Proof of A Group Homomorphism is Injective Exactly When its Kernel is Trivial

Suppose first that $f$ is injective. By Group Homomorphisms Preserve the Identity Element and Inverses (i) we have $f(e_{G})=e_{H}$, so $e_{G}$ lies in the kernel of $f$ (Kernel and Image of a Group Homomorphism (i)). Moreover, if $g\in\ker f$, then $f(g)=e_{H}=f(e_{G})$, and the injectivity of $f$ (Injective, Surjective and Bijective Functions between Classes (i)) gives $g=e_{G}$. Hence $\ker f=\{e_{G}\}$.

Suppose now that $\ker f=\{e_{G}\}$, and let $g,h\in G$ with $f(g)=f(h)$. Since $f$ is a homomorphism, Group Homomorphisms Preserve the Identity Element and Inverses (ii) gives

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

so $g^{-1}h\in\ker f=\{e_{G}\}$. Therefore $h=g(g^{-1}h)=ge_{G}=g$, and $f$ is injective (Injective, Surjective and Bijective Functions between Classes (i)).