For two subgroups $A, B$ in $G$,
$[A,A] \cap [B,B] = [A\cap B, A \cap B]$?
At least, if $G$ is free, is the left contained in the right?
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2$\begingroup$ For $G$ non-free clearly not (take $G=A_6$ and $A\neq B$ two standard copies of $A_5$: then both $A,B$ are perfect, and the left-hand term is $A\cap B$, a standard copy of $A_4$. While $A\cap B$ is the same standard copy of $A_4$ (of order 12), so its derived subgroup $[A\cap B,A\cap B]$ is the derived subgroup in this copy of $A_4$, so is smaller (of order 4). $\endgroup$– YCorCommented Mar 10, 2018 at 9:01
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1$\begingroup$ Second, since $\supset$ is clear, equality and "left is contained in the right" mean the same. $\endgroup$– YCorCommented Mar 10, 2018 at 9:30
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$\begingroup$ @YCor Thank you for your comments. Then I only ask whether the above equality holds for subgroups $A,B$ in a free group $G$. (Is it better to fix the original writing?) $\endgroup$– qkqhCommented Mar 10, 2018 at 9:48
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$\begingroup$ (Also, why "commutator functor"?) $\endgroup$– LSpiceCommented Mar 10, 2018 at 22:48
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$\begingroup$ @LSpice between the categories of groups, $[-,-]$ is a functor, isn't it? $\endgroup$– qkqhCommented Mar 11, 2018 at 5:52
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1 Answer
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Let $G$ be free of rank $2$, and choose $A,B \lhd G$ such that $G/A \cong C_m$, $G/B \cong C_n$ and $G/A \cap B \cong C_m \times C_n$ for $m,n > 1$.
Then $A/[A,A]$ is free abelian of rank $m+1$, so $G/[A,A]$ is virtually free abelian of rank $m+1$. Similarly $G/[B.B]$ is virtually free abelian of rank $n+1$. So $G/([A,A] \cap [B,B]) \le G/[A,A] \times G/[B,B]$ is virtually abelian of rank at most $m+n+2$.
But $G/[A \cap B,A \cap B]$ is virtually free abelian of rank $mn+1$, so by choosing $m,n$ such that $mn+1 > m+n+2$, we get examples in which $[A,A] \cap [B,B] \not\le [A \cap B,A \cap B]$.
answered Mar 10, 2018 at 11:15
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$\begingroup$ Thank you!! If I restrict $A$ as a subgroup of $F$ which is generated by a subset of freely generating set of $F$, then can the equality hold? I guess that in that case, the equality holds, and moreover, $A\cap [B,B]=[A,A] \cap [B,B]$. Do you think is it true? $\endgroup$– qkqhCommented Mar 11, 2018 at 6:00
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$\begingroup$ @qkqh an intersection of finitely many free factors of $F$ is a free factor, cf the Magnus-Karrass-Solitar book. $\endgroup$– YCorCommented Mar 11, 2018 at 16:25