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

We write both operations multiplicatively and use $\varphi(e_G)=e_H$ and $\varphi(a^{-1})=\varphi(a)^{-1}$ from Group Homomorphisms Preserve the Identity Element and Inverses. Recall from Group Homomorphism and Isomorphism that $\varphi$ is injective means: $\varphi(a)=\varphi(b)$ implies $a=b$, for all $a,b\in G$.

Necessity. Suppose $\varphi$ is injective. Since $\varphi(e_G)=e_H$, we have $e_G\in\ker\varphi$, so $\{e_G\}\subseteq\ker\varphi$. Conversely, if $a\in\ker\varphi$ then $\varphi(a)=e_H=\varphi(e_G)$, and injectivity gives $a=e_G$. Hence $\ker\varphi=\{e_G\}$.

Sufficiency. Suppose $\ker\varphi=\{e_G\}$ and let $a,b\in G$ satisfy $\varphi(a)=\varphi(b)$. Using the homomorphism property from Group Homomorphism and Isomorphism and then Group Homomorphisms Preserve the Identity Element and Inverses,

the last equality by Uniqueness of the Identity Element and of Inverses in a Group applied in $H$. Hence $ab^{-1}\in\ker\varphi=\{e_G\}$, that is $ab^{-1}=e_G$.

Multiplying this equation on the right by $b$ gives $(ab^{-1})b=e_Gb=b$. On the other hand, associativity (condition 1 of Group and Abelian Group) applied to the triple $a,b^{-1},b$ gives

Therefore $a=b$, and $\varphi$ is injective.