How to prove that $f$ is injective if $\exists h \in H: |f^{-1}(\{h\})| = 1$? Let $G,H$ be groups and $f: G \to H$ a group homomorphism. I want to show that the following statements are equivalent:


*

*$f$ is injective.

*$\ker f = \{e_G\}$.

*$\exists h \in H: |f^{-1}(\{h\})|=1$.


The standard solution suggests to show the equivalence of 1 and 2, and then the equivalence of 2 and 3 by proving both directions in each case. Therefore, it proves four implications. When I started to prove this, I tried to show only three implications, namely $1 \implies 2 \implies 3 \implies 1$. The first two implications were pretty easy, however I am stuck with the last implication.
Can anyone give me a hint?

Hint: Show that $|f^{-1}({h})|$ is the same size for any $h$ in the image of $f$.
