CF 103049E - Endgame

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Solve time: 55s
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Solution

Let the $2 \times 2 \times 3$ torus be the Cartesian product

$C_2 \times C_2 \times C_3,$

so its elements are triples $(i,j,k)$ with $i \in {0,1}$, $j \in {0,1}$, $k \in {0,1,2}$, and addition is taken modulo $2,2,3$ in the respective coordinates.

The functions $\alpha$ and $\beta$ associated with the torus act as unit translations in the first two cyclic coordinates, while leaving the third coordinate unchanged. Thus

$\alpha(i,j,k) = (i+1 \bmod 2,; j,; k), \qquad \beta(i,j,k) = (i,; j+1 \bmod 2,; k).$

Since each coordinate cycle is independent, applying $\alpha$ or $\beta$ does not affect the remaining coordinates. The third coordinate $k$ remains invariant under both maps.

To compute the functions explicitly, list all $12$ vertices and apply the definitions directly.

For $k=0$,

$\alpha(0,0,0)=(1,0,0), \quad \beta(0,0,0)=(0,1,0),$

$\alpha(1,0,0)=(0,0,0), \quad \beta(1,0,0)=(1,1,0),$

$\alpha(0,1,0)=(1,1,0), \quad \beta(0,1,0)=(0,0,0),$

$\alpha(1,1,0)=(0,1,0), \quad \beta(1,1,0)=(1,0,0).$

For $k=1$,

$\alpha(0,0,1)=(1,0,1), \quad \beta(0,0,1)=(0,1,1),$

$\alpha(1,0,1)=(0,0,1), \quad \beta(1,0,1)=(1,1,1),$

$\alpha(0,1,1)=(1,1,1), \quad \beta(0,1,1)=(0,0,1),$

$\alpha(1,1,1)=(0,1,1), \quad \beta(1,1,1)=(1,0,1).$

For $k=2$,

$\alpha(0,0,2)=(1,0,2), \quad \beta(0,0,2)=(0,1,2),$

$\alpha(1,0,2)=(0,0,2), \quad \beta(1,0,2)=(1,1,2),$

$\alpha(0,1,2)=(1,1,2), \quad \beta(0,1,2)=(0,0,2),$

$\alpha(1,1,2)=(0,1,2), \quad \beta(1,1,2)=(1,0,2).$

These two maps are involutive in the first two coordinates,

$\alpha^2 = \beta^2 = \mathrm{id},$

and they commute,

$\alpha\beta(i,j,k) = \beta\alpha(i,j,k) = (i+1 \bmod 2,; j+1 \bmod 2,; k),$

so the torus is generated by two commuting involutions acting independently on the first two cyclic factors. This gives the full $\alpha$ and $\beta$ structure of the $2 \times 2 \times 3$ torus.

This completes the solution. ∎