**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 (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**

**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

(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) È K(4).

Then,
G(4) is the set of all

**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

**Example 3.5.** Let S_{o} be the following Type D1 Conway square
and consider finding H(4)( S_{o}), i.e. the
H(4) orbit of the square S_{o}.

**Base 2 and Base
10 Coding of S _{o}**

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)( S_{o})
we just add each h in H(4) to each entry of S_{o} as follows:

Notice
that since S_{o} is a Type D1 Conway square,
each square h + S_{o} is also a Type D1 Conway square:

**A Type D1 ****Conway**** square: h + S _{o} =
[0, 0, 0, 0] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[0, 0, 1, 1] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[0, 1, 0, 1] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[0, 1, 1, 0] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[1, 0, 0, 1] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[1, 0, 1, 0] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[1, 1, 0, 0] + S_{o} **

**Another Type D1 ****Conway**** square: h + S _{o} =
[1, 1, 1, 1] + S_{o} **

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 + S_{o},
and the similarity classes corresponding to K(4) (though scrambled) stayed in
the lower left and upper right in h + S_{o}. Furthermore, the pattern
of complements above (though scrambled) was preserved for the similarity
classes corresponding to both H(4) and K(4) in h + S_{o}.

**Exercise 3.2**. Using the _{o}
in Example 3.5, for all k in K(4) investigate k + S_{o}.
Note that you may use the following MapleÔ worksheet [insert link] to
generate the various representations of the new _{o} 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

(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

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

**Exercise 3.4.**** **

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

(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 S_{o}, M(S_{o}), contains only Conway squares
geometrically equivalent to S_{o}. How do the cosets
of M relate to the essentially different _{o})?

**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 = [g_{1}, g_{2}, g_{3}, g_{4}]
and consider the automorphism [g_{1}, g_{2},
g_{3}, g_{4}] → [g_{2}, g_{1}, g_{3},
g_{4}] which switches the first and second components of g. Let S_{o}
be the D1 Conway square in Example 3.5, and call the extension of this switch
mapping applied to every entry of S_{o}, Sw_{1,2}, then Sw_{1,2}(S_{o})
is the image of S_{o} after switching the first and second components
of every entry of S_{o}.

**Base 2 and Base
10 Coding of S _{o}**

**Base 2 and Base
10 Coding of Sw _{1,2}( S_{o})**

Notice
that since S_{o} was a _{1,2} ( S_{o}).
For example, consider the effect of switching the first two components of each
entry (base 2 coding) in the first column of S_{o}
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 Sw_{1,2} ( S_{o}),
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 S_{o}, 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 [g_{1},
g_{2}, g_{3}, g_{4}] → [g_{σ}_{(1)},
g_{σ}_{(2)}, g_{σ}_{(3)},
g_{σ}_{(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] = [g_{σ}_{(1)}, g_{σ}_{(2)},
g_{σ}_{(3)}, g_{σ}_{(4)}]
for each σ in S(4) and g = [g_{1}, g_{2}, g_{3}, g_{4}]
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 S_{o} in Example 3.5. This
time we find the S(4) orbit of S_{o}.

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{o}_{
}= Id**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{1 }= Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{2 }= Sw_{1,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{3 }= Sw_{1,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{4 }= Sw_{2,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{5 }= Sw_{2,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{6 }= Sw_{3,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{7 }= Sw_{2,3} ο
Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{8 }= Sw_{2,4} ο
Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{9 }= Sw_{1,3} ο
Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{10 }= Sw_{1,4} ο
Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{11 }= Sw_{1,4} ο
Sw_{1,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{12 }= Sw_{1,3} ο
Sw_{1,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{13 }= Sw_{2,4} ο
Sw_{2,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{14 }= Sw_{2,3} ο
Sw_{2,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{15 }= Sw_{3,4} ο
Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{16 }= Sw_{2,4} ο
Sw_{1,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{17 }= Sw_{2,3} ο
Sw_{1,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{18 }= Sw_{1,4} ο
Sw_{1,3} ο Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{19 }= Sw_{1,4} ο
Sw_{1,2} ο Sw_{1,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{20 }= Sw_{1,3} ο
Sw_{1,2} ο Sw_{1,4}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{21 }= Sw_{2,4} ο
Sw_{2,3} ο Sw_{1,2}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{22 }= Sw_{3,4} ο
Sw_{2,3} ο Sw_{1,3}**

**Another Type D1 ****Conway**** square via the S(4) orbit of S _{o}: σ_{23 }= Sw_{1,2} ο
Sw_{1,3} ο Sw_{1,4}**

**Exercise 3.6.** Show that there are only twelve essentially different _{o} in Example 3.10 above. Note that you may
use the following MapleÔ worksheet [insert link] to help check for essentially
different _{o}, S(4)( S_{o}).

**Exercise 3.7.** Determine the total number of essentially different _{o}) È S(4)( S_{o}).

**Exercise 3.8.** Explain why we must have G(4)( S_{o}) Ç S(4)( S_{o}) = { S_{o} } 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

(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

(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

**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 MapleÔ 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 MapleÔ 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

**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 MapleÔ 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

**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.