第一同构定理（Noether）

English draft

Translation of the statement above · machine-drafted, pending author review · the Chinese text is authoritative.

Abstract

For a group homomorphism $\varphi : G \to G'$, the assignment $g\ker\varphi \mapsto \varphi(g)$ is an isomorphism $G/\ker\varphi \cong \operatorname{im}\varphi$; in particular a surjective homomorphism has the quotient as image. The proof is three steps: well-definedness, multiplicativity, bijectivity. The same proof works verbatim for any "set + operations + axioms" structure, $A/\ker f \cong \operatorname{im} f$, and the set-level prototype is that every map induces a bijection from the quotient by $a \sim a' \iff f(a) = f(a')$ onto its image.

Statement

Let $\varphi : G \to G'$ be a group homomorphism. Then $\varphi$ induces an isomorphism

In particular: the image of a surjective homomorphism is the quotient group ($G/\ker\varphi \cong G'$ when $\varphi$ is surjective).

Notes

The three steps of the proof. Write $K = \ker\varphi$.

• Well defined: if $gK = g'K$ then $g' = gk$ for some $k \in K$, hence $\varphi(g') = \varphi(g)\varphi(k) = \varphi(g)e' = \varphi(g)$ — elements of the same coset are sent to the same value.
• Homomorphism: $\overline{\varphi}\bigl((gK)(g'K)\bigr) = \overline{\varphi}(gg'K) = \varphi(gg') = \varphi(g)\varphi(g') = \overline{\varphi}(gK)\,\overline{\varphi}(g'K)$.
• Bijection: injective, because $\overline{\varphi}(gK) = e'$ means $\varphi(g) = e'$, i.e. $g \in K$, so $gK = K$; surjective, because the image is $\varphi(G)$ by construction. ∎

⭐ It is really a theorem about the format "set + operations + axioms", not a privilege of groups. The same proof carries over verbatim, provided one replaces "homomorphism" by the corresponding notion:

⭐ The set-level prototype: for any map $f : A \to B$, put $a \sim a' \iff f(a) = f(a')$ on $A$; then $f$ induces a bijection $A/{\sim} \;\to\; \operatorname{im} f$ (see "quotient set and equivalence classes"). The group version adds exactly one sentence: after dividing by the kernel, what remains is a group. The first two steps (well defined, bijection) are the set-level content; only the third (homomorphism) is new.

What question does it answer? "Where does a homomorphism send a group?" has a two-part answer: the kernel (what is collapsed) and the image (what survives), locked together by a single isomorphism. So studying homomorphisms is studying normal subgroups — which is why group theory turns into the study of the lattice of subgroups.