Parabolic double cosets of the infinite dihedral group

The parabolic subgroups. By [F2] the parabolic subgroups are the conjugates of $\{1\}$, $W_I=\{1,s\}$, $W_J=\{1,t\}$ and $W_S=W$; here $\{1\}$ and $W$ are their own conjugates, and the conjugates of $\{1,s\}$ or of $\{1,t\}$ are the two-element subgroups $\{1,r\}$ with $r\in T$, while conversely every $r\in T$ is a conjugate of $s$ or of $t$ by [F3]. Hence the parabolic subgroups are exactly $\{1\}$, the $\{1,r\}$ with $r\in T$, and $W$. By [F8], $W_I\cap W_J=\{1\}$, and the two standard parabolics are not conjugate to each other: the assignment $s\mapsto 0$, $t\mapsto 1$ sends the relators $s^2,t^2$ to $0$, so by [F6] it defines a homomorphism $\phi \text{\hspace{0.17em}⁣}:W\to \mathbb{Z}\mathrm{/}2$, and if $w\{1,s\}{w}^{-1}=\{1,t\}$, then $ws{w}^{-1}=t$ and [F7] gives $0=\phi (s)=\phi (ws{w}^{-1})=\phi (t)=1$, a contradiction.

The one-sided quotients. First $su{s}^{-1}=s(st)s=ts=u^{-1}$, so $s{u}^{m}s=u^{-m}$ for every integer $m$. Every element of $W$ is $u^m$ or $u^{m}s$ with $m\in \mathbb{Z}$: any word in $s,t$ can be shortened by cancelling adjacent equal letters until it alternates, and an alternating word is $(st{)}^m$, $(ts{)}^m$, $(st{)}^{m}s$ or $(ts{)}^{m}s$, which is $u^m$, $u^{-m}$, $u^{m}s$ or $u^{-m}s$. The two families are disjoint: if $u^m=u^{m^{\mathrm{\prime }}}s$ then $s=u^{m-m^{\mathrm{\prime }}}\in ⟨u⟩$, and $s=u^k$ implies $u^{-k}=s{u}^{k}s=s\cdot s\cdot s=s=u^k$, where the first equality uses $su{s}^{-1}=u^{-1}$ and $s=s^{-1}$, so $u^{2k}=1$ and $k=0$ by the infinite order of $u$, whence $s=1$ and $st=t$ of order $2$, contradicting [F1]. Using the reduced alternating forms of [F1]: for $m\ge 1$ one has $s{u}^m=s(st{)}^m=t(st{)}^{m-1}$ of length $2m-1<2m$, so $u^m\mathrm{\notin }{}^{I}W$; for $m<0$ the element $s{u}^m=(st{)}^{\mathrm{\mid }m\mathrm{\mid }}s$ is alternating of length $2\mathrm{\mid }m\mathrm{\mid }+1>2\mathrm{\mid }m\mathrm{\mid }=\mathrm{\ell }(u^m)$, and for $m=0$ one has $s\cdot 1=s$ of length $1>0$, so no left descent occurs for $m\le 0$; for $m\ge 0$ one has $s(u^{m}s)=u^{-m}$ of length $2m<2m+1=\mathrm{\ell }(u^{m}s)$, so $u^{m}s\mathrm{\notin }{}^{I}W$; and for $m\le -1$ the element $s(u^{m}s)=u^{-m}$ has length $2\mathrm{\mid }m\mathrm{\mid }>2\mathrm{\mid }m\mathrm{\mid }-1=\mathrm{\ell }(u^{m}s)$, so no left descent occurs. Hence ${}^{I}W=\{u^m:m\le 0\}\cup \{u^{m}s:m\le -1\}=\{u^{-k}:k\ge 0\}\cup \{u^{-k}s:k\ge 1\}$. Dually, for $m\ge 1$ one has $u^{m}t=(st{)}^{m-1}s$ of length $2m-1<2m$, so $u^m\mathrm{\notin }{W}^J$; for $m\le 0$ the element $u^{m}t=(ts{)}^{\mathrm{\mid }m\mathrm{\mid }}t$ is alternating of length $2\mathrm{\mid }m\mathrm{\mid }+1>2\mathrm{\mid }m\mathrm{\mid }=\mathrm{\ell }(u^m)$, so no right descent occurs; for $m\ge 0$ one has $u^{m}st=u^{m+1}$ of length $2m+2>2m+1=\mathrm{\ell }(u^{m}s)$, so $u^{m}s\in W^J$; and for $m\le -1$ the element $u^{m}st=u^{m+1}$ has length $2\mathrm{\mid }m\mathrm{\mid }-2<2\mathrm{\mid }m\mathrm{\mid }-1=\mathrm{\ell }(u^{m}s)$, so $u^{m}s\mathrm{\notin }{W}^J$. Hence $W^J=\{u^m:m\le 0\}\cup \{u^{m}s:m\ge 0\}$.

