===== recycled on Wed Dec 19 16:24:13 EST 2012 by drorbn on Debian-1207 ====== \def\Proof{{\raisebox{2mm}{\parbox[t]{3.8in}{ {\color{red}Theorem.} $G=M_1$. \hfill {\color{red} \footnotesize $G^{-1}$ is more fun!} \vskip -8mm \[ G\!=\!M_1\!:=\!\left\{ \sigma_{1,j_1}\sigma_{2,j_2}\cdots\sigma_{n,j_n} \colon \forall i, j_i\geq i\text{ and }\sigma_{i,j_i}\in T \right\}. \] \vskip -2mm \par\noindent {\color{red}Proof.} The inclusions $M_1\subset G$ and $\{g_1,\ldots,g_\alpha\}\subset M_1$ are obvious. The rest follows from the following \par\noindent{\color{red}Lemma.} $M_1$ is closed under multiplication. \par\noindent{\color{red}Proof.} By backwards induction. Let \vskip -8mm \[ M_k:=\left\{ \sigma_{k,j_k}\cdots\sigma_{n,j_n} \colon \forall i\geq k, j_i\geq i\text{ and }\sigma_{i,j_i}\in T \right\}. \] \vskip -2mm Clearly $M_nM_n\subset M_n$. Now assume that $M_5M_5\subset M_5$ and show that $M_4M_4\subset M_4$. Start with $\sigma_{8,j}M_4\subset M_4$: \vskip -3mm \[ \sigma_{8,j}(\sigma_{4,j_4}M_5) \overset{1}{=}(\sigma_{8,j}\sigma_{4,j_4})M_5 \overset{2}{\subset}M_4M_5 \] \vskip -2mm \[ \overset{3}{=} \sigma_{4,j_4}(M_5M_5) \overset{4}{\subset}\sigma_{4,j_4}M_5\subset M_4 \] \vskip -1mm (1: associativity, 2: thank the twist, 3: associativity and tracing $i_4$, 4: induction). Now the general case \vskip -4mm \[ (\sigma_{4,j'_4}\sigma_{5,j'_5}\cdots) (\sigma_{4,j_4}\sigma_{5,j_5}\cdots) \] \vskip -1mm falls like a chain of dominos. \hfill {\color{red} \footnotesize Problem Solved!} }}}}