[] (4.700000, 0.000000) arc (180.000000:0.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta^{2}\langle 4, 5 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:2.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:2.000000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:1.250000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 2, 3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:360.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (2.500000, 0.000000) arc (0.000000:-180.000000:1.500000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (0.500000, 0.000000) arc (180.000000:360.000000:0.400000);
[] (1.300000, 0.000000) arc (180.000000:0.000000:0.700000);
[] (2.700000, 0.000000) arc (180.000000:360.000000:1.900000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.850000);
[] (2.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:1.250000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta_{\mathbb{R}I_{2}}^{1/2}\langle 3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:360.000000:1.250000);
[] (2.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (3.500000, 0.000000) arc (0.000000:-180.000000:2.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (0.500000, 0.000000) arc (180.000000:360.000000:0.900000);
[] (2.300000, 0.000000) arc (180.000000:0.000000:0.700000);
[] (3.700000, 0.000000) arc (180.000000:360.000000:1.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.350000);
[] (3.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (4.500000, 0.000000) arc (180.000000:0.000000:0.750000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta \langle 1,2,3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (2.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (3.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (3.700000, 0.000000) arc (180.000000:360.000000:1.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.350000);
[] (3.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (4.500000, 0.000000) arc (180.000000:0.000000:0.750000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta_{I_{2}I_{4}}^{1/2}\langle 3 \rangle $  (6.000000, 0.000000) ;

[] (3.000000, 0.000000) arc (0.000000:-180.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:1.000000);
[] (3.500000, 0.000000) arc (180.000000:360.000000:1.500000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.000000);
[] (4.500000, 0.000000) arc (180.000000:360.000000:0.750000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) $ L_2 $;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) $ L_4 $;
[draw, rectangle, color=red, fill=white] (vert_2) at (2.000000,0.000000) $ C_2 $;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) $ C_1 $;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) $ C_1' $;
[draw, rectangle, color=red, fill=white] (vert_5) at (5.000000,0.000000) $ C_2' $;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) $ L_1 $;

\begin{equation*}\begin{split}
\left[\Gamma_{C_2'}^{-1}\Gamma_{L_1}^{-1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{L_1}\Gamma_{C_2'} , \Gamma_{L_1} \right]=e
\end{split}
\end{equation*}

§.§.§ Vertex Number 15

[scale=1 , xshift=-100]
[] (3.000000, 0.000000) – (4.000000, 0.000000) ;
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 1, 2 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (2.500000, 0.000000) arc (180.000000:0.000000:0.750000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta_{I_{2}\mathbb{R}}^{1/2}\langle 2 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (2.500000, 0.000000) arc (180.000000:0.000000:0.750000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta^{2}\langle 3, 4 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:1.000000);
[] (3.500000, 0.000000) arc (0.000000:180.000000:0.900000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:0.400000);
[] (2.500000, 0.000000) arc (180.000000:0.000000:0.400000);
[] (3.300000, 0.000000) arc (0.000000:-180.000000:0.850000);
[] (1.600000, 0.000000) arc (180.000000:0.000000:1.200000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta \langle 4,5,6 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:2.000000);
[] (5.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (4.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (2.500000, 0.000000) arc (0.000000:180.000000:0.400000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:1.500000);
[] (4.700000, 0.000000) arc (180.000000:0.000000:0.300000);
[] (5.300000, 0.000000) arc (0.000000:-180.000000:1.850000);
[] (1.600000, 0.000000) arc (180.000000:0.000000:0.550000);
[] (2.700000, 0.000000) arc (180.000000:360.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 3, 4 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:2.000000);
[] (5.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (4.500000, 0.000000) arc (0.000000:-180.000000:0.500000);
[] (3.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:1.100000);
[] (4.700000, 0.000000) arc (180.000000:0.000000:0.300000);
[] (5.300000, 0.000000) arc (0.000000:-180.000000:1.500000);
[] (2.300000, 0.000000) arc (180.000000:0.000000:0.850000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 6, 7 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:2.500000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (5.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (3.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:1.600000);
[] (5.700000, 0.000000) arc (180.000000:0.000000:0.300000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:2.000000);
[] (2.300000, 0.000000) arc (180.000000:0.000000:0.850000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta^{2}\langle 4, 5 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:360.000000:2.500000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (5.500000, 0.000000) arc (0.000000:-180.000000:0.500000);
[] (4.500000, 0.000000) arc (0.000000:180.000000:1.000000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:0.400000);
[] (4.300000, 0.000000) arc (180.000000:360.000000:0.700000);
[] (5.700000, 0.000000) arc (180.000000:0.000000:0.300000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:1.150000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta^{2}\langle 1, 2 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:2.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (5.500000, 0.000000) arc (0.000000:-180.000000:0.500000);
[] (4.500000, 0.000000) arc (0.000000:180.000000:1.000000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:0.400000);
[] (4.300000, 0.000000) arc (180.000000:360.000000:0.700000);
[] (5.700000, 0.000000) arc (180.000000:0.000000:0.300000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:1.150000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 5, 6 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:2.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.000000);
[] (4.500000, 0.000000) arc (0.000000:-180.000000:0.500000);
[] (3.500000, 0.000000) arc (0.000000:180.000000:0.500000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:1.100000);
[] (4.700000, 0.000000) arc (180.000000:0.000000:0.800000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta^{2}\langle 4, 5 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (1.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (1.700000, 0.000000) arc (180.000000:360.000000:2.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:2.000000);
[] (2.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:1.400000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta\langle 2, 3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:360.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (2.500000, 0.000000) arc (0.000000:-180.000000:1.500000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (0.500000, 0.000000) arc (180.000000:360.000000:0.400000);
[] (1.300000, 0.000000) arc (180.000000:0.000000:0.700000);
[] (2.700000, 0.000000) arc (180.000000:360.000000:1.900000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.850000);
[] (2.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (3.500000, 0.000000) arc (180.000000:0.000000:1.400000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta_{\mathbb{R}I_{2}}^{1/2}\langle 3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (0.000000, 0.000000) arc (180.000000:360.000000:1.250000);
[] (2.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (3.500000, 0.000000) arc (0.000000:-180.000000:2.000000);
[] (-0.500000, 0.000000) arc (180.000000:0.000000:0.500000);
[] (0.500000, 0.000000) arc (180.000000:360.000000:0.900000);
[] (2.300000, 0.000000) arc (180.000000:0.000000:0.700000);
[] (3.700000, 0.000000) arc (180.000000:360.000000:1.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.350000);
[] (3.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (4.500000, 0.000000) arc (180.000000:0.000000:0.900000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta \langle 1,2,3 \rangle $  (6.000000, 0.000000) ;

