Hello! Given a surjective group homomorphism $\displaystyle \phi: A \rightarrow B$. Why is the restriction to the commutator subgroup $\displaystyle \phi': [A,A] \rightarrow [B,B]$ also surjective?

Greetings

Banach

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- May 25th 2009, 09:34 AMBanachderived subgroup
Hello! Given a surjective group homomorphism $\displaystyle \phi: A \rightarrow B$. Why is the restriction to the commutator subgroup $\displaystyle \phi': [A,A] \rightarrow [B,B]$ also surjective?

Greetings

Banach - May 25th 2009, 09:54 AMNonCommAlg
you only need to show that the generators of [B,B], the commutators, have preimage in [A,A]. so let $\displaystyle b=b_1b_2b_1^{-1}b_2^{-1},$ where $\displaystyle b_1,b_2 \in B.$ since $\displaystyle \phi$ is surjective, $\displaystyle b_1=\phi(a_1), \ b_2=\phi(a_2),$ for some

$\displaystyle a_1,a_2 \in A.$ let $\displaystyle a=a_1a_2a_1^{-1}a_2^{-1} \in[A,A].$ then it's clear that $\displaystyle \phi(a)=b.$ - May 25th 2009, 09:59 AMSwlabr
Quite simply, because it's a homomorphism.

Let $\displaystyle [g,h] \in [B,B]$. Then, $\displaystyle g$ and $\displaystyle h$ are homomorphic images of elements from $\displaystyle A$, because the mapping is onto. So substitute in these elements, $\displaystyle g=a \phi$, $\displaystyle h=b \phi$ and then use the fact that $\displaystyle \phi$ is a homomorphism to see that it is in the image of $\displaystyle [A,A]$. Thus the generators of $\displaystyle [B,B]$ are contained in the image, and so the whole subgroup is.

$\displaystyle [g,h] = [a \phi, b \phi] = [a,b] \phi \in [A,A]\phi$. - May 25th 2009, 10:04 AMBanach
Well, yes of course... Did not see the obvious.

Thank you for helping me.

Greetings

Banach - May 25th 2009, 10:08 AMBanach
Sorry, I am a bit confused today: My actual question is why the induced homomorphism between the abelianizations $\displaystyle \phi^{ab}: A^{ab} \rightarrow B^{ab}$ is then surjective?

Greetings

Banach - May 25th 2009, 10:44 AMNonCommAlg
the induced map $\displaystyle \phi^{ab}$ is defined by $\displaystyle \phi^{ab}([A,A]a)=[B,B]\phi(a).$ for any $\displaystyle [B,B]b \in B^{ab}$ choose $\displaystyle a \in A$ such that $\displaystyle \phi(a)=b.$ then $\displaystyle \phi^{ab}([A,A]a)=[B,B]\phi(a)=[B,B]b.$

- May 25th 2009, 10:55 AMBanach
Well, it is clear: Let $\displaystyle [b] \in B^{ab}$, then there is a $\displaystyle b\in B: p_B(b)=[b]$. But there is an $\displaystyle a \in A: \phi(a)=b. $. Since $\displaystyle \phi^{ab}(p_A(a))=p_B(\phi(a))=[b]$, we are there.

Greetings

Banach - May 25th 2009, 10:56 AMBanach
You were quicker :).

Thank you for your help! - May 25th 2009, 10:58 AMNonCommAlg
- May 25th 2009, 11:16 AMBanach
Sorry, as said before i am completely confused today. I will try to be in a fitter state when i post my next question (Rofl).

By the way p_A and p_B are the projections to the quotient.

Have a nice day.

Banach