Injective implies trivial kernel. Let $f : G \longrightarrow G^\prime$  be a group homomorphism. Then $f$ is injective iff $\ker(f) = \{1\}$ where $1$ represents the identity of $G.$
First I tried to prove  that if $f$ is injective then $\ker(f) = \{1\}.$
Since $f$ is homomorphism, $f(1)=1^\prime$ (identity elt of $G^\prime$)
then $1\in  \ker(f),$ so $\{1\} \subset \ker(f),$ and f is injective $\ker(f) = \{1\}$ 
For this part, I got help from professor but I can't exactly understand the bold and italic part.
This is my proof for other side ( if $\ker(f) = \{1\}$, then $f$ is injective. )
let $f(a)=f(b)$, then $f(a) * f(b)^{-1} = 1^\prime$
and $f(ab^{-1}) = 1^\prime$ so $ab^{-1}$ is in $\ker(f) = \{1\}$
then $a=b.$

I expect, there is no confusion about 1 and $e$. Let $f$ is injective and $x$ $\in$ Ker $f$, then $f$(x)=$e^{'}$ and as 
$f$($e$)=$e^{'}$, you have $f$(x) = $f$($e$); so injective will imply $x$ = $e$; Thus Ker $f$ = {$e$}
