Commutator subgroup a subgroup of any Abelian quotient group?

  • #1
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I am new to group theory, and read about a "universal property of abelianization" as follows: let G be a group and let's denote the abelianization of G as Gab (note, recall the abelianization of G is the quotient G/[G,G] where [G,G] denotes the commutator subgroup). Now, suppose we have a homomorphism p: G --> H with H being an Abelian group. Then this homomorphism has to factor through the Abelianization. (what I mean by factor through is there exists another map q such that q ° θ = p, where I'm letting θ denote the cannonical quotient map G --> G/[G,G])


Ok, that property makes sense. The issue is, there are a couple proofs I'm reading that jump straight from using this property to the fact that then if we have G and H a normal subgroup of G and G/H Abelian, then this means [G,G] is a subgroup of H. I guess this doesn't surprise me, but at the same time I just don't see this quick connection. There is no elaboration of this "next step" in the proofs I'm reading so I assume this is a trivial corollary of the universal property mentioned, unfortunately I don't see it. Why does this fact obviously follow?


This is what I do understand: I know I have my cannonical quotient map θ: G --> G/[G,G]. I also have another this other quotient map λ: G --> G/H with G/H is Abelian by assumption, so that means λ factors through G/[G,G], meaning (by the universal property) there is another map ρ: G/[G,G] --> G/H with ρ°θ=λ, so this is a map going from G --> G/[G,G] --> G/H. I guess my question might be, is it always the case that if I have a group G with subgroups A and B, and if I have a "Factoring through" like this G --> G/A --> G/B, does it means A must be a subgroup of B? Forgive this question if its unbearably obvious, I can't stress enough how poor my algebra skills are
 

Answers and Replies

  • #2
Office_Shredder
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If a is in A, then when you map to G/A a maps to the identity, so when you push forward to G/B a gets mapped to the identity. Therefore a is in B for all a in A
 
  • #3
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oh man. i feel so dumb. But I thank you so much for pointing this out.
 

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