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I'm quite new to this whole topic and so I don't know how get a grip on this question:

Let $G$ be a group and $U,V$ two subgroups. Denote by $[U,V]$ the subgroup of $G$ generated by $[u,v]=uvu^{-1}v^{-1}$ with $u \in U, v \in V$. Show: $U\times V \rightarrow G$, $(u,v) \mapsto [u,v]$ is bilinear $\Leftrightarrow$ [U,V] is contained in the centre of the subgroup of $G$, that is generated by $U$ and $V$

I'd be happy for any hint and help :)

2 Answers 2

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Hint: $[a,bc] = [a,b][a,c]^b$ and $[ab,c] = [b,c]^a [a,c]$. Here, $x^y = yxy^{-1}$ denotes conjugation.

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Hint:

$(u,v)\to [u,v]\,\,\text{bilinear}\,\,\Longleftrightarrow (u_1u_2,v)=(u_1,v)(u_2,v)\,\,,\,\,(u,v_1v_2)=(u,v_1)(u,v_2)$

where we're considering the operation in $\,G\,$ to be multiplicative.

But using the basic property of commutators ($\,[ab,c]=[a,c]^b[b,c]\,$), we get:

$[u_1,v][u_2,v]=[u_1u_2,v]=[u_1,v]^{u_2}[u_2,v]$

Try now to take it from here...

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    ah, sorry saw your comment too late!2012-10-07