Let $k$ be a field of characteristic zero, and let $A_1=A_1(k)$ be the first Weyl algebra.

It is well known (first proved by Dixmier, if I am not wrong) that the group of automorphisms of $A_1$, denote it by $H$, is the free amalgamated product of two of its subgroups (the affine and the triangular automorphisms), see for example, Theorem 2.

Denote the group of automorphisms and anti-automorphisms of $A_1$ by $G$. It is easy to see that $H$ is a normal subgroup of $G$ (of index $2$).

My question: Is it true that $G$ is the free amalgamated product of two of its subgroups, namely, the affine and the triangular automorphisms and anti-automorphisms?

It seems to me that my question should have a positive answer, with a proof for $G$ similar to a proof for $H$ (am I missing something?).

Thank you very much for your help.

Let $k$ be a field of characteristic zero, and let $A_1=A_1(k)$ be the first Weyl algebra.

It is well known (first proved by Dixmier, if I am not wrong) that the group of automorphisms of $A_1$, denote it by $H$, is the free amalgamated product of two of its subgroups (the affine and the triangular automorphisms), see for example, Theorem 2.

Denote the group of automorphisms and anti-automorphisms of $A_1$ by $G$. It is easy to see that $H$ is a normal subgroup of $G$ (of index $2$).

My question: Is it true that $G$ is the free amalgamated product of two of its subgroups, namely, the affine and the triangular automorphisms and anti-automorphisms?

It seems to me that my question should have a positive answer, with a proof for $G$ similar to a proof for $H$ (am I missing something?).

Thank you very much for your help.

Edit: I do not know if Karrass and Solitar's paper may help answer my question, since it deals with subgroups of a free amalgamated group, while in my case a free amalgamated group appears as an index $2$ (normal) subgroup of another group (can one prove that a group is free amalgamated by knowing that it has an index $2$ subgroup which is free amalgamated?).