# Introducing $n$-Magic Groups and Characterizing $3$-Magic Finitely Generated
Abelian Groups

Danielle Bowerman<EMAIL_ADDRESS>— supported by the Chancellor’s
Distinguished Fellows Program Missouri University of Science and Technology
Nicholas Fleece<EMAIL_ADDRESS>— supported by the Kummer Institute for
Student Success, Research and Economic Development Missouri University of
Science and Technology  Matt Insall<EMAIL_ADDRESS>Missouri University of
Science and Technology

###### Abstract

In this paper, we define an $n$-magic square in a group to be an $(n\times n)$
array of group elements whose rows, columns, and diagonals have the same
product. This definition is akin to the idea of magic squares in the integers.
Groups that have an $n$-magic square are said to be $n$-magic. We begin with
some preliminary results and focus much of our attention on $3$-magic groups.
Through a series of propositions, we ultimately prove a characterization
theorem for $3$-magic finitely generated abelian groups. We then discuss some
additional results about non-abelian groups as well as $n$-magic groups where
$n>3$.

##  1 Introduction

In this section, we review previous work on magic squares, provide fundamental
definitions, and present the manner in which this work is organized.

###  1.1 Previous Work

Magic squares are one of the most well-known topics in recreational
mathematics. The notion initially referred to a square array of distinct
positive integers in which every row, column, and diagonal sum to the same
number. It is well known that magic squares exist for every size larger than a
$(2\times 2)$ array. In fact, many examples of these were found in antiquity.

This paper originates in our work on the problems posed in the Numberphile
video “The Parker Square—Numberphile” [3]. While searching for a magic square
of squares, we began considering squares in settings other than the integers.
In [1] magic squares in finite fields were discussed. The work in this paper
began with the goal of defining magic squares in the realm of group theory, at
which point we learned that [4] had initiated an investigation of this idea.
The definition we created independently of their work is very similar.
However, here, we take the idea further and characterize the finitely
generated abelian groups with $(3\times 3)$ magic squares.

The previously mentioned 1997 paper by Sun and Yihui [4] touches on this topic
and begins to address the topics on which our work is founded. Lower bounds
for the orders of the groups that admit magic squares were established in that
paper, but the authors also provide results that determine several classes of
groups which have $(n\times n)$ magic squares, leaving open the question of
classifications of groups which do not have $(n\times n)$ magic squares. It
was proved in the paper that for any positive integer $n\geq 3$, any abelian
group of order $n^{2}$ admits an $(n\times n)$ magic square. They also showed
that for any positive prime number $p$ and any positive integer $n$, any
elementary abelian $p-$group of size $p^{2n}$ admits a magic square of size
$(2n\times 2n)$.

###  1.2 Definitions and Notation

We begin this section with the fundamental definition of this paper.

Definition 1.1. We say that group $G$ is $n$-magic if it has an $n\times n$
magic square. That is, there exist distinct
$g_{1,1},g_{1,2},\ldots,g_{1,n},g_{2,1},g_{2,2},\ldots,g_{2,n},\ldots,g_{n,1},g_{n,2},\ldots,g_{n,n}\in
G$ such that

$\displaystyle g_{1,1}g_{1,2}\ldots g_{1,n}$
$\displaystyle=g_{2,1}g_{2,2}\ldots g_{2,n}$ $\displaystyle=\ldots$
$\displaystyle=g_{n,1}g_{n,2}\ldots g_{n,n}$
$\displaystyle=g_{1,1}g_{2,1}\ldots g_{n,1}$
$\displaystyle=g_{1,2}g_{2,2}\ldots g_{n,2}$ $\displaystyle=\ldots$
$\displaystyle=g_{1,n}g_{2,n}\ldots g_{n,n}$
$\displaystyle=g_{1,1}g_{2,2}\ldots g_{n,n}$
$\displaystyle=g_{n,1}g_{n-1,2}\ldots g_{1,n}.$

We call this common product the magic product of the magic square.

