Quotient the domain by the kernel and what remains is exactly the image.Suggest a slogan
Let \(\phi\colon G\to H\) be a group homomorphism. Then \(\ker \phi\) is a normal subgroup of \(G\), \(\im \phi\) is a subgroup of \(H\), and \[ G/\ker \phi \to \im \phi,\qquad g\ker\phi\mapsto\phi(g), \] is a well-defined group isomorphism.
For \(x\in\ker\phi\) and \(g\in G\), \(\phi(gxg^{-1})=\phi(g)\phi(g)^{-1}=e\), so \(\ker\phi\normal G\); \(\im\phi\) is the image of a group under a homomorphism, hence a subgroup of \(H\). The map is well defined and injective because \(\phi(g)=\phi(g')\) if and only if \(g^{-1}g'\in\ker\phi\), it is surjective by construction, and it is a homomorphism because \(\phi\) is.