Parabolic Subgroups and Double Coset Geometry — Examples

The subgroup $H$. The one-line displays $(13)=3214$ and $(24)=1432$ record that these permutations swap $1\leftrightarrow 3$ and $2\leftrightarrow 4$ respectively and fix the remaining points. Both are involutions with disjoint supports, so they commute and $H=\{1,(13),(24),(13)(24)\}$, of order $4$; the product is computed by applying the right factor first, $(13)(24)\text{\hspace{0.17em}⁣}:1\mapsto 3$, $2\mapsto 4$, $3\mapsto 1$, $4\mapsto 2$, hence $(13)(24)=3412$ and $H=\{1234,1432,3214,3412\}$. Further $s_2{s}_1{s}_2=(23)(12)(23)=(13)$ and $s_2{s}_3{s}_2=(24)$, as direct computations in cycle notation; the two generators $s_1,s_3$ commute, so $W_{\{s_1,s_3\}}=\{1,s_1,s_3,s_1{s}_3\}=\{1234,2134,1243,2143\}$, and conjugating by $s_2$ gives $H=⟨s_2{s}_1{s}_2,s_2{s}_3{s}_2⟩=s_2{W}_{\{s_1,s_3\}}{s}_2$, so $H$ is a parabolic subgroup by [F2]. Finally $w=(12)(34)=2143$ satisfies $w(1)=2$, $w(3)=4$ and $w^{-1}(2)=1$, $w^{-1}(4)=3$, so $w(13)w^{-1}$ swaps $2$ and $4$ and fixes $1,3$, that is, $w(13)w^{-1}=(24)$.

The subgroup $H^{\mathrm{\prime }}$. Put $u:=st$. Every word in $s,t$ can be shortened by cancelling adjacent equal letters $ss$ or $tt$ until it alternates, so an alternating word is $(st{)}^k$, $(ts{)}^k$, $(st{)}^{k}s$ or $(ts{)}^{k}s$ for some $k\ge 0$; since $(ts{)}^k=u^{-k}$ and $(ts{)}^{k}s=u^{-k}s$, we get $W=\{u^k:k\in \mathbb{Z}\}\cup \{u^{k}s:k\in \mathbb{Z}\}$, with distinct powers because $u$ has infinite order. The two families are disjoint: the separate homomorphism $\chi :W\to \mathbb{Z}\mathrm{/}2$ defined by $s\mapsto 1$, $t\mapsto 1$ exists by [F5] and takes $u^k$ to $0$ and $u^{k}s$ to $1$. Since $s=s^{-1}$ and $t=t^{-1}$ we have $su{s}^{-1}=s(st)s=ts=u^{-1}$ and hence $s{u}^{2m}s=u^{-2m}$ for all $m$. In particular $tst=t(st)t\cdot t=(ts{)}^{2}s=u^{-2}s$, so $tst$ is a reflection by [F4], and $tst=u^{-2}s\in ⟨s,u^2⟩$; conversely $s\cdot (u^{-2}s)=s{u}^{-2}s=u^2$ and $(u^{-2}s)\cdot s=u^{-2}$, so $⟨s,tst⟩=⟨s,u^2⟩$. The set $S_0:=\{u^{2k}:k\in \mathbb{Z}\}\cup \{u^{2k}s:k\in \mathbb{Z}\}$ is closed under multiplication: the four product types use $s{u}^{2m}s=u^{-2m}$ and give $u^{2k}{u}^{2m}=u^{2(k+m)}$, $u^{2k}s\cdot u^{2m}s=u^{2(k-m)}$, $u^{2k}\cdot u^{2m}s=u^{2(k+m)}s$ and $u^{2k}s\cdot u^{2m}=u^{2(k-m)}s$; it contains $s=u^{0}s$ and $u^2$, and each of its elements is a product of $s$ and $u^2$; hence $S_0=⟨s,u^2⟩=H^{\mathrm{\prime }}$. The assignment $s\mapsto 0$, $t\mapsto 1$ sends the relators $s^2$ and $t^2$ to $0$, so by [F5] it defines a homomorphism $\phi \text{\hspace{0.17em}⁣}:W\to \mathbb{Z}\mathrm{/}2$; by [F6], $\phi (u^k)=k<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$ and $\phi (u^{k}s)=k<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$, so the set $H_0:=\{w:\phi (w)=0\}$ equals $\{u^{2k}:k\in \mathbb{Z}\}\cup \{u^{2k}s:k\in \mathbb{Z}\}=S_0=H^{\mathrm{\prime }}$, and since $\phi (t)=1$ the complement of $H_0$ is the single coset $H_{0}t$; hence $H^{\mathrm{\prime }}$ has exactly two cosets in $W$, of which it is one, so it has index $2$. Since $u$ has infinite order by [F3], $W$ is infinite and $H^{\mathrm{\prime }}$, of index $2$, is infinite as well.

