Here's the question I was trying to prove:
If $f:G\to H$ is a homomorphism with kernel $K$, then $f$ is injective iff $K=\{e_G \}.$
Proving that $K=\{ e_{G} \} \implies f$ is injective is easy. Anyways, here's my attempt for proving the converse:
Assume the contrary that $f$ is injective and $K \ne \{ e_{G} \}$. Notice that $K$ is not empty since $f(e_G)=e_H$ thus $e_G \in K$. Hence there is some element, say $x \in G\setminus \{ e_G \}$ such that $f(x)=e_K $. But $f(e_G)=e_K$ thus $f(x)=f(e_G)$. It follows by the injectivity of $f$ that $x=e_G$. This is a contradiction!
Is my proof correct? Is there any simpler way to carry out this proof?
It is correct, but I would just say that if $f$ is injective, then $\ker f=\{e_G\}$ because $e_G\in\ker f$ and if $x\in\ker F$, then $f(x)=e_H=f(e_G)$ and therefore $x=e_G$.