Note that in [4] a “magic group” refers to the automorphism group on a magic
square. This greatly differs from our notion of an $n$-magic group. It is also
worth noting that natural variants of “magic” objects can be defined
similarly, such as for semi-groups or for magmas. We leave those for a future
investigation.

###  1.3 Organization of This Work

In the remainder of this paper, we first provide some preliminary results on
$3$-magic groups. We then prove a series of propositions that act as lemmas
for the proof of the characterization of $3$-magic finitely generated abelian
groups, which is then presented. Finally, we provide some results related to
nonabelian groups and discuss some ideas for future work on this topic.

##  2 Results

The following two results for a group $G$ are immediate. The first provides a
lower bound on the order of a group for it to be $n$-magic. The second tells
us that if a group is $n$-magic, then any group containing it as a subgroup is
also $n$-magic.

Proposition 2.1. If a group $G$ is $n$-magic, then $|G|\geq n^{2}$.

Proposition 2.2. If $H\leq G$ and $H$ is $n$-magic, then $G$ is $n$-magic.

We can quickly see the following.

Theorem 2.3. No groups are 2-magic.

Proof. Suppose $G$ is 2-magic, and let $\begin{bmatrix}g_{1}&g_{2}\\\
g_{3}&g_{4}\end{bmatrix}$ be a magic square. Then $g_{1}g_{2}=g_{1}g_{3}$, and
through left multiplication by $g_{1}^{-1}$, we have $g_{2}=g_{3}$. This
contradicts the assumption that the entries in a magic square are distinct.
$\square$

Corollary 2.4. Any monobinary algebra with the cancellation property will not
be 2-magic.

Given this, we can begin investigating which groups are $3$-magic. From
Proposition 2, we have that all groups of order $8$ or less cannot be
$3$-magic. We wish to find “fundamental” classes of magic groups for which we
can then apply Proposition 2 to find a much larger class of groups. To this
end, we endeavour to characterize 3-magic finitely generated abelian groups.
The next four propositions are our first steps in this direction.

Proposition 2.5. All infinite, finitely generated abelian groups are
$3$-magic.

Proof. We have that $\mathbb{Z}$ is $3$-magic since

$\displaystyle\begin{bmatrix}8&1&6\\\ 3&5&7\\\ 4&9&2\end{bmatrix}$

is a magic square. If $G$ is an infinite finitely generated abelian group then
$\mathbb{Z}\leq G$ by the Fundamental Theorem of Finitely Generated Abelian
Groups and so $G$ is $3$-magic by Proposition 2. $\square$

Proposition 2.6. The cyclic group $C_{n}$ is $3$-magic if and only if $n\geq
9$.

Proof. Necessity follows immediately from Proposition 2. For sufficiency, we
have that, for $C_{n}=\left<x\right>$,

$\displaystyle\begin{bmatrix}x^{n-3}&x^{2}&x\\\ x^{4}&1&x^{n-4}\\\
x^{n-1}&x^{n-2}&x^{3}\end{bmatrix}$

forms a magic square with a magic product of $1$. The fact that $n\geq 9$
guarantees these entries are distinct. $\square$

Proposition 2.7. The groups $C_{n}^{2}$ and $C_{n}^{3}$ are $3$-magic if and
only if $n\geq 3$.