[scale=1 , xshift=-100]
[] (2.000000, 0.000000) arc (180.000000:0.000000:0.750000);
[] (3.500000, 0.000000) arc (0.000000:-180.000000:1.000000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:1.100000);
[] (3.700000, 0.000000) arc (180.000000:360.000000:1.400000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.350000);
[] (3.800000, 0.000000) arc (180.000000:360.000000:0.350000);
[] (4.500000, 0.000000) arc (180.000000:0.000000:0.900000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:0.650000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_2) at (2.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) ;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_5) at (5.000000,0.000000) ;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) ;

[scale=1 , xshift=100]
[->] (0.000000, 0.000000) – node [above, midway] $ \Delta_{I_{2}I_{4}}^{1/2}\langle 3 \rangle $  (6.000000, 0.000000) ;

[] (3.000000, 0.000000) arc (0.000000:-180.000000:0.750000);
[] (1.500000, 0.000000) arc (180.000000:0.000000:1.000000);
[] (3.500000, 0.000000) arc (180.000000:360.000000:1.500000);
[] (6.500000, 0.000000) arc (0.000000:180.000000:1.000000);
[] (4.500000, 0.000000) arc (180.000000:360.000000:0.500000);
[] (5.500000, 0.000000) arc (180.000000:0.000000:0.400000);
[] (6.300000, 0.000000) arc (0.000000:-180.000000:1.150000);
[draw, circle, color=black, fill=white] (vert_0) at (0.000000,0.000000) $ L_2 $;
[draw, circle, color=black, fill=white] (vert_1) at (1.000000,0.000000) $ L_4 $;
[draw, rectangle, color=red, fill=white] (vert_2) at (2.000000,0.000000) $ C_2 $;
[draw, rectangle, color=red, fill=white] (vert_3) at (3.000000,0.000000) $ C_1 $;
[draw, rectangle, color=red, fill=white] (vert_4) at (4.000000,0.000000) $ C_1' $;
[draw, rectangle, color=red, fill=white] (vert_5) at (5.000000,0.000000) $ C_2' $;
[draw, circle, color=black, fill=white] (vert_6) at (6.000000,0.000000) $ L_1 $;

\begin{equation*}\begin{split}
\Gamma_{L_1}\Gamma_{C_2'}^{-1}\Gamma_{L_1}^{-1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{L_1}\Gamma_{C_2'}\Gamma_{L_1}^{-1}  = \Gamma_{C_1'}
\end{split}
\end{equation*}

§.§.§ Raw relations and simplifications

\begin{equation}\label{arrangement2_auto_calc_vert_1_rel_1}
\begin{split}
\Gamma_{C_1}  = \Gamma_{C_1'} ,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_2_rel_1}
\begin{split}
\left[\Gamma_{C_2}^{-1}\Gamma_{C_1}\Gamma_{C_2} , \Gamma_{L_4}\Gamma_{L_2}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_2_rel_2}
\begin{split}
\left[\Gamma_{L_4} , \Gamma_{L_2}\Gamma_{C_2}^{-1}\Gamma_{C_1}\Gamma_{C_2}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_4_rel_1}
\begin{split}
\Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}^{-1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}^{-1}\Gamma_{L_4}^{-1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_1'}^{-1}  = \Gamma_{C_2'} ,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_5_rel_1}
\begin{split}
\left[\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1} , \Gamma_{C_2}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_6_rel_1}
\begin{split}
\left\{\Gamma_{L_4}^{-1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_1'}^{-1}\Gamma_{C_2'}\Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4} , \Gamma_{L_2}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_7_rel_1}
\begin{split}
\left[\Gamma_{C_2'}\Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}^{-1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_1'}^{-1}\Gamma_{C_2'}^{-1} , \Gamma_{C_1'}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_8_rel_1}
\begin{split}
\left\{\Gamma_{C_1} , \Gamma_{C_2}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_9_rel_1}
\begin{split}
\left\{\Gamma_{C_2'} , \Gamma_{C_1'}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_10_rel_1}
\begin{split}
\left[\Gamma_{C_2'}\Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}^{-1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_1'}^{-1}\Gamma_{C_2'}^{-1} , \Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_11_rel_1}
\begin{split}
\left[\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1} , \Gamma_{C_2'}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_12_rel_1}
\begin{split}
\left[\Gamma_{C_2'}^{-1}\Gamma_{L_1}\Gamma_{C_2'} , \Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_12_rel_2}
\begin{split}
\left[\Gamma_{C_1'} , \Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{C_2'}^{-1}\Gamma_{L_1}\Gamma_{C_2'}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_13_rel_1}
\begin{split}
\left\{\Gamma_{C_2'} , \Gamma_{L_1}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_14_rel_1}
\begin{split}
\Gamma_{L_1}^{-1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}^{-1}\Gamma_{L_1}  = \Gamma_{C_2'} ,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_15_rel_1}
\begin{split}
\left[\Gamma_{C_2'}^{-1}\Gamma_{L_1}^{-1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{L_1}\Gamma_{C_2'} , \Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_vert_16_rel_1}
\begin{split}
\Gamma_{L_1}\Gamma_{C_2'}^{-1}\Gamma_{L_1}^{-1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}^{-1}\Gamma_{C_1}^{-1}\Gamma_{L_1}\Gamma_{C_2'}\Gamma_{L_1}^{-1}  = \Gamma_{C_1'} ,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_auto_calc_projective_rel}
\begin{split}
\Gamma_{L_1}\Gamma_{C_2'}\Gamma_{C_1'}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2} =e.
\end{split}
\end{equation}

Here are the computations for group $ G_6 := \pcpt{\mathcal{B}_6} / \langle \Gamma_{X}^2 \; | \; X\subseteq \mathcal{B}_6 \rangle $:

