homomorphisms準同型じゅんどうけい and isomorphisms同型どうけい: basic exercises

date2026-06-06document_iddoc_490066d11a4ecceed23c93e1b9ef9035description群準同型、環準同型、核、像、同型、第一同型定理を確認する基本演習。prerequisites[群準同型/ぐんじゅんどうけい]と[同型/どうけい] / [準同型/じゅんどうけい]の[基本/きほん] / [第一同型定理/だいいちどうけいていり]の[見取図/みとりず]type[問題/もんだい][演習/えんしゅう]content_typeexercisestatusactive

mathabstract-algebrahomomorphismexercise

1Corresponding lectures

data/lecture/math/abstract-algebra/group-homomorphisms-and-isomorphisms.lecture.n.md data/lecture/math/abstract-algebra/homomorphism-basics.lecture.n.md data/lecture/math/abstract-algebra/homomorphism-theorems-overview.lecture.n.md

2Related exercises

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

3Order note

Problems 1 and 2 can be attempted after "Group homomorphisms and isomorphisms." Problem 4 follows "Basics of homomorphisms," and Problem 3 follows "Overview of the first isomorphism theorem." The proof exercise uses the subgroup criterion and normal subgroups introduced earlier.

4Problem 1: check a group homomorphism群準同型ぐんじゅんどうけい

Define $φ : Z \to Z / 5 Z$ by $φ (k) = [k]$ . Is this a homomorphism準同型じゅんどうけい of additive groups?

4.1Answer

Yes. For any integers $a, b$ ,

φ (a + b) = [a + b] = [a] + [b] = φ (a) + φ (b)

4.2Explanation

For a homomorphism準同型じゅんどうけい, we check whether operating first and then mapping gives the same result as mapping first and then operating.

5Problem 2: find the kernel核かく

Find the kernel核かく of the homomorphism準同型じゅんどうけい in Problem 1.

5.1Answer

The kernel核かく is the set of all integers sent to $[0]$ .

\ker φ = 5 Z

5.2Explanation

The kernel核かく is the part that is collapsed. In this example, multiples of 5 are sent to the zero residue class.

6Problem 3: apply the first isomorphism同型どうけい theorem

Equip the Cartesian product $Z / 6 Z \times Z / 4 Z$ with componentwise addition:

([a]_{6}, [b]_{4}) + ([c]_{6}, [d]_{4}) = ([a + c]_{6}, [b + d]_{4}) .

This makes the product a group. For the group homomorphism

ψ : Z \to Z / 6 Z \times Z / 4 Z, k \mapsto ([k]_{6}, [k]_{4}),

find its kernel and image, and state the isomorphism supplied by the first isomorphism theorem.

6.1Answer

The equality $ψ (k) = ([0]_{6}, [0]_{4})$ holds exactly when both 6 and 4 divide $k$ . Thus

\ker ψ = lcm (6, 4) Z = 12 Z .

The image is the cyclic subgroup generated by $([1]_{6}, [1]_{4})$ . For integers $k, ℓ$ ,

ψ (k) = ψ (ℓ) ⟺ 6 ∣ (k - ℓ) and 4 ∣ (k - ℓ) ⟺ 12 ∣ (k - ℓ) .

Thus $ψ (0), \dots, ψ (11)$ are distinct, and every $ψ (k)$ equals one of them. Hence the image has exactly 12 elements. The first isomorphism theorem therefore gives

Z / 12 Z ≅ ⟨ ([1]_{6}, [1]_{4}) ⟩ [PARSE ERROR: Undefined("Command(\"le\")")] Z / 6 Z \times Z / 4 Z .

6.2Explanation

The codomain has 24 elements, but the image contains only 12. This example makes explicit that quotienting by the kernel produces the image, not necessarily the entire codomain.

7Problem 4: kernel and image of a ring homomorphism

Show that

{ev}_{0} : Z [x] \to Z, f (x) \mapsto f (0)

is a ring homomorphism, and find its kernel and image.

7.1Answer

For all $f, g \in Z [x]$ ,

{ev}_{0} (f + g) = f (0) + g (0), {ev}_{0} (f g) = f (0) g (0), {ev}_{0} (1) = 1,

so ${ev}_{0}$ is a ring homomorphism. The condition $f (0) = 0$ says exactly that the constant term of $f$ is 0, in which case $f (x) = x q (x)$ . Therefore

\ker ({ev}_{0}) = x Z [x] .

Every $n \in Z$ is the image of the constant polynomial $n$ , so

Im ({ev}_{0}) = Z .

7.2Explanation

An evaluation map carries polynomial operations to operations on values. Its kernel $x Z [x]$ is the set of polynomials having 0 as a root, providing a concrete example of a ring-homomorphism kernel that is an ideal.

8Proof exercise: the relation between kernel核かく and injectivity

8.1Problem

Let $φ : G \to H$ be a group homomorphism群準同型ぐんじゅんどうけい. Prove that $\ker φ$ is a normal subgroup正規部分群せいきぶぶんぐん, and prove that $φ$ is injective単射たんしゃ if and only if $\ker φ = {e_{G}}$ .

8.2Answer

Since $φ (e_{G}) = e_{H}$ , we have $e_{G} \in \ker φ$ , so the kernel is nonempty. If $a, b \in \ker φ$ , then

φ (a b^{- 1}) = φ (a) φ (b)^{- 1} = e_{H}

so $a b^{- 1} \in \ker φ$ . By the subgroup criterion, the kernel is a subgroup. Also, for $g \in G$ and $a \in \ker φ$ ,

φ (g a g^{- 1}) = φ (g) φ (a) φ (g)^{- 1} = e_{H}

so the kernel核かく is normal.

If $φ$ is injective and $g \in \ker φ$ , then $φ (g) = e_{H} = φ (e_{G})$ implies $g = e_{G}$ . Hence $\ker φ = {e_{G}}$ . Conversely, suppose $\ker φ = {e_{G}}$ and $φ (g_{1}) = φ (g_{2})$ . Then $φ (g_{1} g_{2^{- 1}}) = e_{H}$ , so $g_{1} g_{2^{- 1}} = e_{G}$ . Hence $g_{1} = g_{2}$ .

8.3Explanation

The kernel核かく is the part collapsed by the homomorphism準同型じゅんどうけい. If the kernel核かく consists only of the identity element単位元たんいげん, then two different elements cannot collapse to the same image像ぞう.