Proof. Necessity follows immediately from Proposition 2. For sufficiency,
given $C_{n}^{2}$, we must cover a few cases. First, note that $C_{n}\leq
C_{n}^{2}$ for all $n$ so that, by Proposition 2, $C_{n}^{2}$ is $3$-magic for
all $n\geq 9$. Next, for $C_{2k+1}^{2}=\left<x,y\right>$,

$\displaystyle\begin{bmatrix}x^{k}&y^{k+1}&x^{k+1}y^{k}\\\
xy^{k}&1&x^{2k}y^{k+1}\\\ x^{k}y&y^{k}&x^{k+1}\end{bmatrix}$

forms a magic square with a common product of $1$. Thus, $C_{3}^{2}$,
$C_{5}^{2}$, and $C_{7}^{2}$ are $3$-magic. Furthermore, since $C_{3}^{2}\leq
C_{6}^{2}$, by Proposition 2, $C_{6}^{2}$ is also $3$-magic. Lastly, for
$C_{4}^{2}=\left<x,y\right>$,

$\displaystyle\begin{bmatrix}x&y^{3}&x^{3}y\\\ x^{2}y&1&x^{2}y^{3}\\\
xy^{3}&y&x^{3}\end{bmatrix}$

forms a magic square with a common product of $1$. Thus, $C_{4}^{2}$ is
$3$-magic. Since $C_{4}^{2}\leq C_{8}^{2}$, by Proposition 2, $C_{8}^{2}$ is
also $3$-magic.

Finally, since $C_{n}^{2}\leq C_{n}^{3}$ for all $n$, Proposition 2 provides
sufficiency for $C_{n}^{3}$. $\square$

Proposition 2.8. The group $C_{n}\times C_{n^{2}}$ is $3$-magic if and only if
$n\geq 3$.

Proof. Necessity again follows from Proposition 2. Sufficiency follows from
Proposition 2, after recognizing that $C_{n^{2}}\leq C_{n}\times C_{n^{2}}$ .
$\square$

To proceed further, we need to define a “normalized” magic square.

Definition 2.9. A normalized $n$-magic square in a group $G$ is an $n$-magic
square in $G$ such that one of the entries is the identity element of $G$.
When $n$ is odd, if the center entry of the magic square is the identity, we
call it a centered normalized $n$-magic square.

It can be shown that in abelian groups we can convert any magic square to a
normalized magic square with the identity in any position that we choose.

Proposition 2.10. An abelian group is $n$-magic if and only if for all
$i,j\in\\{1,2,\ldots,n\\}$ there exists an $n$-magic square in $G$ whose
$g_{i,j}$ entry is the identity. Therefore, an abelian group is $n$-magic if
and only if it admits a normalized magic square.

Proof. Sufficiency is clear. To prove necessity, choose an $n$-magic square in
$G$ with magic product $s$. Let $i,j\in\\{1,2,\ldots,n\\}$ and multiply each
entry by $g_{i,j}^{-1}$. Then the resulting square has $1$ in the $i,j$
position, and each row, column, and diagonal has a product of $g_{i,j}^{-n}s$.
We can guarantee that all these entries are unique since left multiplication
by a group element is an automorphism. $\square$

The next three results answer the remaining question as to whether we can find
an even stronger biconditional that can be used to quickly show that an
abelian group is not $3$-magic.

Theorem 2.11. The elements of any $3$-magic square in an abelian group can be
generated by three elements.

Proof. Sufficiency is immediate. The proof for necessity is adapted from [2]
and translated into the language of groups. If $G$ is $3$-magic, let

$\displaystyle\begin{bmatrix}g_{1}&g_{2}&g_{3}\\\ g_{4}&g_{5}&g_{6}\\\
g_{7}&g_{8}&g_{9}\end{bmatrix}$

be a magic square in $G$ with magic product $s$. We then have

$\displaystyle g_{1}g_{5}g_{9}$ $\displaystyle=s$ $\displaystyle
g_{2}g_{5}g_{8}$ $\displaystyle=s$ $\displaystyle g_{4}g_{5}g_{6}$
$\displaystyle=s$ $\displaystyle g_{7}g_{5}g_{3}$ $\displaystyle=s$
$\displaystyle g_{1}g_{4}g_{7}$ $\displaystyle=s$ $\displaystyle
g_{3}g_{6}g_{9}$ $\displaystyle=s$ $\displaystyle g_{1}g_{2}g_{3}$
$\displaystyle=s$ $\displaystyle g_{7}g_{8}g_{9}$ $\displaystyle=s.$

