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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 :)

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

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:


Try now to take it from here...

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OK, so first I try to show $(u_{1}u_{2},v)=(u_{1},v)(u_{2},v) \Leftrightarrow [U,V]$ contained in centre of the subgroup...(see above) (So the "\Rightarrow" part) (i): $(u_{1}u_{2},v)\mapsto [u_{1}u_{2},v]=[u_{2},v]^{u_{1}}[u_{1},v]=u_{1}u_{2}v u_{2}^{-1}u_{1}^{-1}v^{-1}$. (ii): $(u_{1},v)(u_{2},v)\mapsto [u_{1},v][u_{2},v] = u_{1}v u_{1}^{-1}v^{-1}u_{2}v u_{2}^{-1}v^{-1}$. Now $u_{1}u_{2}v u_{2}^{-1}u_{1}^{-1}v^{-1}$ shell be the same as $u_{1}v u_{1}^{-1}v^{-1}u_{2}v u_{2}^{-1}v^{-1}$. $u_{1}u_{2}v u_{2}^{-1}u_{1}^{-1}v^{-1}=u_{1}v u_{1}^{-1}v^{-1}u_{2}v u_{2}^{-1}v^{-1}$ (no char left) –  Luca Oct 7 '12 at 13:22
I don't think you have to open up the commutator parentheses, as this may make thing messier. Think of this as $$x=x^a\Longrightarrow ax=xa\Longrightarrow x\in C_G(a)\,\,...$$ –  DonAntonio Oct 7 '12 at 13:26
then i multiply both sides from left by $u_{1}$ and from the right with $v$.\\ $u_{2}vu_{2}^{-1}u_{1}^{-1} = vu_{1}^{-1}v^{-1}u_{2}vu_{2}^{-1}$\\ $\Leftrightarrow u_{1} (u_{2}vu_{2}^{-1}v)u_{1}^{-1}(u_{2}v^{-1}u_{2}^{-1}v){-1} = e$\\ So we see that the expression in the brace is an element of [U,V] and lies in the centre of U (?) And this analogous for $v_{1},v_{2}$ shows it lies in the centre of V(?) Here I know I'm missing something... –  Luca Oct 7 '12 at 13:35
ah, sorry saw your comment too late! –  Luca Oct 7 '12 at 13:36

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