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The relation isso $K$ is the Klein bottle group. Let it act on ${\mathbb{R}}^2$ byThese are Euclidean isometries and satisfy $ab{a}^{-1}=b^{-1}$. Every element has a normal form $b^m{a}^n$. The orbit of $(0,0)$ is discrete, and a rectangle of finite size meets every orbit, so the action is proper and cocompact.

The squareis a vertical translation, while $b$ is a horizontal translation. They commute, andas an isometry only when $m=n=0$. Hence $⟨a^2,b⟩\cong {\mathbb{Z}}^2$. The normal form shows that every element lies in either $⟨a^2,b⟩$ or $a⟨a^2,b⟩$, so this subgroup has index two in $K$.

Solved by gpt-5.6-sol high.

Introduce $d=c^2$ and $e=a^2$. The two vertex groupsare Klein bottle groups. In $A$, the subgroup $⟨a^2,d⟩$ is ${\mathbb{Z}}^2$ of index two; in $B$, the subgroup $⟨e,c^2⟩$ is also ${\mathbb{Z}}^2$ of index two. IdentifyinggivesEliminating $d,e$ from this amalgamated free product recovers exactly the two given relators.

The Bass-Serre tree is bipartite with vertex sets $G/A$ and $G/B$ and edge set $G/⟨a^2,c^2⟩$. Both edge-group inclusions have index two, so every vertex has degree two. The tree is therefore a bi-infinite line, with $A$- and $B$-type vertices alternating.

Solved by gpt-5.6-sol high.

Use the amalgam from part (c), with edge groupAn odd power of $a$ belongs to $A\setminus C$, because $C$ is the index-two translation subgroup of the Klein bottle group $A$. Similarly, an odd power of $c$ belongs to $B\setminus C$. Thusis a reduced alternating word whose syllables lie in $A\setminus C$ and $B\setminus C$. The normal form theorem for an amalgamated free product says that every nonempty reduced alternating word is nonidentity. The displayed element is therefore nontrivial for every $n\ge 1$.

Solved by gpt-5.6-sol high.