group homomorphisms群準同型ぐんじゅんどうけい and isomorphisms同型どうけい

date2026-07-14document_iddoc_56e7c273d30060a1591234a6ff558442description群準同型を、演算を保つ写像として定義し、核・像・同型が何を保存するかを説明する。prerequisites[群/ぐん]の[基本/きほん] / [正規部分群/せいきぶぶんぐん]と[商群/しょうぐん] / [写像/しゃぞう]の[基本/きほん]type講義content_typelecturestatusactive

mathabstract-algebragroup-theorylecture

When comparing groups群ぐん, the names of the elements do not need to match. What matters is that the operation structure is preserved. A map that preserves operations in this way is a group homomorphism群準同型ぐんじゅんどうけい.

1Group homomorphisms準同型じゅんどうけい

For groups $G, H$ , a map $φ : G \to H$ is a group homomorphism群準同型ぐんじゅんどうけい if for all $a, b \in G$ ,

φ (a b) = φ (a) φ (b)

holds.

Here the product on the left is the operation in $G$ , and the product on the right is the operation in $H$ . Even if the same symbol is used, it is necessary to distinguish which group's operation is being used.

2What a homomorphism準同型じゅんどうけい automatically preserves

A group homomorphism群準同型ぐんじゅんどうけい $φ : G \to H$ sends the identity element単位元たんいげん to the identity element単位元たんいげん:

φ (e_{G}) = e_{H}

It also sends inverses to inverses:

φ (a^{- 1}) = φ (a)^{- 1}

These facts are not written directly in the definition定義ていぎ, but they follow from preservation of the operation.

3Kernel and image像ぞう

The kernel核かく of a group homomorphism群準同型ぐんじゅんどうけい $φ : G \to H$ is defined by

\ker φ = {g \in G ∣ φ (g) = e_{H}}

The kernel核かく is the part collapsed to the identity element単位元たんいげん by the map.

The image像ぞう is

Im φ = {φ (g) ∣ g \in G}

It is the set of all elements actually reached by the map.

The image is a subgroup of $H$ . It is nonempty because $e_{H} = φ (e_{G})$ lies in the image. If $x = φ (a)$ and $y = φ (b)$ lie in the image, then

x y^{- 1} = φ (a) φ (b)^{- 1} = φ (a b^{- 1})

also lies in the image, so the subgroup criterion applies.

4Isomorphisms

A group homomorphism群準同型ぐんじゅんどうけい $φ : G \to H$ is a group isomorphism同型どうけい when it is bijective. In that case, $G$ and $H$ have the same structure as groups.

G ≅ H

In an isomorphism同型どうけい, the names of elements may change. The operation tables are not literally preserved in their original order; after relabeling every row, column, and entry via $φ$ , the two operation tables agree. This is exactly the law $φ (a b) = φ (a) φ (b)$ .

Consequently, the identity, inverses, inclusion relations among subgroups, the order of the group, the order of each element, and other group-theoretic properties are preserved.

Let us verify these claims. If $K [PARSE ERROR: Undefined("Command(\"le\")")] G$ , then $φ (K)$ is nonempty, and for $a, b \in K$ ,

φ (a) φ (b)^{- 1} = φ (a b^{- 1}) \in φ (K) .

The subgroup criterion therefore gives $φ (K) [PARSE ERROR: Undefined("Command(\"le\")")] H$ . Conversely, if $L [PARSE ERROR: Undefined("Command(\"le\")")] H$ , the preimage

φ^{- 1} (L) = {g \in G ∣ φ (g) \in L}

is a subgroup of $G$ by the same criterion. Since $φ$ is bijective, the assignments $K \mapsto φ (K)$ and $L \mapsto φ^{- 1} (L)$ are inverse to each other, and

K_{1} \subseteq K_{2} ⟺ φ (K_{1}) \subseteq φ (K_{2}) .

Thus inclusion relations among subgroups are preserved.

A bijection also pairs the elements of $G$ and $H$ one-to-one, so $| G | = | H |$ . For $m [PARSE ERROR: Undefined("Command(\"ge\")")] 0$ , induction starting from $φ (e_{G}) = e_{H}$ and using

φ (g^{m + 1}) = φ (g^{m} g) = φ (g^{m}) φ (g)

gives $φ (g^{m}) = φ (g)^{m}$ . If $m = - n < 0$ , preservation of inverses gives

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

Thus $φ (g^{m}) = φ (g)^{m}$ for every $m \in Z$ , and injectivity gives

g^{m} = e_{G} ⟺ φ (g)^{m} = e_{H} .

Thus, when $g$ has finite order, $g$ and $φ (g)$ have the same least positive exponent; if no such exponent exists, both elements have infinite order.

5Concrete example

Define $φ : Z \to Z / n Z$ by

φ (k) = [k]

This is a group homomorphism群準同型ぐんじゅんどうけい because

φ (a + b) = [a + b] = [a] + [b]

The kernel核かく of this map is

\ker φ = n Z

6Proof supplement: what homomorphisms準同型じゅんどうけい preserve

Let $φ : G \to H$ be a group homomorphism群準同型ぐんじゅんどうけい. Then

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

First, $φ (e_{G}) = φ (e_{G} e_{G}) = φ (e_{G}) φ (e_{G})$ . Multiplying by $φ (e_{G})^{- 1}$ from the left gives $e_{H} = φ (e_{G})$ . Next,

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

so $φ (g^{- 1})$ is the inverse of $φ (g)$ . Hence $φ (g^{- 1}) = φ (g)^{- 1}$ .

Moreover, the kernel核かく $\ker φ$ is a normal subgroup正規部分群せいきぶぶんぐん of $G$ . If $a, b \in \ker φ$ , then

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

so $a b^{- 1} \in \ker φ$ , and the subgroup部分群ぶぶんぐん criterion gives that it is a subgroup部分群ぶぶんぐん. Also, for $g \in G$ and $a \in \ker φ$ ,

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

so $g a g^{- 1} \in \ker φ$ . Thus the kernel核かく is normal.

Finally, $φ$ is injective単射たんしゃ if and only if $\ker φ = {e_{G}}$ . If $φ$ is injective単射たんしゃ, then $φ (g) = e_{H} = φ (e_{G})$ implies $g = e_{G}$ . Conversely, suppose $\ker φ = {e_{G}}$ and $φ (g_{1}) = φ (g_{2})$ . Then

φ (g_{1} g_{2^{- 1}}) = e_{H}

so $g_{1} g_{2^{- 1}} \in \ker φ$ , hence $g_{1} g_{2^{- 1}} = e_{G}$ . Therefore $g_{1} = g_{2}$ , and $φ$ is injective単射たんしゃ.

7Exercise link

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

8Summary

A group homomorphism群準同型ぐんじゅんどうけい is a map that preserves operations. The kernel核かく is the part that is collapsed, and the image像ぞう is the part that is reached. An isomorphism同型どうけい is a homomorphism準同型じゅんどうけい that preserves structure completely, expressing the idea of being essentially the same in abstract algebra抽象代数ちゅうしょうだいすう.