Multiplying the first four equations gives us

$\displaystyle(g_{1}g_{2}g_{3})(g_{4}g_{5}g_{6})(g_{7}g_{8}g_{9})g_{5}^{3}$
$\displaystyle=s^{4}$ $\displaystyle s^{3}g_{5}^{3}$ $\displaystyle=s^{4}$
$\displaystyle g_{5}^{3}$ $\displaystyle=s$

Then, define $a:=g_{1}g_{5}^{-1}$, $b:=g_{3}g_{5}^{-1}$, and $c:=g_{5}$. Now
note that since $g_{1}g_{2}g_{3}=g_{5}^{3}$ we know

$\displaystyle g_{2}=g_{1}^{-1}g_{3}^{-1}g_{5}^{3},$

since $g_{7}g_{5}g_{3}=g_{5}^{3}$ we have

$\displaystyle g_{7}=g_{3}^{-1}g_{5}^{2},$

and since $g_{1}g_{5}g_{9}=g_{5}^{3}$ we can state that

$\displaystyle g_{9}=g_{1}^{-1}g_{5}^{2}.$

Combining these statements allows us to say that since
$g_{1}g_{4}g_{7}=g_{5}^{3}$ we know

$\displaystyle
g_{4}=g_{1}^{-1}g_{5}^{3}g_{7}^{-1}=g_{1}^{-1}g_{5}^{3}(g_{3}^{-1}g_{5}^{2})^{-1}=g_{1}^{-1}g_{3}^{-1}g_{5},$

since $g_{2}g_{5}g_{8}=g_{5}^{3}$ we have

$\displaystyle
g_{8}=g_{2}^{-1}g_{5}^{2}=(g_{1}^{-1}g_{3}^{-1}g_{5}^{3})^{-1}g_{5}^{2}=g_{1}g_{3}g_{5}^{-1},$

and since $g_{3}g_{6}g_{9}=g_{5}^{3}$ we have

$\displaystyle
g_{6}=g_{3}^{-1}g_{5}^{3}g_{9}^{-1}=g_{3}^{-1}g_{5}^{3}(g_{1}^{-1}g_{5}^{2})^{-1}=g_{1}g_{3}^{-1}g_{5}.$

Combining these lets us see that

$\displaystyle\begin{bmatrix}ac&a^{-1}b^{-1}c&bc\\\ a^{-1}bc&c&ab^{-1}c\\\
b^{-1}c&abc&a^{-1}c\end{bmatrix}$
$\displaystyle=\begin{bmatrix}g_{1}g_{5}^{-1}g_{5}&(g_{1}g_{5}^{-1})^{-1}(g_{3}g_{5}^{-1})^{-1}g_{5}&g_{3}g_{5}^{-1}g_{5}\\\
(g_{1}g_{5}^{-1})^{-1}g_{3}g_{5}^{-1}g_{5}&g_{5}&g_{1}g_{5}^{-1}(g_{3}g_{5}^{-1})^{-1}g_{5}\\\
(g_{3}g_{5}^{-1})^{-1}g_{5}&g_{1}g_{5}^{-1}g_{3}g_{5}^{-1}g_{5}&(g_{1}g_{5}^{-1})^{-1}g_{5}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}g_{1}&g_{1}^{-1}g_{3}^{-1}g_{5}^{3}&g_{3}\\\
g_{1}^{-1}g_{3}g_{5}&g_{5}&g_{1}g_{3}^{-1}g_{5}\\\
g_{3}^{-1}g_{5}^{2}&g_{1}g_{3}g_{5}^{-1}&g_{1}^{-1}g_{5}^{2}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}g_{1}&g_{2}&g_{3}\\\ g_{4}&g_{5}&g_{6}\\\
g_{7}&g_{8}&g_{9}\end{bmatrix}$

$\square$

This leads to the biconditional result that will be used throughout the
remainder of the paper.

