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Let $A$ be an abelian normal subgroup of $G$ and $x\in G$. How can we prove the following?

(a) The mapping $A\mapsto A$ given by $a \mapsto [a,x]$ is a homomorphism.

(b) $[A,\langle x\rangle]=\{[a,x]|a\in A\}$.

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We probably need $A$ abelian. – peoplepower Dec 24 '12 at 11:41
My bad, it's my first time to use this. – Song Dec 24 '12 at 11:43
No problem. ${}{}$ – anon Dec 24 '12 at 11:44
up vote 2 down vote accepted

Hint: $$[ab,x]=[a,x]^b[b,x]$$

Stronger Hint: $[a,x^2]\in A$ because $A$ is normal. What can you say about $[a,x^2]$ when $a\in \text{Ker}(a\mapsto [a,x])$?

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Of course, we're mapping $A\to A$, so we should be calculating $[ab,x]$. – peoplepower Dec 24 '12 at 12:02
I could prove the (a) ,whether we use the same method to prove (b) – Song Dec 24 '12 at 12:11
@user53587 Use (a) to prove (b). Hint: What are the relevant groups of a homomorphism? E.g. the kernel is one. – peoplepower Dec 24 '12 at 12:16

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