Fundamental Isomorphism Theorem

theoremgroup-theory

Intuition

Formal Statement

For morphism $f:(G,\cdot )\to (H,\circ )$, we have,

$$\mathrm{im}f\simeq {}^G{/}_{\ker f},$$

and there exists an isomorphism $F:{}^G{/}_{\ker f}\to \mathrm{im}f$, defined by $F(g\cdot \ker f)=f(g)$.

Corollaries

• Any cyclic group is isomorphic to $\mathbb{Z}$ or ${\mathbb{Z}}_n={}^{\mathbb{Z}}{/}_{n\mathbb{Z}}$ for some $n\ge 1$.

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