First Isomorphism Theorem (Groups)

A homomorphism factors through the quotient by its kernel, giving G/ker(f) ≅ im(f)

First Isomorphism Theorem (Groups)

First Isomorphism Theorem (Groups). Let $G$ and $H$ be groups , and let $f: G \to H$ be a group homomorphism . Let $K = \ker(f)$ be the kernel of $f$ and let $I = \operatorname{im}(f)$ be the image of $f$ , i.e.

K = \{g \in G : f(g) = e_H\}, \qquad I = \{f(g) : g \in G\}.

Then $K$ is a normal subgroup of $G$ (see kernels are normal subgroups ), and the induced map

\bar f: G/K \to I, \qquad \bar f(gK) = f(g),

is a well-defined isomorphism . In particular, if $f$ is surjective then $G/K \cong H$ .

This result is the basic “quotient = image” principle and is the prototype for the second and third isomorphism theorems . It is often packaged as an exact sequence $1 \to K \to G \to I \to 1$ .

Proof sketch. Because $K \subseteq \ker(f)$ , the map $g \mapsto f(g)$ is constant on cosets of $K$ , so $\bar f$ is well-defined. The universal property of quotient groups gives the induced homomorphism $G/K \to H$ with image $I$ . Injectivity of $\bar f$ follows from $\bar f(gK)=e_H \iff g \in K$ , and surjectivity onto $I$ is immediate from the definition of $I$ .