* Group $ G_6 := \pcpt{\mathcal{B}_6} / \langle \Gamma_{X}^2 \; | \; X\subseteq \mathcal{B}_6 \rangle $ is generated by $ \Gamma_{C_1},\Gamma_{C_2}, \Gamma_{L_1} $, and $ \Gamma_{L_2} $, subject to the following relations:

\begin{equation}\label{arrangement2_second_final_1}
\Gamma_{C_1}^2 = \Gamma_{C_2}^2 = \Gamma_{L_1}^2 = \Gamma_{L_2}^2 = e,
\end{equation}

\begin{equation}\label{arrangement2_second_final_2}
[\Gamma_{L_1}, \Gamma_{L_2} ] = [\Gamma_{C_1}, \Gamma_{L_1}] = [\Gamma_{C_1}, \Gamma_{L_2}] = [\Gamma_{L_1}\Gamma_{L_2}, \Gamma_{C_2}] = e,
\end{equation}

\begin{equation}\label{arrangement2_second_final_3}
\{ \Gamma_{C_1}, \Gamma_{C_2} \} = \{ \Gamma_{C_2}, \Gamma_{L_1} \} = \{ \Gamma_{C_2}, \Gamma_{L_2} \} = e,
\end{equation}

\begin{equation}\label{arrangement2_second_final_4}
[\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}] = e.
\end{equation}

* Group $ G_6 $ is isomorphic to the direct product $ G_6'\times \mbb{Z}_2 $, where $ G_6' $ is generated by $ \Gamma_{C_1},\Gamma_{C_2} $, and $ \Gamma_{L_1} $, subject to the following relations:

\begin{equation}\label{arrangement2_without_L2_1}
\Gamma_{C_1}^2 = \Gamma_{C_2}^2 = \Gamma_{L_1}^2 = e,
\end{equation}

\begin{equation}\label{arrangement2_without_L2_2}
[\Gamma_{C_1}, \Gamma_{L_1}] = e,
\end{equation}

\begin{equation}\label{arrangement2_without_L2_3}
\{ \Gamma_{C_1}, \Gamma_{C_2} \} = \{ \Gamma_{C_2}, \Gamma_{L_1} \} = e,
\end{equation}

\begin{equation}\label{arrangement2_without_L2_4}
[\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}] = e.
\end{equation}

* The element $ \Gamma_{L_4} \in \pcpt{\mathcal{B}_6} $ is  not contained in the center of $ \pcpt{\mathcal{B}_6} $.

Substituting $ \Gamma_{C_1'}=\Gamma_{C_1} $ and $ \Gamma_{C_2'} = \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} $ we get that $ G_6 $ is generated by $ \Gamma_{C_1}, \Gamma_{C_2}, \Gamma_{L_1},\Gamma_{L_2} $ and $ \Gamma_{L_4} $ subject to:

\begin{equation}\label{arrangement2_simpl_1}
\begin{split}
(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})  \Gamma_{L_4}\Gamma_{L_2} = \Gamma_{L_4}\Gamma_{L_2}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}) =\Gamma_{L_2}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4},
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_3}
\begin{split} \Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}  = \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} ,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_4}
\begin{split}
\left[\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1} , \Gamma_{C_2}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_5}
\begin{split} \left\{\Gamma_{L_4}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4} , \Gamma_{L_2}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_6}
\begin{split} \left[\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{C_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_7}
\begin{split}
\left\{\Gamma_{C_1} , \Gamma_{C_2}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_8}
\begin{split}
\left\{\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{C_1}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_9}
\begin{split} \left[\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_10}
\begin{split}
\left[\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1} , \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_11}
\begin{split} \left[\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_12}
\begin{split}
\left[\Gamma_{C_1} , \Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_13}
\begin{split}
\left\{\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{L_1}  \right\}=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_14}
\begin{split}
\left[\Gamma_{L_1}\Gamma_{C_1}\Gamma_{L_1} , \Gamma_{L_1}  \right]=e,
\end{split}
\end{equation}

\begin{equation}\label{arrangement2_simpl_15}
\begin{split}
\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2} =e.
\end{split}
\end{equation}

From (<ref>) we get that $ \Gamma_{C_1} $ and $ \Gamma_{L_1} $ commute, substituting this into (<ref>) we get that $ \{ \Gamma_{C_2}, \Gamma_{L_1}\}=e $.
Using those relations in (<ref>) we get that

\begin{align*}
e & = [ \Gamma_{L_4}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}
\Gamma_{C_2} ] = \\
& = [ \Gamma_{L_4}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}
\Gamma_{C_2} ] = [ \Gamma_{L_4}, \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} ].
\end{align*}

Now, from (<ref>), using (<ref>), we get

\[ e = [\Gamma_{L_4}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}] = [\Gamma_{L_4}, \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}].  \]

Using the above, we can simplify (<ref>) as follows:

\begin{align*} \Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2} & = \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} \\ \Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2} & = \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} \\ \Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1} & = \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2} \\ \Gamma_{L_4}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{L_4} & = \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\\
\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2} = \Gamma_{L_4}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{L_4} & = \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} \\
[\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{L_2}] &= e,
\end{align*}

where we first  used $ \{\Gamma_{C_1}, \Gamma_{C_2} \}=e $ and $ [\Gamma_{L_4}, \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}]=e $; then $ [\Gamma_{C_1}, \Gamma_{L_1}]=e $ and $ \{\Gamma_{C_2}, \Gamma_{L_1}\}=e $; next we applied (<ref>); and finally used $ \{\Gamma_{C_1},\Gamma_{C_2}\}=e $ followed by $ [\Gamma_{L_4}, \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}]=e $.

We now turn to (<ref>). By using derived commutation relations and rearranging we get that it is equivalent to

\[ \left[\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1} , \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}  \right]=e. \]

By applying $ \{ \Gamma_{C_2}, \Gamma_{L_1}\} = e $ and rearranging again,

\[ \left[\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} , \Gamma_{L_1}  \right]=e \]

which, by (<ref>), is equivalent to $ [\Gamma_{L_1}, \Gamma_{L_2}]=e $.

Using $ [\Gamma_{C_1}, \Gamma_{L_1}]=e $ and $ \{\Gamma_{C_2}, \Gamma_{L_1}\}=e $ we get that (<ref>) is equivalent to

\[ \{ \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} \} = e \]

which is, by (<ref>) and $ [\Gamma_{L_1}, \Gamma_{L_2}]=e $, equivalent to $ \{ \Gamma_{C_2}, \Gamma_{L_2}\}=e $.

