TITLE: Existence of Homomorphism so that Diagram Commutes
QUESTION [6 upvotes]: Let $G, H$ be groups and let $\phi: G \to H, \psi: G \to K$ be homomorphism such that $\ker \phi \subseteq \ker \psi.$ Prove that there exists a homomorphism $\theta: H\to K$ such that $\theta\circ\phi = \psi.$ 
Obviously one can find a homomorphism $\theta$ on $\phi(G)$ which satisfies the properties, but I don't see why we can extend this to $H,$ if we even can.
Edit: If someone can come up with a counterexample, that would be appreciated as well. I honestly don't think the statement is true as is...

REPLY [9 votes]: Here should be a counterexample: Let $G, H, K = \mathbb{Z}$. Define $\phi, \psi$ to be multiplication by 4, 2, respectively. Then there can be no map $\theta : H \to K$ satisfying that property, since we would need to have $\theta(1) = \frac{1}{2}$.