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Let $S \leq G$ and $N \lhd G$ two subgroups of $G$.

Is the commutator $[SN,SN]$ equal to $[S,S]N$ ?

The first is generated by commutators $[ax,by]$ (where $a,b\in S$ and $x,y \in N$), and each generator (after inserting $e$'s in the forms $cc^{-1}$ or $c^{-1}c$ in the product $(ax)^{-1}(by)^{-1}axby$ and using the fact that $N$ is normal) can be expressed as $[a,b]n$ for a suitable $n$. Is this correct?

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2 Answers 2

up vote 7 down vote accepted

No. Take $G$ to be a nontrivial abelian group and $S,N = G.$ Then $(SN)' = 0,$ whereas $S'N = G.$

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Thanks for the counterexample. I'm pretty sure that every $[ax,by]$ can be put in the form $[a,b]n$ for some $n \in N$; I think the inverse is not true then. –  user14174 Jan 16 '12 at 17:20
    
@Lmn6: If $[a,b]=a^{-1}b^{-1}ab$ and $x^y = y^{-1}xy$, then the general identity is $$[xy,zt]=[x,t]^y[y,t][x,z]^{yt}[y,z]^t.$$ If $y,t\in N$, then $[x,y]^y$, $[y,t]$, $[y,z]^t$ are in $N$. Use $ab=ba[a,b]$ to replace $[x,y]^y[y,t][x,z]^{yt}$ with $[x,z]^{yt}n$ with $n\in N$; then use $a^b = a[a,b]$ to replace $[x,z]^{yt}$ with $[x,z]n'$ with $n'\in N$, to get it to the form you want. –  Arturo Magidin Jan 17 '12 at 16:59

To be more precise $[SN,SN] = [S,S][N,N][S,N].$ For the left-hand group must contain the right. On the other hand, the right-hand group is normal in $SN$ and the corresponding quotient group is Abelian. Here $[S,N]= \langle [s,n]: s \in S, n \in N \rangle,$ which is normal in $SN.$

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It is explained carefully –  Geoff Robinson Mar 26 at 20:19
    
$[SN,SN]$ contains $[a,b]$ for all $a,b \in SN.$ This applies when $a,b \in S,$ when $a,b \in N$ and when $a \in S, b \in N.$ Hence the LHS contains the RHS. On the other hand, $[N,N]$ is characteristic in $N,$ so normal in $G,$ and $[S,N] \lhd G.$ Letting ${\bar G}$ denote $G/[N,N][S,N],$ we see that ${\bar G} = {\bar S} {\bar N}$ and that $ [{\bar S}, {\bar N}] = [{\bar N},{\bar N}] = 1.$ Then $[{\bar S, {\bar S}] {\bar G}].$ The full preimage in $G$ of $[{\bar S},{\bar S}] $ is clearly $[S,S][N,N][S,N]$ which is therefore normal in $G$ by the isomorphism theorem. –  Geoff Robinson Mar 26 at 20:37

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