First isomorphism theorem

Statement

Let $G$ and $H$ be groups and let $\varphi \colon G \to H$ be a group homomorphism. Then the map $g \ker \varphi \mapsto \varphi(g)$ is a well-defined isomorphism from the quotient group $G / \ker \varphi$ to the image $\operatorname{im} \varphi$.

Remarks

Here $\ker \varphi$ is normal by this fact, so the quotient exists. If $\varphi$ is surjective, then $H \cong G / \ker \varphi$. For example, the determinant gives $\mathrm{GL}_n(\mathbb{R}) / \mathrm{SL}_n(\mathbb{R}) \cong \mathbb{R}^\times$. For finite $G$, a consequence is $|G| = |\ker \varphi| \cdot |\operatorname{im} \varphi|$.

Explanations

L1 human

A homomorphism collapses each coset of its kernel to one point of the image. After this collapse, nothing else is lost: the quotient by the kernel and the image are the same group. For example, reduction modulo $n$ shows that $\mathbb{Z}/n\mathbb{Z}$ is the image of $\mathbb{Z}$.