Proof of Kernel and Image of a Group Homomorphism are Subgroups

Routine facts about nonempty classes are used without comment (Toolkit for AL1).

For (i), the kernel contains $e_{G}$ by Kernel and Image of a Group Homomorphism (i), because $f(e_{G})=e_{H}$ (Group Homomorphisms Preserve the Identity Element and Inverses (i)), so it is nonempty. For $g,h\in\ker f$, since $f$ is a homomorphism, Group Homomorphisms Preserve the Identity Element and Inverses (ii) and $e_{H}^{-1}=e_{H}$ (Cancellation Laws and Basic Inverse Identities in a Group (ii)) give

$$f(gh^{-1})=f(g)f(h^{-1})=f(g)f(h)^{-1}=e_{H}e_{H}^{-1}=e_{H}.$$

Hence $gh^{-1}\in\ker f$ by Kernel and Image of a Group Homomorphism (i), and $\ker f\le G$ by Subgroup Criterion and Basic Examples (i).

For (ii), the image contains $f(e_{G})=e_{H}$ by Kernel and Image of a Group Homomorphism (ii) and Group Homomorphisms Preserve the Identity Element and Inverses (i), so it is nonempty. Its elements are the values $f(g)$ with $g\in G$ (Kernel and Image of a Group Homomorphism (ii)). For two of them, $f(g)$ and $f(h)$, using Group Homomorphisms Preserve the Identity Element and Inverses (ii) and that $f$ is a homomorphism,

$$f(g)f(h)^{-1}=f(g)f(h^{-1})=f(gh^{-1})\in f(G),$$

so $f(G)\le H$ by Subgroup Criterion and Basic Examples (i).

For (iii), let $g\in G$ and $n\in\ker f$. Using that $f$ is a homomorphism, that $f(n)=e_{H}$, and Group Homomorphisms Preserve the Identity Element and Inverses (ii), we get

$$f\bigl((gn)g^{-1}\bigr)=f(g)f(n)f(g^{-1})=f(g)\,e_{H}\,f(g)^{-1}=e_{H},$$

so $(gn)g^{-1}\in\ker f$. Together with (i), this shows that $\ker f$ is normal (Normal Subgroups (i)).