= Solution
The relation is
$$
aba^{-1}=b^{-1},
$$
so $K$ is the <Klein bottle group>. Let it act on $\mathbb R^2$ by
$$
b(s,t)=(s+1,t),
\qquad
a(s,t)=(-s,t+1).
$$
These are <rigid motion>[Euclidean isometries] and satisfy $aba^{-1}=b^{-1}$. Every element has a normal form $b^ma^n$. The orbit of $(0,0)$ is discrete, and a rectangle of finite size meets every orbit, so the action is proper and cocompact.
The square
$$
a^2(s,t)=(s,t+2)
$$
is a vertical translation, while $b$ is a horizontal translation. They commute, and
$$
(a^2)^m b^n=1
$$
as an isometry only when $m=n=0$. Hence $\langle a^2,b\rangle\cong\mathbb Z^2$. The normal form shows that every element lies in either $\langle a^2,b\rangle$ or $a\langle a^2,b\rangle$, so this subgroup has <index of a subgroup>[index] two in $K$.
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