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This is a follow-up question to Is there a non-free group $G$ whose subgroups are all freely decomposable?

In the answer to that question, Cornulier gives the following example (due to Kurosh) of a group $G$ which is not free, yet isomorphic to $G*\mathbb{Z}$: $$G=\langle (a_n)_{n\geq 0},(b_n)_{n\geq 1},\mid \left[a_n,b_n\right]=a_{n-1},\forall n\geq 1\rangle.$$

Of course $G\cong G*F$ whenever $F$ is a finitely generated free group.

Question: Is the group $$H=\langle (a_n)_{n\geq 0},(b_n)_{n\geq 1},(c_n)_{n\geq 1},(d_n)_{n\geq 1}\mid \left[a_n,b_n\right]\left[c_n,d_n\right]=a_{n-1},\forall n\geq 1\rangle$$ non-isomorphic to $H*F_n$, $n=1,2 \pmod3$? ($H$ is isomorphic to $H*F_3$.) Is $H$ isomorphic to $G$? It seems to me like it shouldn't be.

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  • $\begingroup$ That just kills all the $a$s, so the abelianizations are both countably generated free abelian groups. $\endgroup$
    – seldom seen
    Commented Dec 22, 2020 at 17:54
  • $\begingroup$ Actually I see no reason that $H$ is not residually nilpotent. For instance I see no reason that $a_n\in [G,[G,G]]$ (unlike in $G$). $\endgroup$
    – YCor
    Commented Dec 22, 2020 at 18:47
  • $\begingroup$ That's a good point. I wonder if the question is better if we compare the groups one (!) gets by gluing together a chain of genus $3$ surfaces to $H$, i.e., the group with relations $\left[a_n,b_n\right]\left[c_n,d_n\right]\left[e_n,f_n\right]=a_{n-1}$. (Maybe there is another obstruction to residual nilpotency?) $\endgroup$
    – seldom seen
    Commented Dec 22, 2020 at 18:52
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    $\begingroup$ Let $H_m$ be the group obtained by omitting all generators and relators for $n>m$. $H_m$ is freely indecomposable relative to $a_0$. (The cyclic JSJ relative to $a_0$ is $m$ genus three surfaces with one boundary component glued together in a chain.) If $H$ was free then $a_0$ must live in some finitely generated free factor $H'$, but since $H_m$ is freely indecomposable relative to $a_0$, $H_m<H'$. This is true for all $m$, so $H<H'$, hence $H=H'$. $\endgroup$
    – seldom seen
    Commented Dec 22, 2020 at 19:18
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    $\begingroup$ I think this answers it in general. The idea is the same as the proof that $H$ is not free, but uses the JSJ decompositions of the $H_m$. Define $G_m$ in the same way. If $G_m<H_l$ then $G_m$ inherits a graph of groups decomposition from the cyclic JSJ of $H_l$ relative to $H_{l-1}$. If $a_0$ (the one from $G$) is contained in $H_{l-1}$ then either $G_m$ is contained in $H_{l-1}$ or there is a cover of the surface of genus two with one boundary component contained in some QH vertex group of $G_m$, but there isn't enough room since $G$ is made up out of tori with one boundary component. $\endgroup$
    – seldom seen
    Commented Dec 22, 2020 at 19:57

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