Next we deal with (<ref>). Applying derived commutations, $ \{ \Gamma_{C_1}, \Gamma_{C_2} \}=e $ and rearranging, we get

\[ [\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}] = e. \]

Using $ \{\Gamma_{C_2}, \Gamma_{L_1} \} =e $, $ [\Gamma_{C_1}, \Gamma_{L_1}]=e $ and $ {[\Gamma_{L_4}, \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}] = e} $ yields

\[ [\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{L_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{L_4}\Gamma_{C_2}, \Gamma_{C_1}] = e. \]

Applying $ [\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{L_2}]=e $ and rearranging once more, gives us

\[ [\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_2}, \Gamma_{L_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2} ] \]

which then, by using (<ref>), becomes $ [\Gamma_{L_2}, \Gamma_{C_1}] = e $.

From (<ref>) we get, by conjugating with $ \Gamma_{C_2}\Gamma_{C_1} $ the following:

\begin{align*}
e & = [\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}] = \\
& = [\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}] = \\
& = [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}],
\end{align*}

and from (<ref>) we get:

\begin{align*}
e & =
\left[\Gamma_{C_1} , \Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}  \right] = \\
& =
\left[\Gamma_{C_1} , \Gamma_{C_2}\Gamma_{L_4}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}  \right] = \\
& =
\left[\Gamma_{C_1} , \Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_1}\Gamma_{C_2}  \right].
\end{align*}

Together those relations can be written as

\[ \Gamma_{L_1}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4} = (\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4}\Gamma_{L_1} = \Gamma_{L_4}\Gamma_{L_1}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}). \]

Summarizing those simplifications we get that $ G_6 $ is generated by $ \Gamma_{C_1},\Gamma_{C_2}, \Gamma_{L_1}, \Gamma_{L_2} $, and $ \Gamma_{L_4} $, subject to the following relations:

\begin{equation}\label{arrangement2_final_1}
\Gamma_{C_1}^2 = \Gamma_{C_2}^2 = \Gamma_{L_1}^2 = \Gamma_{L_2}^2 = \Gamma_{L_4}^2 =e,
\end{equation}

\begin{equation}\label{arrangement2_final_2}
[\Gamma_{L_1}, \Gamma_{L_2} ] = [\Gamma_{C_1}, \Gamma_{L_1}] = [\Gamma_{C_1}, \Gamma_{L_2}] = e,
\end{equation}

\begin{equation}\label{arrangement2_final_3}
\{ \Gamma_{C_1}, \Gamma_{C_2} \} = \{ \Gamma_{C_2}, \Gamma_{L_1} \} = \{ \Gamma_{C_2}, \Gamma_{L_2} \} = e,
\end{equation}

\begin{equation}\label{arrangement2_final_4}
\Gamma_{L_1}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4} = (\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4}\Gamma_{L_1} = \Gamma_{L_4}\Gamma_{L_1}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}),
\end{equation}

\begin{equation}\label{arrangement2_final_5}
\Gamma_{L_2}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4} = (\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2})\Gamma_{L_4}\Gamma_{L_2} = \Gamma_{L_4}\Gamma_{L_2}(\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}),
\end{equation}

\begin{equation}\label{arrangement2_final_6}
[\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{L_2}] = [\Gamma_{L_4}, \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}] = [\Gamma_{L_4}, \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}] = e,
\end{equation}

\begin{equation}\label{arrangement2_final_7}
\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_4}\Gamma_{L_2} =e.
\end{equation}

Note that because $ [ \Gamma_{C_2}, \Gamma_{L_1}\Gamma_{L_2} ] = e $, we have

\begin{equation}\label{arrangement2_second_L1_L2}
\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2} = \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1},
\end{equation}

\begin{equation}
\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_1} = \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_2}.
\end{equation}

We now use (<ref>) to eliminate $ \Gamma_{L_4} $. To that end, write

$$ \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2} = \Gamma_{L_4} = \Gamma_{L_4}^{-1} = \Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}, $$

and substitute it in relations (<ref>)-(<ref>).

First, using (<ref>), $ \Gamma_{L_4}^2 = e $ is equivalent to

\begin{equation}\label{arrangement2_second1}
\begin{split}
e = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2} &= \Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2} = \\
= \Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2} &= \Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2} = \\
&= [ \Gamma_{C_2}, \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}].
\end{split}
\end{equation}

Next we deal with (<ref>). After the substitution, the first part reads

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2} = \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_1}
\end{equation*}

and because $ \Gamma_{L_1} $ commutes with both $ \Gamma_{C_1} $ and $ \Gamma_{L_2} $, it is equivalent to $ \{ \Gamma_{C_2}, \Gamma_{L_1} \} = e $, which already appears in  (<ref>).
The second part of (<ref>) reads

\begin{equation*}
\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_1} = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}.
\end{equation*}

Multiplying by $ \Gamma_{C_1}\Gamma_{C_2} $ from the right and using the fact that both $ \Gamma_{C_1} $ and $\Gamma_{C_2}$ commute with $ \Gamma_{L_2}\Gamma_{L_1} $, $ [\Gamma_{L_1},\Gamma_{L_2}]=e $ and $ \{ \Gamma_{C_1},\Gamma_{C_2} \}=e $, we get

\begin{equation*}
\Gamma_{C_2}\Gamma_{L_2} = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}.
\end{equation*}

This equation is equivalent to

\begin{equation}\label{arrangement2_second2}
\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_2} = \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}.
\end{equation}

Using (<ref>) and  (<ref>) we see that

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} = \Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2} = \Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{C_2}
\end{equation*}

so, the left side of (<ref>) is equal to $\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}$.
Using (<ref>) and $ [\Gamma_{C_1}, \Gamma_{L_1} ] =e $, we get that the right side of (<ref>) is equal to

\begin{align*} \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} &	= \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} = \\
= \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}	&= \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1},
\end{align*}

meaning that (<ref>) reads

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} = \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1},
\end{equation*}

which is equivalent to (<ref>).

