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Thread: derived subgroup

  1. #1
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    derived 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
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    Quote Originally Posted by Banach View Post
    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
    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 $\displaystyle \phi(a)=b.$
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    Quote Originally Posted by Banach View Post
    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
    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.

    $\displaystyle [g,h] = [a \phi, b \phi] = [a,b] \phi \in [A,A]\phi$.
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  4. #4
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    Well, yes of course... Did not see the obvious.
    Thank you for helping me.

    Greetings
    Banach
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    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
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    Quote Originally Posted by Banach View Post
    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
    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 $\displaystyle \phi^{ab}([A,A]a)=[B,B]\phi(a)=[B,B]b.$
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    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
    Last edited by Banach; May 25th 2009 at 11:13 AM.
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  8. #8
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    You were quicker .
    Thank you for your help!
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  9. #9
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    Quote Originally Posted by Banach View Post
    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^{ab}: \phi(a)=b. $. Since $\displaystyle \phi^{ab}(p_A(a))=p_B(\phi(a))=[b]$, we are there.

    Greetings
    Banach
    i'm not sure who you're talking to here and what exactly are those scary notations $\displaystyle p_A,$ etc., haha ... but i guess you got your answer anyway.
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  10. #10
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    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
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