Let $f: G \rightarrow G'$ be a group homomorphism with kernel $H$. Then we know as an elementary fact of abstract algebra that there is an injective group homomorphism $f^*:G/H \rightarrow G'$ such that $f^*(x+H) = f(x)$. More generally, let $\psi: G \rightarrow G/H$ be a surjective homomorphism with kernel $H$. Then again we have an injective homomorphism $f_{\psi}^* : G/H \rightarrow G'$ defined by $f^*_{\psi}(x+H)=f(y)$ where $\psi(y) = x+H$.

Could you verify the validity of this reasoning? Thanks.


To give you an example about the motivation of this question, we know that given a homomorphism $f:G \rightarrow G'$ with kernel $H$ there exists always a homomorphism $f^*: G/H \rightarrow G'$ such that $f = f^* \circ \phi$ where $\phi:G \rightarrow G/H$ is the canonical mapping.

Now, what if instead of the canonical mapping $\phi$ we have another surjective homomorphism $\psi:G \rightarrow G/H$ with kernel $H$? Does then exist a homomorphism $f^*_{\psi} : G/ H \rightarrow G'$ such that $f = f^*_{\psi} \circ \psi$?

  • $\begingroup$ You have something mixed up. $x + H$ is not in $G/H$. $\endgroup$ – Karolis Juodelė Aug 8 '12 at 18:36
  • $\begingroup$ @KarolisJuodelė: Why not? $\endgroup$ – Manos Aug 8 '12 at 18:39
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    $\begingroup$ I don't see how the situation of $\psi:G\to G/H$ is "more general". $\endgroup$ – Andrew Aug 8 '12 at 18:41
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    $\begingroup$ First of all, what do you mean by $f(G)$? I just took it to mean the range, and that thus $f$ was necessarily an epimorphism, but it really is not clear. $\endgroup$ – Thomas Andrews Aug 8 '12 at 18:49
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    $\begingroup$ But there is nothing that says $f$ is the canonical quotient map to begin with. $\endgroup$ – Andrew Aug 8 '12 at 18:50

By the first isomorphism theorem there is a unique $\psi^*:G/H\to G/H$ satisfying $\psi^*\circ\phi=\psi,$ which must be an isomorphism since it is injective. Let $f^*_\psi:=f^*\circ(\psi^*)^{-1}.$ Then $$f^*_\psi\circ\psi = \left(f^*\circ(\psi^*)^{-1}\right)\circ\psi = f^*\circ\left((\psi^*)^{-1}\circ\psi\right) = f^*\circ\phi=f.$$

  • $\begingroup$ Nice. Small typo: $\psi^* \circ \phi = \psi$ :-) $\endgroup$ – Manos Aug 8 '12 at 19:29
  • $\begingroup$ Fixed it, thanks. $\endgroup$ – Andrew Aug 8 '12 at 19:31

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