We now turn our attention to (<ref>).
The first part reads

\begin{equation*}
\Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2} = \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}.
\end{equation*}

Because $ \Gamma_{C_1} $ commutes with $ \Gamma_{L_2} $ it is equivalent to $ [\Gamma_{L_2}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}]=e $. We assert that this relation follows from existing ones. Indeed, because $ [\Gamma_{L_1}\Gamma_{L_2}, \Gamma_{C_2}]=e $ and $ [\Gamma_{L_1}\Gamma_{L_2}, \Gamma_{L_1}]=e $, we have $ {[\Gamma_{L_1}\Gamma_{L_2}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}]=e} $, and the relation $ \{ \Gamma_{L_1}, \Gamma_{C_2}\}=e $ is equivalent to $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}] = e $, which together give us the desired $ [\Gamma_{L_2}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}] = e $.

The second part of (<ref>) becomes after the substitution

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1} = \Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}.
\end{equation*}

Obviously $ \Gamma_{L_1} $ cancels out and we get $ \{\Gamma_{C_1}, \Gamma_{C_2}\}=e $.

We are left with the two commutation relations in (<ref>) that contain $ \Gamma_{L_4} $.
First, using $ \{ \Gamma_{C_1}, \Gamma_{C_2} \} = \{ \Gamma_{L_1}, \Gamma_{C_2} \} = e $ and $ [\Gamma_{C_1}, \Gamma_{L_2} ] =e $ (in this order), we get

\begin{equation*}
\begin{split}
[\Gamma_{L_4}, \Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}] & = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1} = \\
& = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1} = \\
& = \Gamma_{L_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1} = \\
& = \Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{L_2}\Gamma_{C_2}\Gamma_{C_1},
\end{split}
\end{equation*}

which is trivial by $ [\Gamma_{L_1}\Gamma_{L_2}, \Gamma_{C_2}] = e $.

\begin{equation*}
\begin{split}
[\Gamma_{L_4}, \Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}] &= \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} = \\
&=  \Gamma_{L_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1} = \\
&=  \Gamma_{L_2}\Gamma_{C_2}\Gamma_{L_2}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{L_1},
\end{split}
\end{equation*}

which is redundant because $ \Gamma_{C_2} $ commutes with $ \Gamma_{L_1}\Gamma_{L_2} $. This finishes the proof of the first assertion in the lemma.

To prove item $ (2) $ of the lemma, define $ \additionalgen = \Gamma_{L_1}\Gamma_{L_2} \in G_6 $.
Because $ \Gamma_{L_2} = \Gamma_{L_1}\additionalgen $, group $ G_6 $ is generated by $ \Gamma_{C_1},\Gamma_{C_2},\Gamma_{L_1},\additionalgen $, we compute the relations on those generators.
By $ [\Gamma_{L_1}, \Gamma_{L_2}] = e $ we immediately get that $ \additionalgen $ and $ \Gamma_{L_1} $ commute.
Thus, $ \Gamma_{L_2}^2 = e $ becomes $ \additionalgen^2 = e $.
Because $ \Gamma_{C_1} $ commutes with both $ \Gamma_{L_1} $ and $ \Gamma_{L_2} $, it commutes with $ \additionalgen $ as well.
We get that $ \additionalgen $ is in the center of $ G_6 $ and so the relation $ \{ \Gamma_{C_2}, \Gamma_{L_2} \} = e $ follows from $ \{ \Gamma_{C_2}, \Gamma_{L_1} \} =e $.
The above discussion implies that the generators $ \Gamma_{C_1}, \Gamma_{C_2} $, and $ \Gamma_{L_1} $ satisfy exactly the relations among (<ref>)-(<ref>) that do not involve $ \Gamma_{L_2} $, which is precisely the assertion of item $ (2) $ of the lemma.

Now, consider group $G_6' = G_6/\langle\Gamma_{\additionalgen} \rangle $.
The group generated by $ \Gamma_{C_1},\Gamma_{C_2} $, and $ \Gamma_{L_1} $, subject to the relations $ [\Gamma_{C_1},\Gamma_{L_1}] = \{ \Gamma_{C_1}, \Gamma_{C_2} \} = \{ \Gamma_{C_2}, \Gamma_{L_1} \} = e $, is the affine Coxeter group of type $ \widetilde{C}_2 $.
Thus, $ G_6' $ is its quotient by the relation $ [\Gamma_{C_2}, \Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}]=e $.
The generators are mapped to this representation of $ \widetilde{C}_2 $ as follows:

\begin{equation*}
\Gamma_{C_1} \mapsto \begin{pmatrix}
1 & 2 \\ \overline{1} & 2
\end{pmatrix} \; ; \; \Gamma_{C_2} \mapsto \begin{pmatrix}
1 & 2 \\ 2 & 1
\end{pmatrix} \; ; \; \Gamma_{L_1} \mapsto \begin{pmatrix}
1 & 2 \\ 1 & \overline{2}^1
\end{pmatrix}
\end{equation*}

where an over line denotes $ \epsilon=-1 $ and we omit positive signs and zero exponents.

We now consider the additional relation, by computing it explicitly in $ \widetilde{C}_2 $, as follows:

\begin{equation*}
\Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1} \mapsto \begin{pmatrix}
1 & 2 \\ 1^{-1} & 2^1
\end{pmatrix}
\end{equation*}

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2} \mapsto \begin{pmatrix}
1 & 2 \\ 1^1 & 2^{-1}
\end{pmatrix}.
\end{equation*}

This means that the elements of $ G_6' $ are signed permutations, and we identify two signed permutations if the underlying regular permutation (in $ S_2 $) is the same, they have the same signs, the same sum of exponents, and the same parity of exponents.

We are now ready to prove that $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}] \ne e $.
For that, compare the images of $ \Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}
\Gamma_{C_2} $ and $ \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1} $ in $ G_6' $:

\begin{equation*}
\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} \mapsto \begin{pmatrix}
1 & 2 \\ 1 & \overline{2}^1
\end{pmatrix}\cdot \begin{pmatrix}
1 & 2 \\ 1 & \overline{2}
\end{pmatrix} = \begin{pmatrix}
1 & 2 \\ 1 & 2^{-1}
\end{pmatrix}
\end{equation*}

\begin{equation*}
\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1} = \begin{pmatrix}
1 & 2 \\ 1 & \overline{2}
\end{pmatrix} \cdot \begin{pmatrix}
1 & 2 \\ 1 & \overline{2}^1
\end{pmatrix} = \begin{pmatrix}
1 & 2 \\ 1 & 2^1
\end{pmatrix}.
\end{equation*}

Because the sums of exponents are different, the images of $ \Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2} $ and $ \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}\Gamma_{L_1} $ in

$$ G_6' \cong \left. \widetilde{C}_2 \middle/ \left\langle \begin{pmatrix}
1 & 2 \\ 1^2 & 2^{-2}
\end{pmatrix} \right\rangle \right. $$

are different, meaning that $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}] \ne e $ in $ G_6 $.

