Wikipedia:Reference desk/Archives/Mathematics/2012 November 19

= November 19 =

Normal subgroups and bijection
Let $$\phi :G \rightarrow G'$$ be a surjective homomorphism. Suppose you have some subgroup $$H'$$ of $$G'$$ and define $$H := \phi^{-1}(H')$$. Let $$N$$ be a subgroup of $$G$$ containing $$H$$ and $$N'$$ be a normal subgroup of $$G'$$ containing $$H'$$. Show that there is a bijection between $$N$$ and $$N'$$.--AnalysisAlgebra (talk) 22:27, 19 November 2012 (UTC)


 * But what you've stated is clearly false.  Sławomir Biały  (talk) 00:06, 20 November 2012 (UTC)
 * I TOTALLY misunderstood the question. I need to show that there is a bijection between the set of subgroups of G and the set of subgroups of G' . I'm not sure if they need to be normal or not.--AnalysisAlgebra (talk) 08:46, 20 November 2012 (UTC)
 * Subgroups containing H and H' respectively, that is.--AnalysisAlgebra (talk) 08:48, 20 November 2012 (UTC)
 * They don't need to be normal. Start by showing that if $$\phi(x) = \phi(y)$$, then $$ x \in N \leftrightarrow y \in N$$.  This depends on $$H \subseteq N$$.  Then use that to show that $$\phi^{-1}$$ is the desired bijection.--149.148.254.207 (talk) 09:53, 20 November 2012 (UTC)
 * Hem. That's harder than it looks. You can get $$\phi(x) = \phi(y) \implies (x \in H \iff y \in H)$$. How does the result follow? How do you use that $$\phi$$ is surjective?--AnalysisAlgebra (talk) 17:31, 20 November 2012 (UTC)
 * Hint:consider $$\phi(x^{-1}y)$$.
 * Surjectivity isn't important; since the image of $$\phi$$ is a subgroup, you could just replace $$G'$$ with image$$\phi$$.--80.109.106.49 (talk) 18:17, 20 November 2012 (UTC)
 * Yes, $$x^{-1}y$$ is in the kernel of $$\phi$$. So what?--AnalysisAlgebra (talk) 18:46, 20 November 2012 (UTC)
 * What's the relationship between the kernel of $$\phi$$ and H?--80.109.106.49 (talk) 19:02, 20 November 2012 (UTC)
 * All I can think of is $$\ker\phi$$ is a suubgroup of $$H$$; did you have something else in mind? How does it relate to $$N$$?--AnalysisAlgebra (talk) 20:43, 20 November 2012 (UTC)
 * Since N contains H, this tells you that $$x^{-1}y \in N$$.--80.109.106.49 (talk) 20:59, 20 November 2012 (UTC)