Corollary 2.12. An abelian group $G$ is $3$-magic if and only if there is a
centered normalized $3$-magic square in $G$ of the form

$\displaystyle\begin{bmatrix}a&a^{-1}b^{-1}&b\\\ a^{-1}b&1&ab^{-1}\\\
b^{-1}&ab&a^{-1},\end{bmatrix}$

where $a,b\in G$.

Proof. This result follows from multiplying every entry in

$\displaystyle\begin{bmatrix}ac&a^{-1}b^{-1}c&bc\\\ a^{-1}bc&c&ab^{-1}c\\\
b^{-1}c&abc&a^{-1}c\end{bmatrix}$

by $c^{-1}$. $\square$

Hence, for an abelian group $G$ to be $3$-magic, there must be two elements
$a,b\in G$ such that $a$, $a^{-1}b^{-1}$, $b$, $a^{-1}b$, $ab^{-1}$, $b^{-1}$,
$ab$, and $a^{-1}$ are distinct. This proves to be especially useful in
showing the impossibility of having a $3$-magic square in certain classes of
groups.

Corollary 2.13. The elements of any normalized $3$-magic square in an abelian
group can be generated by two elements of the group.

Proof. From the previous proposition, every $3$-magic square in an abelian
group is of the form

$\displaystyle\begin{bmatrix}ac&a^{-1}b^{-1}c&bc\\\ a^{-1}bc&c&ab^{-1}c\\\
b^{-1}c&abc&a^{-1}c\end{bmatrix}.$

If this were normalized, then one of the entries would be $1$. This would give
us an equation in which we could solve for $c$ in terms of $a$ and $b$.
$\square$

We can now make use of these results to continue determining which abelian
groups are $3$-magic.

Proposition 2.14. The group $C_{n}^{k}$, with $k\geq 4$, is $3$-magic if and
only if $n\geq 3$.

Proof. Since $C_{n}^{3}\leq C_{n}^{k}$, Proposition 2 gives sufficiency for
$n\geq 3$. Now let $n=2$. Assume there is a centered normalized $3$-magic
square in $C_{n}^{3}\leq C_{n}^{k}$ given Corollary 2, and use the notation in
that corollary for its form. Since every nontrivial element in $C_{2}^{k}$ is
an involution, $a=a^{-1}$, which violates distinctness. Hence, $C_{2}^{k}$ is
not $3$-magic for any $k$. $\square$

Proposition 2.15. The group $C_{n}\times C_{n^{k}}$, with $k\geq 3$, is
$3$-magic if and only if $n\geq 2$.

Proof. The fact that $C_{n}\times C_{n^{3}}$ is $3$-magic for $n\geq 3$
follows from Proposition 2 after recognizing that $C_{n^{3}}\leq C_{n}\times
C_{n^{3}}$.

For $C_{2^{3}}\times C_{2}=\left<x,y\right>$, we have that

$\displaystyle\begin{bmatrix}xy&x^{5}&x^{2}y\\\ x&1&x^{7}\\\
x^{6}y&x^{3}&x^{7}y\end{bmatrix}$

forms a magic square with a common product of $1$. For $n=2$ and $k\geq 4$, we
can then use Proposition 2 in a similar manner as was previously done.
$\square$

The following proposition completes the search for abelian $3$-magic groups of
odd order and leaves us only with those of even order.

Proposition 2.16. All abelian groups of odd order $n$ with $n\geq 9$ are
$3$-magic.

