cosets剰余類じょうよるい, 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: basic exercises

date2026-06-06document_iddoc_7eb1feff44ad9372b330673a4dd12546description群の剰余類、ラグランジュの定理、正規部分群、商群を確認する基本演習。prerequisites[剰余類/じょうよるい]とラグランジュの[定理/ていり] / [正規部分群/せいきぶぶんぐん]と[商群/しょうぐん]type[問題/もんだい][演習/えんしゅう]content_typeexercisestatusactive

mathabstract-algebragroup-theoryexercise

1Corresponding lectures

data/lecture/math/abstract-algebra/cosets-and-lagrange-theorem.lecture.n.md data/lecture/math/abstract-algebra/normal-subgroups-and-quotient-groups.lecture.n.md

2Suggested order

After “Cosets and Lagrange's theorem,” complete Problems 1, 2, and 4 and the proof exercise on partitioning by cosets. After “Normal subgroups and quotient groups,” continue with Problem 3. The page prerequisites describe what is needed to complete the whole page.

3Related exercises

data/exercise/math/abstract-algebra/groups-and-subgroups.exercise.n.md data/exercise/math/abstract-algebra/homomorphisms-and-isomorphisms.exercise.n.md

4Problem 1: find a coset剰余類じょうよるい

For the subgroup部分群ぶぶんぐん $H = {[0], [3]}$ of $(Z / 6 Z, +)$ , find $[1] + H$ .

4.1Answer

[1] + H = {[1] + [0], [1] + [3]} = {[1], [4]}

4.2Explanation

A coset剰余類じょうよるい is the set obtained by shifting the whole subgroup部分群ぶぶんぐん by one element.

5Problem 2: use Lagrange's theoremラグランジュの定理

Can a finite group of order位数いすう 12 have a subgroup部分群ぶぶんぐん of order位数いすう 5?

5.1Answer

No. By Lagrange's theoremラグランジュの定理, the order位数いすう of a subgroup部分群ぶぶんぐん must divide the order位数いすう of the group. But 5 does not divide 12.

5.2Explanation

Lagrange's theoremラグランジュの定理 narrows down possible subgroup部分群ぶぶんぐん orders. Divisibility is a necessary condition必要条件ひつようじょうけん, not a sufficient condition十分条件じゅうぶんじょうけん.

6Problem 3: condition for forming a quotient group商群しょうぐん

For a subgroup $N [PARSE ERROR: Undefined("Command(\"le\")")] G$ , state the necessary and sufficient condition for the coset product $(a N) (b N) = a b N$ to be independent of representatives, and prove both directions.

6.1Answer

The necessary and sufficient condition is that $N$ be a normal subgroup正規部分群せいきぶぶんぐん.

First suppose $N [PARSE ERROR: Undefined("Command(\"trianglelefteq\")")] G$ and let $a N = a^{'} N$ and $b N = b^{'} N$ . Write $a^{'} = a n_{1}$ and $b^{'} = b n_{2}$ with $n_{1}, n_{2} \in N$ . Then

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

Normality gives $(b^{- 1} n_{1} b) n_{2} \in N$ , so $a^{'} b^{'} N = a b N$ .

Conversely, suppose the product is independent of representatives. For arbitrary $g \in G$ and $n \in N$ , we have $(g n) N = g N$ . Multiplying these equal cosets by $g^{- 1} N$ must give the same result, so

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

Hence $g n g^{- 1} \in N$ , and therefore $N [PARSE ERROR: Undefined("Command(\"trianglelefteq\")")] G$ .

6.2Explanation

Normality makes the product well-defined; conversely, well-definedness forces $g n g^{- 1} \in N$ and hence normality.

7Proof exercise: partition into cosets剰余類じょうよるい

7.1Problem

Prove that the left cosets剰余類じょうよるい of $H [PARSE ERROR: Undefined("Command(\"le\")")] G$ partition $G$ .

7.2Answer

Since the identity $e$ lies in $H$ , every $g \in G$ satisfies $g = g e \in g H$ , so the union of the left cosets剰余類じょうよるい covers $G$ . If $a H \cap b H \neq [PARSE ERROR: Undefined("Command(\"varnothing\")")]$ , then for some $x$ we have $x = a h_{1} = b h_{2}$ . Then $b^{- 1} a = h_{2} h_{1^{- 1}} \in H$ . For every $a h \in a H$ ,

a h = b (b^{- 1} a) h \in b H,

so $a H \subseteq b H$ . The same argument gives $b H \subseteq a H$ , and hence $a H = b H$ . Thus any two cosets剰余類じょうよるい that intersect are equal.

7.3Explanation

Cosets always form a partition as sets, but normality is needed to put a group operation on the set of cosets剰余類じょうよるい.

8Problem 4: from Lagrange's theorem to element order

For an element $g$ of a finite group $G$ , prove that the order of $g$ divides $| G |$ . Then show that if $| G | = p$ is prime, the group $G$ is cyclic.

8.1Answer

The subgroup $⟨ g ⟩$ generated by $g$ is a subgroup of $G$ , and the order of $g$ is $| ⟨ g ⟩ |$ . Applying Lagrange's theorem to this subgroup gives

| ⟨ g ⟩ | ∣ | G | .

If $| G | = p$ is prime, choose a nonidentity element $g \in G$ . Its order is not 1 and divides $p$ , so its order is $p$ . Hence $| ⟨ g ⟩ | = | G |$ and therefore $⟨ g ⟩ = G$ . Thus $G$ is cyclic.