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Suppose that $a$ acted as a hyperbolic isometry of a tree, with translation length $\tau >0$. Nonzero powers have the same axis andTranslation length is invariant under conjugacy, whereas the defining relation in the Baumslag-Solitar group says that $a^m$ and $a^n$ are conjugate. Hence $m\tau =n\tau$, contradicting $m\ne n$. Therefore $a$ acts elliptically in every combinatorial tree action.

Solved by gpt-5.6-sol high.

Suppose $G=A\ast B$ were a nontrivial free product. Its Bass-Serre tree action has trivial edge stabilizers and no global fixed vertex. Part c makes $a$ elliptic. Since $a$ has infinite order, the fixed set of every nonzero power $a^r$ is a single vertex: it is nonempty, while fixing two vertices would fix the intervening edge and put the infinite-order element $a^r$ in a trivial edge stabilizer.

Let this vertex be $v$. The relation $b{a}^m{b}^{-1}=a^n$ givesand both sides are the singleton $\{v\}$. Thus $b$ also fixes $v$. Since $a$ and $b$ generate the Baumslag-Solitar group, the entire group fixes $v$, contradicting the Bass-Serre action of a nontrivial free product. Hence no such decomposition exists.

Solved by gpt-5.6-sol high.

The HNN extension is the Baumslag-Solitar groupIts Bass-Serre tree has vertices $G/⟨u⟩$ and oriented edges $G/⟨u⟩$, with the two endpoint maps induced by the identity embedding and the index-two embedding $n\mapsto 2n$. At each vertex there is one incident edge on the identity side and two on the index-two side. The underlying unoriented tree is therefore infinite and $3$-regular, with an orientation in which every vertex has one incoming and two outgoing edges, up to reversing the convention.

Solved by gpt-5.6-sol high.