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Begin with $BS(n,n)=\langle t,x\mid tx^nt^{-1}=x^n\rangle$. Let $L$ be the kernel of the homomorphism $BS(n,n)\to\mathbf{Z}$, $t\mapsto 0$, $x\mapsto 1$. Using the action of $BS(n,n)$ on its standard Bass-Serre tree shows that $L$ is free. It is generated by the $x^mtx^{-m}$ when $m$ ranges over $\mathbf{Z}$, but since this sequence is $n$-periodic, $m$ ranging over $\{0,\dots,n-1\}$ is enough. It can be shown that this is a free family but I won't check it since this is enough to show that $L$ is free of rank $\le n$, which is enough for your purposes. Now $x^n$ is central and $L\times\langle x^n\rangle$ has finite index.

For $BS(n,-n)=\langle t,x:tx^nt^{-1}=x^n\rangle$$BS(n,-n)=\langle t,x:tx^nt^{-1}=x^{-n}\rangle$, we need to consider the subgroup $H$ of index 2 kernel of the homomorphism mapping $x$ to $0$ and $t$ to 1 mod 2. It has the presentation, denoting $u=t^2$ and $y=tx^{-1}t^{-1}$: $\langle u,x,y\mid x^n=y^n, tx^nt^{-1}=x^n,ty^nt^{-1}=y^n\rangle$. But the latter is also isomorphic to a subgroup of index 2 in $BS(n,n)$ (kernel of the same homomorphism, but with $y$ defined as $txt^{—1}\in BS(n,n)$ instead). This yield a (free $\times$ infinite cyclic) subgroup of finite index, but rather $F_{2n-1}\times\mathbf{Z}$ than $F_n\times\mathbf{Z}$ (although I don't claim there's no $F_n\times\mathbf{Z}$ of finite index).

Begin with $BS(n,n)=\langle t,x\mid tx^nt^{-1}=x^n\rangle$. Let $L$ be the kernel of the homomorphism $BS(n,n)\to\mathbf{Z}$, $t\mapsto 0$, $x\mapsto 1$. Using the action of $BS(n,n)$ on its standard Bass-Serre tree shows that $L$ is free. It is generated by the $x^mtx^{-m}$ when $m$ ranges over $\mathbf{Z}$, but since this sequence is $n$-periodic, $m$ ranging over $\{0,\dots,n-1\}$ is enough. It can be shown that this is a free family but I won't check it since this is enough to show that $L$ is free of rank $\le n$, which is enough for your purposes. Now $x^n$ is central and $L\times\langle x^n\rangle$ has finite index.

For $BS(n,-n)=\langle t,x:tx^nt^{-1}=x^n\rangle$, we need to consider the subgroup $H$ of index 2 kernel of the homomorphism mapping $x$ to $0$ and $t$ to 1 mod 2. It has the presentation, denoting $u=t^2$ and $y=tx^{-1}t^{-1}$: $\langle u,x,y\mid x^n=y^n, tx^nt^{-1}=x^n,ty^nt^{-1}=y^n\rangle$. But the latter is also isomorphic to a subgroup of index 2 in $BS(n,n)$ (kernel of the same homomorphism, but with $y$ defined as $txt^{—1}\in BS(n,n)$ instead).

Begin with $BS(n,n)=\langle t,x\mid tx^nt^{-1}=x^n\rangle$. Let $L$ be the kernel of the homomorphism $BS(n,n)\to\mathbf{Z}$, $t\mapsto 0$, $x\mapsto 1$. Using the action of $BS(n,n)$ on its standard Bass-Serre tree shows that $L$ is free. It is generated by the $x^mtx^{-m}$ when $m$ ranges over $\mathbf{Z}$, but since this sequence is $n$-periodic, $m$ ranging over $\{0,\dots,n-1\}$ is enough. It can be shown that this is a free family but I won't check it since this is enough to show that $L$ is free of rank $\le n$, which is enough for your purposes. Now $x^n$ is central and $L\times\langle x^n\rangle$ has finite index.

For $BS(n,-n)=\langle t,x:tx^nt^{-1}=x^{-n}\rangle$, we need to consider the subgroup $H$ of index 2 kernel of the homomorphism mapping $x$ to $0$ and $t$ to 1 mod 2. It has the presentation, denoting $u=t^2$ and $y=tx^{-1}t^{-1}$: $\langle u,x,y\mid x^n=y^n, tx^nt^{-1}=x^n,ty^nt^{-1}=y^n\rangle$. But the latter is also isomorphic to a subgroup of index 2 in $BS(n,n)$ (kernel of the same homomorphism, but with $y$ defined as $txt^{—1}\in BS(n,n)$ instead). This yield a (free $\times$ infinite cyclic) subgroup of finite index, but rather $F_{2n-1}\times\mathbf{Z}$ than $F_n\times\mathbf{Z}$ (although I don't claim there's no $F_n\times\mathbf{Z}$ of finite index).

Source Link
YCor
YCor
  • 63.9k
  • 5
  • 187
  • 286

Begin with $BS(n,n)=\langle t,x\mid tx^nt^{-1}=x^n\rangle$. Let $L$ be the kernel of the homomorphism $BS(n,n)\to\mathbf{Z}$, $t\mapsto 0$, $x\mapsto 1$. Using the action of $BS(n,n)$ on its standard Bass-Serre tree shows that $L$ is free. It is generated by the $x^mtx^{-m}$ when $m$ ranges over $\mathbf{Z}$, but since this sequence is $n$-periodic, $m$ ranging over $\{0,\dots,n-1\}$ is enough. It can be shown that this is a free family but I won't check it since this is enough to show that $L$ is free of rank $\le n$, which is enough for your purposes. Now $x^n$ is central and $L\times\langle x^n\rangle$ has finite index.

For $BS(n,-n)=\langle t,x:tx^nt^{-1}=x^n\rangle$, we need to consider the subgroup $H$ of index 2 kernel of the homomorphism mapping $x$ to $0$ and $t$ to 1 mod 2. It has the presentation, denoting $u=t^2$ and $y=tx^{-1}t^{-1}$: $\langle u,x,y\mid x^n=y^n, tx^nt^{-1}=x^n,ty^nt^{-1}=y^n\rangle$. But the latter is also isomorphic to a subgroup of index 2 in $BS(n,n)$ (kernel of the same homomorphism, but with $y$ defined as $txt^{—1}\in BS(n,n)$ instead).