normal subgroups正規部分群せいきぶぶんぐん and quotient groups商群しょうぐん

date2026-07-14document_iddoc_439dabe4b5d3745363a37869e69b53a5description正規部分群を、剰余類の集合に群構造を入れるための条件として説明し、商群の意味を整理する。prerequisites[剰余類/じょうよるい]とラグランジュの[定理/ていり]type講義content_typelecturestatusactive

mathabstract-algebragroup-theorylecture

To merely form the set of cosets剰余類じょうよるい, any subgroup部分群ぶぶんぐん is enough. But if we want to multiply cosets剰余類じょうよるい and obtain cosets剰余類じょうよるい again, the result must not depend on the choice of representatives. A subgroup部分群ぶぶんぐん satisfying this condition is a normal subgroup正規部分群せいきぶぶんぐん.

1Definition of a normal subgroup正規部分群せいきぶぶんぐん

A subgroup部分群ぶぶんぐん $N [PARSE ERROR: Undefined("Command(\"le\")")] G$ is a normal subgroup正規部分群せいきぶぶんぐん if for every $g \in G$ ,

g N = N g

holds. In this case we write

N [PARSE ERROR: Undefined("Command(\"trianglelefteq\")")] G

The element $g n g^{- 1}$ is called the conjugate共役きょうやく of $n$ by $g$ . An equivalent condition is that for every $g \in G$ and $n \in N$ ,

g n g^{- 1} \in N

holds.

To verify equivalence, first suppose $g N = N g$ . Since $g n \in g N = N g$ , there is $n^{'} \in N$ with $g n = n^{'} g$ , so $g n g^{- 1} = n^{'} \in N$ . Conversely, assume the conjugation condition. Then $g n = (g n g^{- 1}) g \in N g$ , giving $g N \subseteq N g$ . Applying the condition with $g^{- 1}$ gives $g^{- 1} n g \in N$ and hence $n g = g (g^{- 1} n g) \in g N$ , so $N g \subseteq g N$ . Thus $g N = N g$ .

2Why normality is necessary

We want to define the product of cosets剰余類じょうよるい by

(g N) (h N) = (g h) N

For this definition定義ていぎ to be independent of representatives, replacing $g$ or $h$ by another element in the same coset剰余類じょうよるい must produce the same resulting coset剰余類じょうよるい.

Normality is the condition that guarantees this well-definedness.

3Quotient groups

When $N [PARSE ERROR: Undefined("Command(\"trianglelefteq\")")] G$ , the set of all cosets剰余類じょうよるい

G / N = {g N ∣ g \in G}

becomes a group群ぐん under the product

(g N) (h N) = (g h) N

Well-definedness is proved below. The group axioms follow from

((a N) (b N)) (c N) = ((a b) c) N = (a (b c)) N = (a N) ((b N) (c N)),

N (g N) = (g N) N = g N, (g N) (g^{- 1} N) = (g^{- 1} N) (g N) = N .

Thus the identity is $N = e N$ and the inverse of $g N$ is $g^{- 1} N$ . This group is called a quotient group.

In a quotient group商群しょうぐん, differences lying inside $N$ are collapsed as if they were 0. In other words, we ignore movement inside $N$ and study the remaining structure.

4Concrete example: quotient groups商群しょうぐん of integers

Let $n [PARSE ERROR: Undefined("Command(\"ge\")")] 1$ . In $(Z, +)$ , the subgroup部分群ぶぶんぐん $n Z$ is normal because $Z$ is an abelian group, and every subgroup of an abelian group is normal.

The quotient group商群しょうぐん

Z / n Z

is the group obtained by classifying integers by their remainders modulo $n$ .

5What changes and what is preserved

In a quotient group商群しょうぐん, differences inside $N$ are no longer distinguished. What changes is the granularity of elements: the elements are cosets剰余類じょうよるい rather than individual elements. What is preserved is the group operation structure in a well-defined form.

6Proof supplement: the condition for quotient-group operations to be well-defined

Let $N [PARSE ERROR: Undefined("Command(\"le\")")] G$ . We want to define multiplication of cosets剰余類じょうよるい by

(a N) (b N) = (a b) N

For this definition定義ていぎ to be independent of the chosen representatives, $N$ must be a normal subgroup正規部分群せいきぶぶんぐん.

First suppose $N$ is normal. Let $a N = a^{'} N$ and $b N = b^{'} N$ . Then there exist $n_{1}, n_{2} \in N$ such that $a^{'} = a n_{1}$ and $b^{'} = b n_{2}$ . Hence

a^{'} b^{'} = a n_{1} b n_{2} = a b (b^{- 1} n_{1} b) n_{2}

Since $N$ is normal, $b^{- 1} n_{1} b \in N$ , and therefore $(b^{- 1} n_{1} b) n_{2} \in N$ . Thus $a^{'} b^{'} N = a b N$ . The product is independent of representatives.

Conversely, suppose this product is always well-defined. Take any $g \in G$ and $n \in N$ . As left cosets剰余類じょうよるい, $(g n) N = g N$ , so changing the representative of the first factor from $g$ to $g n$ must give the same product with $g^{- 1} N$ . Therefore

(g N) (g^{- 1} N) = N, ((g n) N) (g^{- 1} N) = g n g^{- 1} N

must be the same coset剰余類じょうよるい. Hence $g n g^{- 1} N = N$ , so $g n g^{- 1} \in N$ . Since this holds for every $g, n$ , the subgroup $N$ is normal.

Thus normality is precisely the condition that makes it consistent to multiply cosets剰余類じょうよるい as elements.

7Exercise link

data/exercise/math/abstract-algebra/cosets-normal-subgroups-and-quotient-groups.exercise.n.md

8Counterexample: not every subgroup部分群ぶぶんぐん is normal

In the symmetric group $S_{3}$ , consider $H = {e, (12)}$ . This is a subgroup部分群ぶぶんぐん. However, if $g = (13)$ , then

g H = {(13), (13) (12)}

while

H g = {(13), (12) (13)}

Here $(13) (12)$ and $(12) (13)$ are different permutations. Therefore $g H \neq H g$ , so $H$ is not a normal subgroup正規部分群せいきぶぶんぐん.

In this example, the set of cosets can be formed, but their multiplication depends on representatives. Indeed, $H = e H = (12) H$ . Computing $H \cdot g H$ with representative $e$ gives $g H$ , while representative $(12)$ gives $(12) g H$ . If these cosets were equal, then $g^{- 1} (12) g = (23)$ would lie in $H = {e, (12)}$ , a contradiction. Thus in noncommutative groups, checking normality is essential.

9Summary

Normality is the necessary and sufficient condition for defining multiplication of cosets independently of representatives and hence for putting a group structure on their set. A quotient group is the structure that remains after collapsing differences inside a normal subgroup.