1. ## derived subgroup

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

Greetings
Banach

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

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

$a_1,a_2 \in A.$ let $a=a_1a_2a_1^{-1}a_2^{-1} \in[A,A].$ then it's clear that $\phi(a)=b.$

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

Greetings
Banach
Quite simply, because it's a homomorphism.

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

$[g,h] = [a \phi, b \phi] = [a,b] \phi \in [A,A]\phi$.

4. Well, yes of course... Did not see the obvious.
Thank you for helping me.

Greetings
Banach

5. Sorry, I am a bit confused today: My actual question is why the induced homomorphism between the abelianizations $\phi^{ab}: A^{ab} \rightarrow B^{ab}$ is then surjective?

Greetings
Banach

6. Originally Posted by Banach
Sorry, I am a bit confused today: My actual question is why the induced homomorphism between the abelianizations $\phi^{ab}: A^{ab} \rightarrow B^{ab}$ is then surjective?

Greetings
Banach
the induced map $\phi^{ab}$ is defined by $\phi^{ab}([A,A]a)=[B,B]\phi(a).$ for any $[B,B]b \in B^{ab}$ choose $a \in A$ such that $\phi(a)=b.$ then $\phi^{ab}([A,A]a)=[B,B]\phi(a)=[B,B]b.$

7. Well, it is clear: Let $[b] \in B^{ab}$, then there is a $b\in B: p_B(b)=[b]$. But there is an $a \in A: \phi(a)=b.$. Since $\phi^{ab}(p_A(a))=p_B(\phi(a))=[b]$, we are there.

Greetings
Banach

8. You were quicker .

9. Originally Posted by Banach
Well, it is clear: Let $[b] \in B^{ab}$, then there is a $b\in B: p_B(b)=[b]$. But there is an $a \in A^{ab}: \phi(a)=b.$. Since $\phi^{ab}(p_A(a))=p_B(\phi(a))=[b]$, we are there.

Greetings
Banach
i'm not sure who you're talking to here and what exactly are those scary notations $p_A,$ etc., haha ... but i guess you got your answer anyway.

10. Sorry, as said before i am completely confused today. I will try to be in a fitter state when i post my next question .

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

Have a nice day.
Banach