Visual Magic Squares and Group Orbits I

3. Generating Conway Squares via Group Orbits

In this section of the paper we will assume that the reader is familiar with basic finite group theory at the level of Slomson [7]: Chapter 7. Permutations and Groups, and Chapter 8. Group Actions. We will continue to pursue the task of finding all Conway squares as in Section 2, however here we will show how to generate all Conway squares using group orbits. In particular, we will show that each of the six types of Conway squares defined in Section 2, can be generated by the transitive action of the Conway (square) group, i.e. the group of symmetries of the hypercube (also know as the hyperoctahedral group, the monomial group, and the group of signed permutations).

Definition 3.1. Let (G, , e) be a group and let X be a non-empty set. Then the group G acts on the set X just in case there is a mapping (a group action) [. , .]: G X → X such that for all g, h in G and x in X: (1) [e, x] = x and (2) [g, [h, x]] = [g h, x]. Note that in order to simplify notation we will often write g(x) for [g, x].

Base 2 and Base 10 Coding of Conway Items

Illustration 3.1 ``

Definition 3.2. As specified in Definition 2.6, the two main diagonals of Conway items in Illustration 3.1 represent one similarity class and the rest of the Conway items represent the other similarity class. Since we will be especially interested in the base 2 representation in this section, we define:

(1) H(4) = {[0, 0, 0, 0], [0, 1, 0, 1], [1, 0, 1, 0], [1, 1, 1, 1], [0, 0, 1, 1], [0, 1, 1, 0], [1, 0, 0, 1], [1, 1, 0, 0]},

(2) K(4) = {[0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 1, 1], [1, 0, 1, 1], [1, 1, 1, 0], [1, 1, 0, 1], [1, 0, 0, 0], [0, 1, 0, 0]},

(3) G(4) = H(4) K(4).

Then, G(4) is the set of all Conway items (in base 2), and H(4) and K(4) are the two similarity classes of Conway items. Further, for all a, b in G(4), define a + b to be addition (mod 2) in each coordinate, e.g. [0, 1, 0, 1] + [0, 1, 1, 1] = [0, 0, 1, 0].

Theorem 3.3.

(1) (G(4), +) is an abelian group isomorphic to C2 C2 C2 C2, where C2 is the cyclic group with two elements.

(2) (H(4), +) is a subgroup of (G(4), +) isomorphic to C2 C2 C2.

(3) For each k in K(4), k + H(4) = K(4), i.e. K(4) is the other coset of (H(4), +) in (G(4), +).

Exercise 3.1. Prove Theorem 3.3.

Definition 3.4. Let CS denote the set of all Conway squares, and define the following group action of G(4) on CS. Let g in G(4) and S in CS, then [g , S]: G(4) CS → CS is defined by [g , S] = g + S, where g + S is the square obtained by component wise addition of g (mod 2) to every entry of S (base 2 coding). Further, for each subgroup M of G(4) and for each S in CS define M(S) = {g + S: g in M}, then M(S) is called the M orbit of S.  

Example 3.5. Let So be the following Type D1 Conway square and consider finding H(4)( So), i.e. the H(4) orbit of the square So.

Base 2 and Base 10 Coding of So ``

Note that H(4) is easily recognized as the members of G(4) having an even number of 1's. So, to find H(4)( So) we just add each h in H(4) to each entry of So as follows:

 Notice that since So is a Type D1 Conway square, each square h + So is also a Type D1 Conway square:

A Type D1 Conway square:  h + So = [0, 0, 0, 0] + So

 

Another Type D1 Conway square:  h + So = [0, 0, 1, 1] + So

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Another Type D1 Conway square:  h + So = [0, 1, 0, 1] + So

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Another Type D1 Conway square:  h + So = [0, 1, 1, 0] + So

 

Another Type D1 Conway square:  h + So = [1, 0, 0, 1] + So

 

Another Type D1 Conway square:  h + So = [1, 0, 1, 0] + So

 

Another Type D1 Conway square:  h + So = [1, 1, 0, 0] + So

 

Another Type D1 Conway square:  h + So = [1, 1, 1, 1] + So

Next, recall from Definition 2.7 that a Type D1 Conway square has the items in the two similarity classes organized as follows:

Notice that in Example 3.5, for each h in H(4) the similarity classes corresponding to H(4) (though scrambled) stayed in the upper left and lower right in h + So, and the similarity classes corresponding to K(4) (though scrambled) stayed in the lower left and upper right in h + So. Furthermore, the pattern of complements above (though scrambled) was preserved for the similarity classes corresponding to both H(4) and K(4) in h + So.  