Finally, we deduce that $ L_4 $ does not commute with $ L_1 $ in $ G_6 $ and thus, it is not contained in the center of $ G_6 $, meaning that it is not in the center of $ \pi_1(\mathbb{CP}^2 - \mathcal{B}_6) $, as well.
Indeed, because $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{C_2}] \ne e $ and $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{L_1}\Gamma_{C_2}] = e $ we get that $ [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}] \ne e $, that is,

$$ [\Gamma_{L_1}, \Gamma_{L_4} ] = [\Gamma_{L_1}, \Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_1}\Gamma_{C_2}\Gamma_{C_1}\Gamma_{L_2}] \ne e. $$

§.§ $\mathcal{B}_5$ and $\mathcal{B}_6$ are Zariski pair with non-isomorphic groups

$ \left(\mathcal{B}_5, \mathcal{B}_6\right) $ is a Zariski pair with non-isomorphic fundamental groups.

It is a direct and easy verification that $ \mathcal{B}_5 $ and $ \mathcal{B}_6 $ has the same combinatorics. Assume, towards a contradiction, that $ \varphi: \left(\mathbb{CP}^2, \mathcal{B}_5\right) \rightarrow \left(\mathbb{CP}^2, \mathcal{B}_6\right) $ is an homeomorphism. Then $ \varphi\left(\left\{P_1,P_3\right\}\right)=\left\{P_1, P_4\right\} $ so $ \varphi\left(L_3\cup C_1\right)=L_4\cup C_1 $ meaning that $ \varphi\left(C_1\right)=C_1 $ and $ \varphi\left( L_3 \right) = L_4  $.

Now, consider the induced map on fundamental groups $ \varphi_* : \pcpt{\mathcal{B}_5} \to \pcpt{\mathcal{B}_6} $.
It must send a loop around $ L_3 $ to a loop around $ L_4 $, meaning that $ \varphi_*(\Gamma_{L_3}) $ is conjugate either to $ L_4 $ or to $ L_4^{-1} $.
Moreover, $ \varphi_* $ sends squares of the generators to squares of the generators, that is, it induces isomorphism $
\varphi_*:G_5\to G_6 $ and this isomorphism sends $ \Gamma_{L_3} $ to a conjugate of $ \Gamma_{L_4} $.
Now, because $ L_3 $ is in the center of $ G_5 $ by Lemma <ref>, $ \varphi_*(L_3) $ must be in the center of $ G_6 $ and thus, also $ \Gamma_{L_4} $, in contradiction to Lemma <ref>.


# Solving Atari Games Using Fractals And Entropy

Sergio Hernandez Cerezo
HCSoft Programación, S.L., 30007 Murcia, Spain
<EMAIL_ADDRESS>
Guillem Duran Ballester
<EMAIL_ADDRESS>
Spiros Baxevanakis
<EMAIL_ADDRESS>
###### Abstract

In this paper we introduce a novel MCTS based approach that is derived from
the laws of the thermodynamics. The algorithm, coined Fractal Monte Carlo
(FMC), allows us to create an agent that takes intelligent actions in both
continuous and discrete environments while providing control over every aspect
of the agent’s behavior. Results show that FMC is several orders of magnitude
more efficient than similar techniques, such as MCTS, in the Atari games
tested.

##  1 Introduction

Artificial intelligence methods are currently limited by the lack of a
concrete definition of intelligence that would assist in creating agents that
exhibit intelligent behavior. Fractal AI theory (FAI) (Hernández et al., 2018)
is inspired by the work of Wissner-Gross and Freer who proposed the concept of
Causal Entropic Forces and showed that an agent exhibits intelligent behavior
when it tries to maximize its Causal Path Entropy or equivalently, maximize
its future freedom of action. In order to achieve that, the agent directly
modifies its degrees of freedom in such a way that it assumes the state with
the highest number of possible, different futures. To find the available
futures, the agent will need to scan the action space and recreate the Causal
Cone that contains the paths of all possible future internal configurations
that start from its initial state. This scanning process is called a Scanning
Policy. Available actions are then assigned a probability of being chosen by
the Deciding Policy. In Fractal AI, intelligence is defined as the ability to
minimize a sub-optimallity coefficient based on the similitude of two
probability distributions created from the scanning and decision policies.

Using the principles described in FAI theory we developed a Monte Carlo
approach coined Fractal Monte Carlo (FMC) that is based on the second law of
thermodynamics. The algorithm develops a swarm of walkers that evolve in an
environment while balancing exploitation and exploration by means of a
mechanism named cloning. This process generates a "fractal tree" that will
tend to fill up all the causal cone, from the interconnected paths of the
walkers. The algorithm can be applied to both continuous and discrete decision
spaces while remaining extremely efficient.

In addition, FMC provides a suite of parameters that control computational
resources as well as agent reaction time. To test our algorithm we put it up
against 55 different Atari environments and compared our results to state of
the art algorithms like A3C (Mnih et al., 2016), NoisyNet (Fortunato et al.,
2017), DQN (Mnih et al., 2013, 2015) and variants.

##  2 Related work

Recently there have been numerous breakthroughs in reinforcement learning most
of them originating from Deepmind. They created an end to end model free
reinforcement learning technique, named Deep Q Learning (Mnih et al., 2013,
2015), that scored astonishingly well and outperformed previous approaches in
Atari games. Their Deep Q Learner achieves such feats by estimating Q-values
directly from images and stabilizes learning by means such as experience
replay and frame skipping. Later in 2016 they created AlphaGo which went to
beat the world champion in the game of Go (Silver et al., 2016) and AlphaChem
which was shown to outperform hardcoded heuristics used in retrosynthesis
(Segler et al., 2017). Both AlphaGo and AlphaChem use some form of deep
reinforcement learning in conjunction with a MCTS variant, UCP (Isasi et al.,
2014).

