Let $G$ and $H$ be groups and let $f :G \to H$ be a group homomorphism. Show that $\ker(f)$ is a normal subgroup of $G$.
In order for $\ker(f)$ to be normal subgroup I have to have that $\forall a \in G : a\ker(f) = \ker(f)a$?
So isn't $a\ker(f) = \{ag \in G \mid f(ag) = e_H \}$? If so then since $f$ is a homomorphism I have that $f(ag) = f(a) \cdot f(e_H) = f(a)$ and similarly $\ker(f)a = \{ga \in G \mid f(ga) = e_H \}$, but $f(ga) = f(g) \cdot f(a) = f(a)$ so $a\ker(f) = \ker(f)a$?
I'm not sure I'm going in the right direction here, what am I missing?