Proof. Let $G$ be a finite abelian group of odd order with $|G|=n\geq 9$. By
the Fundamental Theorem of Finite Abelian Groups, we know that
$G=C_{n_{1}}\times C_{n_{2}}\times\ldots\times C_{n_{\ell}}$, where $\ell$ is
the rank of $G$. If $\ell=1$, then $G$ is $3$-magic by Proposition 2. Suppose
then that $\ell>1$. Since $\ell$ is the rank of $G$, by the Chinese Remainder
Theorem, there must be a $p|n_{i}$ for $i\in\\{1,2,\ldots,\ell\\}$ where $p$
is an odd prime. Thus $C_{p}^{\ell}\leq G$ and since $p\geq 3$, $G$ is
$3$-magic by Propositions 2, 2, and 2. $\square$

We now turn our attention to abelian groups of even order.

Proposition 2.17. The group $C_{2}^{k}\times C_{4}$ is not $3$-magic for any
$k$.

Proof. Let $C_{2}^{k}\times C_{4}=<x_{1},x_{2},\ldots,x_{k},y>$, where
$|x_{i}|=2$ (that is, the order of $x_{i}$ is 2) for all $i$ and $|y|=4$.
Assume there is a centered normalized $3$-magic square in $C_{2}^{k}\times
C_{4}$ by Corollary 2, and use the notation in that corollary for its form.
Since $a$ and $b$ cannot be involutions, $|a|=|b|=4$. Then $a=X_{1}y^{\alpha}$
and $b=X_{2}y^{\beta}$, where $X_{1}=x_{i_{1}}x_{i_{2}}\ldots x_{i_{l}}$,
$X_{2}=x_{j_{1}}x_{j_{2}}\ldots x_{j_{m}}$, and $\alpha,\beta\in\\{1,3\\}$.

Case 1 ($\alpha=\beta$):

Without loss of generality, we may assume that $\alpha=\beta=1$. Then
$a^{-1}b=(X_{1}y^{3})(X_{2}y)=X_{1}X_{2}=(X_{1}y)(X_{2}y^{3})=ab^{-1}$,
contradicting distinctness.

Case 2 ($\alpha\neq\beta$):

Without loss of generality, we may assume that $\alpha=1$ and $\beta=3$. Then
$a^{-1}b^{-1}=(X_{1}y^{3})(X_{2}y)=X_{1}X_{2}=(X_{1}y)(X_{2}y^{3})=ab$,
contradicting distinctness. $\square$

In a very similar manner as was previously done, we can determine another
family of groups that can be omitted from the search.

Proposition 2.18. The group $C_{2}^{k}\times C_{3}$ is not $3$-magic for any
$k$.

Proof. This is true for $k=1$ since $|C_{6}|<9$. Let $k\geq 2$, and let
$C_{2}^{k}\times C_{3}\cong C_{2}^{k-1}\times
C_{6}=<x_{1},x_{2},\ldots,x_{k-1},y>$, where $|x_{i}|=2$ for all $i$ and
$|y|=6$. Assume there is a centered normalized $3$-magic square in
$C_{2}^{k-1}\times C_{6}$ by Corollary 2, and use the notation used in that
corollary for its form. Since $a$ and $b$ cannot be involutions,
$|a|,|b|\in\\{3,6\\}$. Then $a=X_{1}y^{\alpha}$ and $b=X_{2}y^{\beta}$, where
$X_{1}=x_{i_{1}}x_{i_{2}}\ldots x_{i_{l}}$, $X_{2}=x_{j_{1}}x_{j_{2}}\ldots
x_{j_{m}}$, and $\alpha,\beta\in\\{1,2,4,5\\}$.

Case 1 ($\alpha=\beta$):

Then
$a^{-1}b=(X_{1}y^{-\alpha})(X_{2}y^{\alpha})=X_{1}X_{2}=(X_{1}y^{\alpha})(X_{2}y^{-\alpha})=ab^{-1}$,
contradicting distinctness.

Case 2 ($\beta\equiv_{6}-\alpha$):

Then
$a^{-1}b^{-1}=(X_{1}y^{-\alpha})(X_{2}y^{\beta})=X_{1}X_{2}=(X_{1}y^{\alpha})(X_{2}y^{-\alpha})=ab$,
contradicting distinctness.