##  3 Background

FMC is a robust path-search algorithm that efficiently approximates path
integrals formulated as a Markov decision process by exploiting the deep link
between intelligence and entropy maximization (Wissner-Gross and Freer, 2013)
that naturally produces an optimal decision-making process. FMC formulates
agents that exhibit intelligent behavior in Atari game emulators. Such agents
create a swarm of walkers that explores the Causal Cone and eventually, when
the time horizon is met, select an action based on the walkers’ distribution
over the action space.

###  3.1 Causal cones

In order to find the best path, Fractal Monte Carlo scans the space of
possible future states thereby constructing a tree which consists of potential
trajectories that describe the future evolution of the system. We define a
Causal Cone $X(x_{0},\tau)$ as the set of all possible paths the system can
take starting from an initial state $x_{0}$ if allowed to evolve over a time
interval of length $\tau$, the ’time horizon’ of the cone. A Causal Cone can
be divided into a set of Causal Slices defined as $X_{H}(x_{0},t)$, where each
Causal Slice contains all the possible future states of the paths at a given
time $t$. If $t=\tau$ the Causal Slice is called the cone’s ‘horizon’ and
contains the final states. The rest of the cone, where $t<\tau$, is usually
referred to as the cone’s ‘bulk’.

Figure 1: (from Wissner-Gross and Freer (2013)): A Causal Cone visualization.
On the left, the Causal Cone expands from the initial $x(0)$ point and all the
possible future paths are expanded upwards in time. On the right the entropic
force compels the walkers to avoid the grey excluded volume and therefore
evolve on the remaining space.

###  3.2 Dead and alive statuses

The death condition is a flag set by the programmer that lets us incorporate
arbitrary boundary conditions to the behaviour of the agent and helps the
swarm avoid undesired regions of the state space. We will assume an external
death condition is defined over $E$ so a portion of the space can be forbidden
for the system, as shown in Figure 1 . We will consider a state inside this
excluded region "dead" while all other states are "alive".

###  3.3 Reward function

Agents make decisions based on a non-negative reward function that (we assume)
is defined over the state space. For every slice $X_{H}(x_{0},t)$ of the
causal cone, we can calculate the total reward $R_{TOT}(x_{0},t)$ of the slice
as the integral of the reward over the slice. We may then convert the reward
into a probability density $P_{R}$ over the slice as follows:

$P_{R}(x\mid x_{0},t)=R(x)/R_{TOT}(x_{0},t)$ (1)

The general idea behind the algorithm will be that, the density distribution
of the scanning should match the reward density distribution of the state
space.

###  3.4 Policies

The proposed algorithm uses a set of two policies to choose and score actions.
First we define a scanning policy $\pi_{s}$ that, given a swarm of initially
identical states, defines its possible evolution over time as a stochastic
process.
After the scanning is finished we need a deciding policy $\pi_{D}$ that will
assign a probability of being chosen to each action:

$\pi_{D}:A\rightarrow[0,1]$ (2)

In order to measure how different two probability distributions are we will
use a modified version of the Kullback-Leibler divergence:

$D_{H}(P\mid\mid Q)=log(\Pi(2-p_{i}^{p_{i}})/\Pi(2-q_{i}^{p_{i}}))$ (3)

This divergence is well defined for any possible distributions $p$ and $q$,
including the problematic case when $(p_{i}>0,q_{i}=0).$

##  4 Fractal Monte Carlo

Fractal Monte Carlo is a path-search algorithm derived from Fractal AI theory
(Hernández et al., 2018) that produces intelligent behavior by maximizing an
intrinsic reward represented as Causal Path Entropy (Wissner-Gross and Freer,
2013). When making a decision, Fractal Monte Carlo (FMC) establishes a tree
that describes the future evolution of the system. This tree is expanded by a
swarm of walkers that populates its leaf nodes. The swarm will undergo an
interactive cellural automaton like process in order to make the tree grow
efficiently. When a maximum amount of computation has been reached, the
utility of each action will be considered proportional to the number of
walkers that populate leaf nodes originating from the same action.

###  4.1 The algorithm

FMC steps are outlined below:

STEP 1: Initialize the walkers to the root state.

STEP 2: _Perturb_ the swarm of walkers.

STEP 3: _Evaluate_ the position of each walker with respect to the whole
swarm.

STEP 4: _Recycle_ the walkers that are in a dead state or have been poorly
valued against the rest of the swarm.

STEP 5: Repeat phases 2-4 until we reach maximum computational resources.

STEP 6: _Choose_ the action with the highest utility.

In more detail:

Perturb:

  1. 1.

Every walker chooses a random action and acts in the environment.

Evaluate:

  1. 1.

For every walker $A$ select an alive walker $B$ at random and measure the
euclidean distance $D_{i}$ between their observations.

  2. 2.

Normalize distances and rewards using the "Relativize" function

  3. 3.

Calculate the virtual reward of each walker. We define virtual reward $VR$ at
a state $W_{i}$ as:

$VR_{i}=R_{i}*D_{i}$ (4)

Where $R_{i}$ is the reward value at state $W_{i}$. Virtual reward is a
stochastic measure of the importance of a given walker in respect to the whole
swarm.

Recycle:

  1. 1.

Each walker A is compared to another randomly selected walker C and gets
assigned a probability of cloning to the leaf node of walker C.

  2. 2.

Determine if the walker A will clone to C based on the cloning probability and
the death condition.

  3. 3.

Finally transfer the walkers that are set for cloning to their target leaf
node.

Choose:

  1. 1.

When assigning a utility value to an option, FMC counts how many walkers took
each option at the root state. To choose an action in the continuous or
general case we calculate the average of the actions weighted by their
normalized utilities or scores. In the discrete case, the action that
approximates better the aforementioned average is chosen.

Relativize :

  1. 1.

Given $R$ real values we calculate the mean $R_{m}$ and the standard deviation
$R_{d}$.

  2. 2.

Normalize the values using:

$R_{i}^{\prime}=(R_{i}-R_{m})/R_{d}$ (5)

  3. 3.

Reshape the values into a Gaussian N(0,1) distribution.

  4. 4.

