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This is a problem inspired from the Tomb Raider Anniversary video game (2007?) at the beginning of the Sanctuary of Scion in Egypt when Lara needed to solve the puzzle with 4 columns where the columns have the correct pictures facing each other (see this youtube video). This could be similar to the whirligig puzzle (see http://www.jaapsch.net/puzzles/circle.htm). I translated the problem into a problem in Group theory. I figured the solution but now I want an isomorphism to $(\mathbb Z/16\mathbb Z,+)$. Here is my interpretation of it:

For simplicity I will denote the additive group $(\mathbb Z/4 \mathbb Z, +)$ as $G$. Consider a 4 tuple $(1,3,3,1)$ in $G^4$ generating other 4-tuples in $G^4$ by finite number of the following operations:

For $i=1,\dots,4$ let $\pi_i: G^4 \rightarrow G^4$ be defined by $\pi_i(x):=y$ where $y_j = x_j+1$ if $j\neq i$ and otherwise $y_i=x_i$.

Let $S$ be all the tuples that $(1,3,3,1)$ generates from finite compositions of $\pi_1,\pi_2,\pi_3$ and $\pi_4$. So an element $x\in S$ can be expressed as

$ \pi_1^{m_1}\circ\pi_2^{m_2}\circ\pi_3^{m_3}\circ\pi_4^{m_4}((1,3,3,1)) $

for some integers $m_1$, $m_2$,$m_3$ and $m_4$.

Question: Do the mappings $\pi_i|S$ generate $(\mathbb Z/16\mathbb Z,+)$? Is there a natural isomorphism? How can I get $(3,1,3,1)$ from $(1,3,3,1)$?

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You can get the map $$(a,b,c,d) \mapsto (a-1,b,c,d)$$ as the composition of $\pi_1^2 \circ \pi_2 \circ \pi_3 \circ \pi_4$: this has the effect $$(a,b,c,d) \mapsto (a+1,b+1,c+1,d) \mapsto (a+2,b+2,c+1,d+1) \mapsto (a+3,b+2,c+2,d+2) \mapsto (a+3,b+3,c+3,d+3) \mapsto (a+3,b,c,d).$$ The maps $\pi_2^2 \circ \pi_1 \circ \pi_3 \circ \pi_4$, etc. are similar and it is clear that you can get from $(1,3,3,1)$ to any configuration this way.

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  • $\begingroup$ So these maps make up a group under composition which is isomorphic to $G^4$ (the group has 256 elements) $\endgroup$ – quantum Nov 17 '15 at 15:54
  • $\begingroup$ @quantum That seems right $\endgroup$ – user290996 Nov 17 '15 at 16:12

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