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This configuration appear as problem 3845 in Crux Mathematicorum. I see it is very beautiful. This configuration are generalization of Pascal theorem and Brianchon theorem:

Consider six points $A_1$, $A_2$, $\cdots$ , $A_6$ on a circle $(O_A)$ and a point $B_1$ on another circle $(O_B)$. Let the circle $(A_iA_{i+1}B_i)$ meet $(O_B)$ again at $B_{i+1}$ for $i=1,\cdots, 5$. Then the four points $A_6$, $A_1$, $B_1$, $B_6$ lie on a circle. Denoting by $(O_i)$ the circle $(A_iA_{i+1}B_{i+1}B_i)$ for $i=1,\cdots,6$, taking subscripts modulo 6. If $(O_1)$ meets $(O_4)$ at two points $C_3, C_6$, $(O_2)$ meets $(O_5)$ at two points $C_4, C_1$ and $(O_3)$ meets $(O_6)$ at two points $C_2, C_5$, then the six points $C_1$, $C_2$, $C_3$, $C_4$, $C_5$, $C_6$ lie on a circle. The configuration in Figure as follows:

My question: what is the symmetric group of this configuration ?

enter image description here

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    $\begingroup$ One has to rectify what kind of symmetries are allowed. I tend to view this as a configuration of 18 points and 12 circles; there are 3 circles with 4 points of them, the rest have 6 point on them, this is limits the possibilites already... $\endgroup$ – Dima Pasechnik Jul 19 at 15:52
  • $\begingroup$ Dear Dr. @DimaPasechnik Thank you very much, I have deleted "the 3 circles with 4 points of them" (because these circles were adding to prove). Now I have updated the true configuration which I ask $\endgroup$ – Đào Thanh Oai Jul 20 at 4:17
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    $\begingroup$ It still isn't clear what transformations you want. Rigid transformations of the plane? Inversions around arbitrary circles? Are these in general position, or do you want a maximally symmetric configuration? $\endgroup$ – Watson Ladd Jul 20 at 15:52
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    $\begingroup$ It's still unclear. "What is the symmetric group" is not well-defined. You probably want to know the automorphism group of something. It could mean many things. Isometries of the plane preserving these subsets? Affine? Homographies? Automorphisms of this set of points and circles with the relation of incidence?... $\endgroup$ – YCor Jul 20 at 16:06
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    $\begingroup$ What about taking the symmetry of the incidence relations, like this one (The Desargues configuration has a symmetry group $S_5$)? This is the most general notion of symmetry. $\endgroup$ – Bullet51 Jul 20 at 16:17
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Observations.

  1. Every automorphism fixes the pair $\{C_n,C_{n+3}\}$ (indicies are taken modulo $6$), as $C_{n+3}$ is the only point sharing $3$ points with $C_n$.

  2. Every automorphism fixes the circle in magenta, as it's the only circle with points sharing $3$ circles.

  3. The action of the automorphism group on the circle in magenta can be described as $C_2 ^3 ⋊S_3$, where the $C_2$ acts by exchanging $C_n$ and $C_{n+3}$, and the $S_3$ acts by permuting the blocks $\{C_1,C_4\},\{C_2,C_5\},\{C_3,C_6\}$.

  4. Consider the normal subgroup stabilizing all $C_n$'s. One symmetry is exchanging $O_A$ with $O_B$ and $A_n$ with $B_n$. If $O_A$ and $O_B$ are both fixed, the only symmetry remains is exchanging $O_n$ with $O_{n+3}$, $A_n$ with $A_{n+3}$ and $B_n$ with $B_{n+3}$. So the structure of the group is $C_2 \times C_2$.

Conclusion.

The whole group can be described as $(C_2 \times C_2)⋊(C_2^3⋊S_3)$. The elements of the subgroup stabilizing all the $C_n$'s and the automorphism mapping $C_n$ to $C_{n+3}$ while stabilizing everything else commutes with all the elements of the group. So the group can also be described as $C_2 \times C_2 \times C_2 \times (C_2^2⋊S_3)$.

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  • $\begingroup$ Your (3) seems to show that $C_2^3$ doesn't commute with $S_3$; or is it a different $C_2^3$ in the two decompositions $(C_2 \times C_2) \rtimes (C_2^3 \rtimes S_3)$ and $C_2^3 \times (C_2^2 \rtimes S_3)$? $\endgroup$ – LSpice Jul 20 at 17:55
  • $\begingroup$ There should be a different $C_2^3$ in $C^3_2×(C_2^2⋊S_3)$, not the $C_2^3$ in (3). $\endgroup$ – Bullet51 Jul 20 at 17:59

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