First isomorphism consequence for groups

For a homomorphism f, the quotient G/ker(f) is isomorphic to im(f)

First isomorphism consequence for groups

Proposition (Quotient by the kernel). Let $f:G\to H$ be a group homomorphism . Then there exists a unique group isomorphism

\overline f:\; G/\ker(f)\ \longrightarrow\ \mathrm{im}(f)

such that $\overline f(g\,\ker(f))=f(g)$ for all $g\in G$ . In particular,

G/\ker(f)\ \cong\ \mathrm{im}(f)

as groups, where $G/\ker(f)$ is the quotient group and $\mathrm{im}(f)$ is a subgroup of $H$ .

Context. This is the standard “hands-on” form of the first isomorphism theorem . It identifies precisely what information about $G$ is “lost” under $f$ : exactly the kernel .

Proof sketch. Define $\overline f(g\ker(f)):=f(g)$ .

Well-defined: if $g\ker(f)=g'\ker(f)$ then $g^{-1}g'\in\ker(f)$ so $f(g)=f(g')$ .
Homomorphism: follows from $f(gg')=f(g)f(g')$ .
Bijective onto $\mathrm{im}(f)$ : surjectivity is by definition of image; injectivity is $\ker(\overline f)=\{\,\ker(f)\,\}$ . Thus $\overline f$ is an isomorphism.