Exercise 3.2. Using the Conway square So in Example 3.5, for all k in K(4) investigate k + So. Note that you may use the following Maple worksheet [insert link] to generate the various representations of the new Conway squares k + So as in Example 3.5.

Motivated by Example 3.5, Exercise 3.2, and our observations above, we have the following theorem.

Theorem 3.6. Let S be a Conway square in CS, then

(1) for all g in G(4), [g , S] = g + S is in CS,

(2) the mapping [ . , S]: G(4) → CS is 1-1,

(3) |G(4)(S)| = |G(4)| = 16,

(4) for all h in H(4), [h , S] = h + S preserves the original location in S of the two similarity classes determined by H(4) and K(4), though scrambling occurs within the locations of both H(4) and K(4).

(5) for all k in K(4), [k , S] = k + S interchanges the original locations in S of the two similarity classes determined by H(4) and K(4).

(6) for all g in G(4), if S is of Type D1, D2, D3, R1, R2, or R3, then [g , S] = g + S  is of Type D1, D2, D3, R1, R2, or R3, respectively.

Exercise 3.3. Prove Theorem 3.6.

Definition 3.7. Let D(4) = {Id, Rot90, Rot180, Rot270, FlipH, FlipV, FlipLD, FlipRD} be the group of symmetries of the geometrical square (the dihedral group), where the rotations Rot90, Rot180, Rot270 are measured in the clockwise direction, and where FlipH, FlipV, FlipLD, FlipRD are reflections in the horizontal, vertical, left main diagonal, right main diagonal axes, respectively. Note that D(4) determines a group action on the set of Conway squares CS by simply applying each transformation of D(4) to each S in CS, viewed geometrically as a decorated square.

In addition, two Conway squares S1 and S2 in CS will be called geometrically equivalent if and only if there is a T in D(4) such that T(S1) = S2. Further, two Conway squares will be called essentially different if and only if they are not geometrically equivalent.

Exercise 3.4.

(1) Beginning with the same Conway square So as in Example 3.5, determine how many essentially different Conway squares are in G(4)( So), i.e. first determine which of these Conway squares are geometrically equivalent using the group D(4). Note that you may use the following Maple worksheet [insert link] to help check for essentially different Conway squares in the G(4) orbit of So, G(4)( So).

(2) Find a subgroup M of G(4) isomorphic to the subgroup {Id, Rot180, FlipH, FlipV} of D(4) such that the M orbit of So, M(So), contains only Conway squares geometrically equivalent to So. How do the cosets of M relate to the essentially different Conway squares in G(4)( So)?  

Remark 3.8. From Theorem 3.6 and Exercise 3.4 we see that the orbit of G(4) on a specific S in CS produces only 16 of the 384 Type D1 Conway squares, and only 4 of the 48 essentially different Type D1 Conway squares (see Theorem 2.8. and Corollary 2.9). Furthermore, it's not hard to see that the same results hold for Type D2 and D3 Conway squares.

In order to generate still more Type D1 Conway squares, we can again try to find other automorphisms of the group G(4) whose extensions preserve Type D1 Conway squares. Instead of G(4) acting on itself via left translation using the group operation, we can investigate the extension of a simple permutation of the components of g in G(4). For example, let g = [g1, g2, g3, g4] and consider the automorphism [g1, g2, g3, g4] → [g2, g1, g3, g4] which switches the first and second components of g. Let So be the D1 Conway square in Example 3.5, and call the extension of this switch mapping applied to every entry of So, Sw1,2, then Sw1,2(So) is the image of So after switching the first and second components of every entry of So.

Base 2 and Base 10 Coding of So ``

Base 2 and Base 10 Coding of Sw1,2( So) ``

Notice that since So was a Conway square to begin with, the Conway property must be preserved in Sw1,2 ( So). For example, consider the effect of switching the first two components of each entry (base 2 coding) in the first column of So above. Proceeding from top to bottom this column has the following sequence of first components: 0, 1, 1, 0, and the following sequence of second components: 0, 0, 1, 1. Similarly, proceeding from top to bottom in Sw1,2 ( So),   we see that these two sequences are switched, i.e. the first column has the following sequence of first components: 0, 0, 1, 1, and the following sequence of second components: 0, 1, 1, 0. Hence, switching the first and second components of all the entries in So, merely switches one sequence of four bits satisfying the Conway property with another sequence of four bits satisfying the Conway property, and this holds for each column, row, and both main diagonals. Moreover, if σ is an arbitrary permutation in S(4), the symmetric group on {1, 2, 3, 4}, then the automorphism [g1, g2, g3, g4] → [gσ(1), gσ(2), gσ(3), gσ(4)] of G(4) will also have a corresponding extension to CS that preserves the Conway property, and thus it will map Conway squares to Conway squares.

