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...

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    $\begingroup$ Shouldn't be $\phi$ surjective? In that case you can (well)define $\theta$ by pulling back $\phi$ and map through $\psi$ to $K$, $\endgroup$ – Nicky Hekster Dec 13 '13 at 20:37
  • $\begingroup$ that's what I thought, but the problem statement contains no such condition $\endgroup$ – cats Dec 13 '13 at 20:37
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    $\begingroup$ @NickyHekster could you please elaborate on the pull-back construction? Is it something like $\pi:G/\ker\psi\to G: g\ker\psi\mapsto g, g\in G$? $\endgroup$ – msd15213 Apr 21 '19 at 15:58
  • $\begingroup$ Assume that $\phi: G \rightarrow H$ is surjective. let $h \in H$. Then we can find a $g \in G$ such that $\phi(g)=h$. Define $\theta: H \rightarrow K$, by $\theta(h)=\psi(g)$. Then obviously $\theta \circ \phi=\psi$. The only thing we need to check is that $\theta$ is well-defined: assume that $\phi(g_1)=h=\phi(g_2)$. Then $g_1g_2^{-1} \in ker(\phi) \subseteq ker(\psi)$, whence $\psi(g_1)=\psi(g_2)$. $\endgroup$ – Nicky Hekster Apr 21 '19 at 20:27

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 $\theta(1) = \frac{1}{2}$.

  • $\begingroup$ It's worth noting that the issue, as said in the comments, is that $\phi$ is not surjective. $\endgroup$ – Simon Rose Dec 13 '13 at 20:44
  • $\begingroup$ yes, thanks. The problem had me confused for at least 10 minutes. It's nice to know I wasn't just missing something obvious $\endgroup$ – cats Dec 13 '13 at 20:46
  • $\begingroup$ @Simon, nice counterexample, one up from me. $\endgroup$ – Nicky Hekster Dec 13 '13 at 20:52

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