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

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November 19

Normal subgroups and bijection

Let ϕ:GG be a surjective homomorphism. Suppose you have some subgroup H of G and define H:=ϕ1(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 ϕ(x)=ϕ(y), then xNyN. This depends on HN. Then use that to show that ϕ1 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 ϕ(x)=ϕ(y)(xHyH). How does the result follow? How do you use that ϕ is surjective?--AnalysisAlgebra (talk) 17:31, 20 November 2012 (UTC)
Hint:consider ϕ(x1y).
Surjectivity isn't important; since the image of ϕ is a subgroup, you could just replace G with imageϕ.--80.109.106.49 (talk) 18:17, 20 November 2012 (UTC)
Yes, x1y is in the kernel of ϕ. So what?--AnalysisAlgebra (talk) 18:46, 20 November 2012 (UTC)
What's the relationship between the kernel of ϕ and H?--80.109.106.49 (talk) 19:02, 20 November 2012 (UTC)
All I can think of is kerϕ 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 x1yN.--80.109.106.49 (talk) 20:59, 20 November 2012 (UTC)