Definition 3.9. In order to define the action of S(4) on CS, we first define the action of S(4) on G(4). Let [. , .]: S(4) G(4) → G(4) be defined by [σ, g] = [gσ(1), gσ(2), gσ(3), gσ(4)] for each σ in S(4) and g = [g1, g2, g3, g4] in G(4). Next, we extend this action to CS. Let [. , .]: S(4) CS → CS be defined by [σ, S] = σ(S), where σ is applied to each entry of S.

Example 3.10. Once again, we begin with the D1 Conway Square So in Example 3.5. This time we find the S(4) orbit of So.

Another Type D1 Conway square via the S(4) orbit of So: σo = Id

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Another Type D1 Conway square via the S(4) orbit of So: σ1 = Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ2 = Sw1,3

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Another Type D1 Conway square via the S(4) orbit of So: σ3 = Sw1,4

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Another Type D1 Conway square via the S(4) orbit of So: σ4 = Sw2,3

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Another Type D1 Conway square via the S(4) orbit of So: σ5 = Sw2,4

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Another Type D1 Conway square via the S(4) orbit of So: σ6 = Sw3,4

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Another Type D1 Conway square via the S(4) orbit of So: σ7 = Sw2,3 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ8 = Sw2,4 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ9 = Sw1,3 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ10 = Sw1,4 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ11 = Sw1,4 ο Sw1,3

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Another Type D1 Conway square via the S(4) orbit of So: σ12 = Sw1,3 ο Sw1,4

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Another Type D1 Conway square via the S(4) orbit of So: σ13 = Sw2,4 ο Sw2,3

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Another Type D1 Conway square via the S(4) orbit of So: σ14 = Sw2,3 ο Sw2,4

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Another Type D1 Conway square via the S(4) orbit of So: σ15 = Sw3,4 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ16 = Sw2,4 ο Sw1,3

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Another Type D1 Conway square via the S(4) orbit of So: σ17 = Sw2,3 ο Sw1,4

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Another Type D1 Conway square via the S(4) orbit of So: σ18 = Sw1,4 ο Sw1,3 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ19 = Sw1,4 ο Sw1,2 ο Sw1,3

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Another Type D1 Conway square via the S(4) orbit of So: σ20 = Sw1,3 ο Sw1,2 ο Sw1,4

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Another Type D1 Conway square via the S(4) orbit of So: σ21 = Sw2,4 ο Sw2,3 ο Sw1,2

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Another Type D1 Conway square via the S(4) orbit of So: σ22 = Sw3,4 ο Sw2,3 ο Sw1,3

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Another Type D1 Conway square via the S(4) orbit of So: σ23 = Sw1,2 ο Sw1,3 ο Sw1,4

 

Exercise 3.6. Show that there are only twelve essentially different Conway squares in the S(4) orbit of So in Example 3.10 above.   Note that you may use the following Maple worksheet [insert link] to help check for essentially different Conway squares in the S(4) orbit of So, S(4)( So).

Exercise 3.7. Determine the total number of essentially different Conway squares in G(4)( So) S(4)( So).

Exercise 3.8. Explain why we must have G(4)( So) S(4)( So) = { So } by examining the two different group actions, without using the listings of members of these orbits (above).

Motivated by Example 3.10, and Exercises 3.6-3.8, we have the following theorem.

Theorem 3.11. Let S be a Conway square in CS, then

(1) for all σ in S(4), [σ, S] = σ(S) is in CS,

(2) the mapping [. , S]: S(4) → CS is 1-1,

(3) |S(4)(S)| = |S(4)| = 24,

(4) for all σ in S(4), σ(S) preserves the original location in S of the two similarity classes determined by H(4) and K(4), though scrambling occurs within the locations of both H(4) and K(4).

(5) for all σ in S(4), if S is of Type D1, D2, D3, R1, R2, or R3, then σ(S) is of Type D1, D2, D3, R1, R2, or R3, respectively.

Exercise 3.9. Prove Theorem 3.11.

Remark 3.12. We have shown that if S is a Conway square in CS then |G(4)(S)| = 16 and |S(4)(S)| = 24, but these still fall far short of the totals for each type found in Theorems 2.8 and 2.12, e.g. there are 384  Type D1 Conway squares. However, it is natural to think about combining these two group actions. For example, suppose for each sigma in S(4), we consider the orbit of G(4) on sigma(S), i.e. for each g in G(4) consider g + sigma(S). Note that we have already done this in the case that sigma = Id, which is just the orbit of G(4) on S. This approach is particularly encouraging, since |G(4)(S)| |S(4)(S)| = (16)(24) = 384.

