Kernel of injective group homomorphism If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?
I think the $Ker\phi = \{x\in G:\phi(x)=1\}$
because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$
but then i realize it is injective how does that affect the kernel?

A homomorphism $ \phi $ is injective iff its kernel is {e}.
To prove this , suppose $\phi$ is injective . Then $\phi (x) = e' =\phi (e) $ means that $x=e$. Thus $ker \phi = {e} $.
Now suppose  $ker \phi ={e}$. Then if $ \phi (a) =\phi (b)$ it means that $\phi (a b^-1) = e' $ ie $a b^-1 = e$ and thus a=b. Thus $\phi $is injective. 
