + B = C. The inverse of a + H is -a + H. Two cosets a + H and b + H agree exactly when a - b belongs to H.

*(drawing: Changing representatives from 1 and 2 to 5 and 6 leaves the sum coset unchanged.)*

**Example.** Build a quotient table by combining each row representative with each column representative. Find the coset containing the result. For the Z8/H cosets just named, the table is:
+  H  A  B  C
H  H  A  B  C
A  A  B  C  H
B  B  C  H  A
C  C  H  A  B
For example, the row C and column C give B because 3 + 3 = 6 belongs to B.

For finite G and normal N, each coset has |N| elements, so |G/N| = |G| / |N|. Vertical bars count elements. Cyclic groups are abelian, so their subgroups are normal. In Z_n, list the subgroup residues, count them and divide n by that count. The centre Z(G) consists of elements commuting with every group element. It is normal because conjugation fixes each central element.

For a homomorphism phi: G -> K, its kernel N is normal: phi(gng^-1) = phi(g)e phi(g)^-1 = e for n in N. The first isomorphism theorem gives an isomorphism gN -> phi(g) from G/N to the image. For finite G, |image| = |G| / |N|. The image need not be all of K. If G is cyclic, its image is generated by the image of a generator.

The sign homomorphism sends even permutations to 1 and odd permutations to -1, with multiplication in {1, -1}. Even means expressible as an even number of swaps; odd means an odd number. This parity does not depend on the chosen expression. The identity is even and a single swap is odd, so both signs occur. Its kernel is the alternating subgroup of even permutations.

For finite G and normal N, the order of a quotient element gN is the least positive k with g^k in N. In additive notation, find the least positive k with ka in N. In Z20 with N = {0, 10}, the coset 5 + N has order 2: 5 is outside N, but 2 times 5 belongs to N. This counts repetitions of one quotient element, not the number of quotient elements.

**Recap.** Normality makes coset multiplication consistent, quotients divide orders, and every image is a quotient by its kernel.

*The lesson, questions, worked solutions and base flashcards were written for this dataset.*

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Homomorphisms, Cosets and Lagrange's Theorem](https://lightmysky.com/learn/mathematics/homomorphisms-cosets-and-lagranges-theorem-mt_yKiZ8sg7yN)

## Optional connections

- [Equivalence Relations and Partitions](https://lightmysky.com/learn/mathematics/equivalence-relations-and-partitions-mt_wgEupYwEUF)

## Opens up

- [The Galois Correspondence](https://lightmysky.com/learn/mathematics/the-galois-correspondence-mt_30kyliQR_L)
- [Decomposing a Representation from Its Character](https://lightmysky.com/learn/mathematics/decomposing-a-representation-from-its-character-mt_ccEcWW9xAa)
- [Group Actions, Orbits and the Class Equation](https://lightmysky.com/learn/mathematics/group-actions-orbits-and-the-class-equation-mt_urut0TFTOR)
- [The Sylow Theorems and Groups of Small Order](https://lightmysky.com/learn/mathematics/the-sylow-theorems-and-groups-of-small-order-mt_vg2VYkLuz8)
