Hello! Given a surjective group homomorphism . Why is the restriction to the commutator subgroup also surjective?

Greetings

Banach

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- May 25th 2009, 09:34 AMBanachderived subgroup
Hello! Given a surjective group homomorphism . Why is the restriction to the commutator subgroup also surjective?

Greetings

Banach - May 25th 2009, 09:54 AMNonCommAlg
- May 25th 2009, 09:59 AMSwlabr
Quite simply, because it's a homomorphism.

Let . Then, and are homomorphic images of elements from , because the mapping is onto. So substitute in these elements, , and then use the fact that is a homomorphism to see that it is in the image of . Thus the generators of are contained in the image, and so the whole subgroup is.

. - 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 is then surjective?

Greetings

Banach - May 25th 2009, 10:44 AMNonCommAlg
- May 25th 2009, 10:55 AMBanach
Well, it is clear: Let , then there is a . But there is an . Since , 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