The blocks of the decomposition. For $k\ge 0$ put $d_k:=(ts{)}^k=u^{-k}$. Then $s(ts{)}^k=(st{)}^{k}s=u^{k}s$, since $sts=sts$ may be rewritten as $(st)s$; further $(ts{)}^{k}t=(ts{)}^{k+1}s=u^{-(k+1)}s$, and $s(ts{)}^{k}t=(st{)}^{k+1}=u^{k+1}$. Hence $W_I(ts{)}^k{W}_J=\{u^{-k},u^{k}s,u^{-(k+1)}s,u^{k+1}\}$, and by the reduced alternating forms of [F1] these four elements have lengths $2k$, $2k+1$, $2(k+1)-1=2k+1$ and $2k+2$; they are pairwise distinct, the two translations having different lengths and the two $s$-multiples being equal only if $u^{2k+1}=1$, impossible by [F1]. The minimum length among them is $2k=\mathrm{\ell }(d_k)$, and $d_k=u^{-k}\in {}^I{W}^J$ by the description of step 1.2, so [F4] shows that $d_k$ is the unique minimum of the double coset $W_I(ts{)}^k{W}_J$.

The two-sided quotient. By step 1.2, ${}^I{W}^J={}^{I}W\cap W^J$ consists of the elements of the list $\{u^m:m\le 0\}\cup \{u^{m}s:m\le -1\}$ that also lie in the list $\{u^m:m\le 0\}\cup \{u^{m}s:m\ge 0\}$. Since the two families are disjoint, an element of the first family lies in the second list precisely when its index already satisfies $m\le 0$, and an element of the second family would have to satisfy $m\le -1$ (for ${}^{I}W$) and $m\ge 0$ (for $W^J$) simultaneously, which is impossible. Hence ${}^I{W}^J=\{u^m:m\le 0\}=\{u^{-k}:k\ge 0\}=\{(ts{)}^k:k\ge 0\}$.

The intersection, the sizes and the decomposition. For $k\ge 0$ one has $d_k^{-1}s{d}_k=u^{k}s{u}^{-k}=u^{2k}s$ because $s{u}^{-k}s=u^k$; the only element of $I$ is $s$, and $u^{2k}s\in J=\{t\}$ would force $u^{2k}s=t=u^{-1}s$, hence $u^{2k}=u^{-1}$ and $u^{2k+1}=1$, impossible since $u$ has infinite order by [F1]. Hence $K=\mathrm{\varnothing }$, $W_K=\{1\}$ and $W_I^K=W_I$; the size formula of [F4] gives $\mathrm{\mid }{W}_I(ts{)}^k{W}_J\mathrm{\mid }=2\cdot 2=4=2\cdot 2\mathrm{/}1$. Finally each element of $W$ is $u^m$ or $u^{m}s$ with a unique $m\in \mathbb{Z}$, and the four families $u^{-k}(k\ge 0)$, $u^{k+1}(k\ge 0)$, $u^{k}s(k\ge 0)$ and $u^{-(k+1)}s(k\ge 0)$ list exactly $u^m$ for $m\le 0$, $u^m$ for $m\ge 1$, $u^{m}s$ for $m\ge 0$ and $u^{m}s$ for $m\le -1$, without repetition by the disjointness of the two families in step 1.2. Hence $W$ is the disjoint union of the double cosets $W_I(ts{)}^k{W}_J$, $k\ge 0$, the first block being $W_I{W}_J=\{1,s,t,st\}$; this completes the example.