Proposition 1   Let $G,H$ be groups, and let $f \colon G \to H$ be a group homomorphism. Then $f$ is injective if and only if $\Ker(f)=\{e_G\}$ where $e_G$ is the identity element of $G$ and $\Ker$ denotes the kernel of $f$ (see also Kernel of a group homomorphism).

Proof. First assume that $f$ is injective (i.e. $f(g_1)=f(g_2) \Rightarrow g_1=g_2$ . Recall that: $$\Ker(f)=\{ g\in G : f(g)=e_H \}$$ where $e_H$ is the identity element of $H$ Since $f$ is a group homomorphism, it follows that $f(e_G)=e_H$ Let $g\in \Ker(f)$ then $f(g)=e_H=f(e_G)$ which implies that $g=e_G$ by the injectivity of $f$ Thus $\Ker(f)=\{ e_G \}$

For the converse, we assume that $\Ker(f)=\{ e_G \}$ and suppose that $f(g_1)=f(g_2)$ for some $g_1,g_2 \in G$ Since $f$ is a homomorphism: $$f(g_1)=f(g_2) \Rightarrow f(g_1)\cdot f(g_2)^{-1}=e_H \Rightarrow f(g_1\cdot g_2^{-1})=e_H$$ Thus $g_1\cdot g_2^{-1} \in \Ker(f)$ and the kernel is trivial so $g_1\cdot g_2^{-1}=e_G$ therefore $g_1=g_2$