Definition 3.13. We now define the Conway (square) group (B(4), ) with underlying set B(4) = G(4) S(4) and group operation defined by (g, σ) (h, t) = (g + σ(h), σ o t). This is a simplified approach to the well known (permutation) wreath product construction of the group of symmetries of the hypercube, also known as the hyperoctahedral group, the monomial group, and the group of signed permutations (see Humphreys [3], p. 170). Further, since this group is usually denoted by B(4), we have followed this convention in our development here.

Note that as usual B(4) acts on itself, but we want to restrict this action to G(4), in order to then extend it to the set of all Conway squares CS. First, we define the restricted action of B(4) on G(4) by [(g, σ), (h, Id)] = (g, σ) (h , Id) = (g + σ(h), h), where we simplify notation and simply write [(g, σ), h] = g + σ(h). Next, we define the action of B(4) on CS by [(g, σ), S] = g + σ(S).

Exercise 3.10. Verify that (B(4), ) is a group.

Theorem 3.14. (Conway Group Orbit Theorem)  

Let S be a Conway square in CS, then

(1) for all σ in B(4), [(g, σ), S] = g + σ(S) is in CS,

(2) the mapping [. , S]: B(4) → CS is 1-1,

(3) |B(4)(S)| = |B(4)| = 384,

(4) for all (g, σ) in B(4), if S is of Type D1, D2, D3, R1, R2, or R3, then g + σ(S) is of Type D1, D2, D3, R1, R2, or R3, respectively.

Corollary 3.15. The action of the Conway group B(4) on the Conway squares CS partitions CS into seven orbits each of size 384. One orbit for each of the Types D1, D2, D3, R1, R2, and two orbits for Type R3, corresponding to the additional symmetry for this type mentioned in the proof of Theorem 2.12.

Exercise 3.11. Prove Theorem 3.14.

Exercise 3.12. Verify Corollary 3.15 using the Euler squares you found in Exercises 1.2 and 1.3 as the seeds to generate the seven B(4) orbits, and the following Maple worksheet [insert link].

Exercise 3.13. Repeat Exercise 3.14 using the Conway squares provided as examples in the proofs of Theorems 2.8 and 2.12, as the seeds to generate the seven B(4) orbits, and the following Maple worksheet [insert link].

Exercise 3.14. Examine each of the orbits of B(4) in Exercise 3.12 and determine which elements of B(4) form a subgroup isomorphic to (a subgroup of) D(4) that determines the geometrically equivalent Conway squares in each orbit. How does your answer to this task depend on the type of square your start with, or on the particular square you choose as a seed for an orbit of a given type? Here is a helpful Maple worksheet [insert link].

Exercise 3.15. Using your results in Exercise 3.14 verify the number of essentially different Conway squares of each type found in Corollaries 2.9 and 2.13. Here is a helpful Maple worksheet [insert link].

Exercise 3.16. (Generating Euler Squares I) We know that there are Euler squares of each of the six different types of Conway squares, but there are far fewer Euler squares, i.e. there are 144 essentially different Euler squares out of a total of 528 essentially different Conway squares. By Exercise 3.12 we know that using Euler squares as seeds and the group action provided by B(4), we can generate the seven B(4) orbits of Conway squares. Try to find a group action on the Euler squares that generates all and only Euler squares. You might begin by trying to use the base 4 representation of Euler squares in a way similar to how we used the base 2 representation of Conway squares to get B(4). Hint: Let (Z4, +) be the group of integers (mod 4), and let F be the group (Z4 Z4, +). Now, construct E(2) by analogy with B(4) = G(4) S(4), letting F play the role of G(4), and letting S(2), the group of permutations of two objects, play the role of S(4). Then let E(2) = F S(2), with group operation defined as in B(4). Find the orbit of E(2), restricted to F, on the smallest set of Euler squares needed to generate them all.

Exercise 3.17. (Generating Euler Squares II) Repeat Exercise 3.16, using the same hint, but let F be the group (S(4) S(4), ο), instead of (Z4 Z4, +). Once again, let E(2) = F S(2), with group operation defined as in B(4), and find the orbit of E(2), restricted to F, on the smallest set of Euler squares needed to generate them all.

Remark 3.15. In a sequel to this paper [5], we extend this approach to the remaining (352) NonConway 4 4 magic squares, developing additional types and using appropriate group orbits to generate them all. In addition, we explore the extension of these ideas to the analysis of higher order magic squares.