I was doing some group theory and this exercise came up.
Let $f: G \to H$ be a group morphism and $N$ a normal subgroup of $G$. Denote $\pi: G \to G/N$ and let $N'$ be the normal closure of $f(N)$. Show that the amalgamated product $H*_G(G/N) \cong H/N'$.
I want to use the universal property for the amalgamated product. Since we can make a homomorphism $\varphi: H \to H/N'$ and a homomorphism $\psi : G/N \to H/N'$ we know there must be a unique homomorphism from $\alpha: H *_G (G/N) \to H/N'$. Now since it holds for $q_H : H \to H *_G (G/N) $ that $\alpha \circ q_H = \varphi$, we obtain that $\alpha$ must also be surjective. I can't seem to figure out however how I should prove that $\alpha$ is injective.
Thank you for your help!