$H$ is not standard. By [F1] the standard parabolic subgroups of $S_4$ are $W_J=\{w:S(w)\subseteq J\}$ for the eight subsets $J\subseteq \{s_1,s_2,s_3\}$, and $W_J\cap S=J$ makes distinct subsets give distinct subgroups. Listing by the defining relations: $W_{\mathrm{\varnothing }}=\{1234\}$, $W_{\{s_1\}}=\{1234,2134\}$, $W_{\{s_2\}}=\{1234,1324\}$, $W_{\{s_3\}}=\{1234,1243\}$, $W_{\{s_1,s_2\}}=\{1234,2134,1324,2314,3124,3214\}$, $W_{\{s_2,s_3\}}=\{1234,1324,1243,1342,1423,1432\}$, $W_{\{s_1,s_3\}}=\{1234,2134,1243,2143\}$ and $W_S=S_4$, of orders $1,2,2,2,6,6,4,24$. By step 1.1, $H$ has order $4$ and contains $3214$; the only standard parabolic of order $4$ is $W_{\{s_1,s_3\}}$, whose displayed list does not contain $3214$, and no other $W_J$ has order $4$. Hence $H\mathrm{\ne }{W}_J$ for every $J$, while $H$ is parabolic by step 1.1: the inclusion "standard parabolic $⊊$ parabolic" is strict.

$H^{\mathrm{\prime }}$ is not parabolic. By [F2] the parabolic subgroups of $W$ are the conjugates of $W_{\mathrm{\varnothing }}=\{1\}$, $W_{\{s\}}=\{1,s\}$, $W_{\{t\}}=\{1,t\}$ and $W_{\{s,t\}}=W$. The subgroups $\{1\}$ and $W$ are their own conjugates; the conjugates of $\{1,s\}$ or of $\{1,t\}$ are the subgroups $\{1,r\}$ with $r\in T$, and conversely every $r\in T$ is a conjugate of the simple reflection $s$ or of the simple reflection $t$ by [F4], so the conjugates of these two standard parabolics are exactly the subgroups $\{1,r\}$, $r\in T$. By step 1.2 the group $H^{\mathrm{\prime }}$ contains $s$, and $s\mathrm{\ne }1$ because $s=1$ would force $st=t$ to have order $2$, against [F3]; further $H^{\mathrm{\prime }}$ is infinite and has index $2$ in $W$, so $t\mathrm{\notin }{H}^{\mathrm{\prime }}$ (indeed $\phi (t)=1$) and $H^{\mathrm{\prime }}\mathrm{\ne }W$. Hence $H^{\mathrm{\prime }}$ is none of $\{1\}$, of the two-element subgroups $\{1,r\}$, or of $W$: $H^{\mathrm{\prime }}$ is a reflection subgroup that is not parabolic, and the inclusion "parabolic $⊊$ reflection subgroup" is strict.

The three classes. By definition every standard parabolic subgroup is parabolic, and by [F2] every parabolic subgroup is a reflection subgroup, so the two inclusions of the statement hold; by step 2.1 the first is strict and by step 2.2 the second is strict, and the examples $H$ (parabolic, not standard) and $H^{\mathrm{\prime }}$ (reflection subgroup, not parabolic) exhibit the strictness. This completes the example.

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.