A group homomorphism is injective if and only if the identity of $G$ is the only element mapped to the identity in $H$ 
A group homomorphism is injective if and only if the identity of $G$ is the only element mapped to the identity in $H$

I have the first direction.
$"\Rightarrow"$: $f$ is an injective homomorphism. Suppose $f(a)=1_H, a \ne 1_G$. Then $f(a1_g)=f(a)f(1_g)=1_H$, since $f$ is given to be a homomorphism. This implies $f(1_G)=1_H$, so $a$ must equal $1_G$ otherwise $f$ is not injective, a contradiction.
But not sure how to show the other direction, is it also by contradiction? 
 A: Here is a direct proof:


*

*If $f$ is a homomorphism, then $f(1_G)=1_H$. If $f$ is also injective, no other $g \in G$ can be sent to $1_H$.

*If $f(g_1)=f(g_2)$, then $1_H = f(g_2)f(g_1)^{-1}= f(g_2 g_1^{-1}) $. If $f$ only sends $1_G$ to $1_H$, then $g_2 g_1^{-1}=1_G$ and so $g_2=g_1$, which means that $f$ is injective.
A: It seems to me that you've shown that if $f$ is an injection, then the identity in $G$ is the only element mapped to the identity in $H$. 
To show the converse, you need to show that if $G$ is the only element mapped to the identity, then the homomorphism $f$ is injective. The way I would do it is directly.
We want to show the definition of injective holds; that is, show $f(a)=f(b)$ implies $a=b$. We'll suppose there were two elements mapped to the same place; say $f(a)=f(b)=x\in H$. Then consider what that tells you about the element $ab^{-1}$, and where it is mapped. This should tell you $a=b$, and we are finished!
A: HINT: Suppose $f(a)=f(b)$. What must $f(ab^{-1})$ be? So if $a\not=b$, what does this mean about $f^{-1}(1_H)$?
A: For the first direction you can just use the injectivity of the homomorphism and the fact that by definition of a group homomorphism we must have that $f(1_G)=1_H$. 
For the other direction assume that $ker(f)$ is more than just the identity, and that the group homomorphism is injective, then reach a contradiction.