Then scale using: if $R_{i}^{\prime}<0$ then
$R_{i}^{\prime}=\exp(R_{i}^{\prime})$ else
$R_{i}^{\prime}=1+ln(R_{i}^{\prime})$

###  4.2 Parameters

Time Horizon sets a limit for how far in the future the walkers of the Swarm
will foresee the aftermath of their initial actions. In other words, the
walkers will seek to meet their set Time Horizon when going deeper in the tree
but never go past it. The ideal Time Horizon value allows an agent to see far
enough in the future to detect which actions lead inevitably to death.

Max samples is an upper bound on computational resources. It limits the number
of times that FMC can make a perturbation to build a causal cone. The
algorithm will try to keep computational resource usage as low as it can
providing it meets the time horizon. A good guide to setting this parameter is
$Number\\_of\\_walkers*time\\_horizon*RepeatActions$, with $RepeatActions=5$ a
number that works well in Atari games but highly depends on the task.

Number of walkers represents the maximum number of paths that FMC will
simulate. This number is related to "how thick" we want the resulting
representation of the causal cone to be. The algorithm will try to use the
maximum number of walkers possible.

Time step (dt) is the time interval the agents keep each decision made. Time
horizon / dt will define the number of steps to be taken by walkers.

###  4.3 Time complexity

Computational time complexity of the algorithm can be shown to be of $O(n)$
with $n=number\\_of\\_steps*number\\_of\\_walkers$.

##  5 Findings

In this section we present our performance results in Atari games and then
compare our approach to MCTS and other state of the art learnin-based
approaches in eight games.

###  5.1 Atari environments

We tested our algorithm in 55 different Atari games using the OpenAI gym
toolkit. In most games the agent used RAM data as observations to make
decisions. As seen in table 1, our results show that Fractal Monte Carlo
outperforms previous state of the art (SOtA) approaches in 49 of those (89%).
In each game we choose the appropriate parameters by experimentation and
intuition. The extensive table containing the exact parameters used for every
game is available in our github repository (Hernández et al., 2017a).
Furthermore, we compared FMC performance when using RAM data versus using
images as observations in eight games and found that there is an overall
$61\%$ performance difference in favor of RAM data(Table 2).

Table 1: Fractal Monte Carlo in Atari games Available games | 55 |
---|---|---
Games played by FMC | 55 | 100.00%
FMC better than avg human | 51 | 92.73%
FMC better than SOtA | 49 | 89.09%
Solved or above human record | 25 | 45.45%
Solved due to the 1M bug | 11 | 20.00%
In 51 out of 55 games FMC scored better than the average human meaning a human that has played for 2 hours. FMC solved or scored higher than the human record in $45\%$ of the games tested. A game is considered ’solved’ when we hit an eventual score limit or we can play it endlessly. SOtA is an abbreviation for "State Of the Art". The term "1M bug" refers to some games having a hardcoded score limit, usually found at 999,999. Table 2: Image data versus RAM dump Environment | IMG | RAM | RAM vs IMG | RAM vs SoTA
---|---|---|---|---
atlantis | 145000 | 139500 | 96.21% | 1.59%
bank heist | 160 | 280 | 175.00% | 0.68%
boxing | 100 | 100 | 100.00% | 100.60%
centipede | immortal | immortal | 100.00% | 2373.87%
ice hockey | 16 | 33 | 206.25% | 311.32%
ms pacman | 23980 | 29410 | 122.64% | 468.09%
qbert | 17950 | 22500 | 125.35% | 358.11%
video pinball | 273011 | 999999 | 366.29% | 105.31%
|  |  | 161.47% | 464.95%
Comparison of FMC scores when using IMG and RAM data as observations in eight
Atari games. Parameters used are: Repeat Actions=5, Time Horizon=15, Max
Samples=300, Number of Walkers=30. We find that, in total, RAM yields
$61.47\%$ better results than IMG scores and outperforms State Of the Art
(SOtA) methods in most games tested.

###  5.2 Comparison against UCT

In table 3 we compare FMC with the state of the art MCTS implementation UCT on
the only eight games we found to be solved in the literature (Guo et al.,
2014). The results presented in table 3 show that our method our method
clearly outperforms MCTS while being three to four orders of magnitude more
efficient.

Table 3: Comparison against MCTS variant, UCT Game | Scores | Samples per step
---|---|---
| MCTS | FMC | % | MCTS | FMC | Efficiency
Asterix | 226,000 | 999,500 | 442% | —– | 241 | —–
Beam rider | 7,233 | 288,666 | 3991% | 3,000,000 | 946 | x 3,171
Breakout | 406 | 864 | 213% | 3,000,000 | 866 | x 3,386
Enduro | 788 | 5,279 | 670% | 4,000,000 | 390 | x 10,256
Pong | 21 | 21 | 100% | 150,000 | 158 | x 949
Q-bert | 18,850 | 999,999 | 3523% | 3,000,000 | 3,691 | x 813
Seaquest | 3,257 | 999,999 | 30703% | 3,000,000 | 964 | x 3,112
Space invaders | 2,354 | 17,970 | 763% | 3,000,000 | 1,830 | x 1,639
Comparison of MCTS versus FMC performance in eight Atari Games. We also
compare the number of simulations or "samples" each algorithm used per step.
The metric "efficiency" is calculated as MCTS samples per step / FMC samples
per step for each game

##  6 Conclusions

In this work, we propose a new thinking framework called Fractal AI theory
that we used to define intelligent behavior. FAI principles provided a basis
for creating a new Monte Carlo approach based on maximizing Causal Path
Entropy. We put up our algorithm against Atari environments and our results
showed that it performs better than state of the art algorithms, like DQN and
its variants, in most games.

Our algorithm has many potential applications especially for improving methods
that use MCTS. Furthermore FMC can produce high amount and high quality
training data for use in training reinforcement learning agents. For example,
deep Q agents learn to associate reward expectations with states after being
trained on a huge amount of data mainly consisting of random rollouts. FMC
high quality rollouts can be fed into a DQN in the training stage which might
result in a boost in training performance. Another promising idea for
improving FMC is to add learning capabilities to walkers using a neural
network. The network would be trained on correct decisions made by the agent
and output a probability distribution over the action space. Then walkers
would sample that distribution instead of picking a random action thus
transitioning into an informed search model.

Fractal Monte Carlo is a just one of the possible algorithms inspired by FAI
theory. Much research is still needed to explore other possible
implementations of this new concept and their many potential applications in
the real world.

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