Case 3 ($\\{\alpha,\beta\\}=\\{1,2\\}$):

Without loss of generality, we may assume $\alpha=1$ and $\beta=2$. Then
$a^{-1}b^{-1}=(X_{1}y^{5})(X_{2}y^{4})=X_{1}y^{3}X_{2}=(X_{1}y)(X_{2}y^{2})=ab$,
contradicting distinctness.

Case 4 ($\\{\alpha,\beta\\}=\\{1,4\\}$):

Without loss of generality, we may assume $\alpha=1$ and $\beta=4$. Then
$a^{-1}b=(X_{1}y^{5})(X_{2}y^{4})=X_{1}y^{3}X_{2}=(X_{1}y)(X_{2}y^{2})=ab^{-1}$,
contradicting distinctness.

Case 5 ($\\{\alpha,\beta\\}=\\{2,5\\}$):

Without loss of generality, we may assume $\alpha=2$ and $\beta=5$. Then
$a^{-1}b=(X_{1}y^{4})(X_{2}y^{5})=X_{1}y^{3}X_{2}=(X_{1}y^{2})(X_{2}y)=ab^{-1}$,
contradicting distinctness.

Case 6 ($\\{\alpha,\beta\\}=\\{4,5\\}$):

Without loss of generality, we may assume $\alpha=4$ and $\beta=5$. Then
$a^{-1}b^{-1}=(X_{1}y^{2})(X_{2}y)=X_{1}y^{3}X_{2}=(X_{1}y^{4})(X_{2}y^{5})=ab$,
contradicting distinctness. $\square$

The next proposition may seem less impactful than the previous ones, but its
usefulness will be seen in the next section. It turns out knowing whether
$C_{4}\times C_{8}$ is $3$-magic or not is an exceptional case in the proving
of the characterization theorem; the next proposition handles this case.

Proposition 2.19. The group $C_{4}\times C_{8}$ is $3$-magic.

Proof. Let $C_{4}\times C_{8}=\left<x,y\right>$, where $|x|=4$ and $|y|=8$.
Then

$\displaystyle\begin{bmatrix}y&x^{3}y^{6}&xy\\\ x&1&x^{3}\\\
x^{3}y^{7}&xy^{2}&y^{7}\end{bmatrix}$

is a magic square with common product $1$. $\square$

##  3 A Characterization Theorem

We now present the characterization of finitely generated abelian $3$-magic
groups, which is built upon many of the previous results of narrower scope.

Theorem 3.1. (Characterization of the of Finitely Generated Abelian $3$-magic
Groups)

Let $G$ be a finitely generated abelian group. If $G$ is infinite, then it is
$3$-magic. If $|G|=n$, then we have the following:

  1. 1.

If $n$ is odd, we know $G$ is $3$-magic if and only if $n\geq 9$.

  2. 2.

If $n$ is even, we can write its Sylow-2-subgroup in the form
$C_{2}^{\alpha_{1}}\times C_{2^{2}}^{\alpha_{2}}\times
C_{2^{3}}^{\alpha_{3}}\times\ldots\times C_{2^{l}}^{\alpha_{l}}$, which lets
us say that:

    1. (a)

If $G$ is a $2$-group, then

      1. i.

if $\alpha_{i}\neq 0$ for some $i\geq 4$, we have $G$ is $3$-magic, else

      2. ii.

if $\alpha_{2}\geq 2$ or if $\alpha_{3}\geq 2$, we have $G$ is $3$-magic, or
else

      3. iii.

$G$ is $3$-magic if and only if ($\alpha_{1}\neq 0$ or $\alpha_{2}=1$) and
$\alpha_{3}=1$.

    2. (b)

If $G$ is not a $2$-group, then

      1. i.

if $p|n$, where $p\geq 5$, we have that $G$ is $3$-magic or else

      2. ii.

if there is no prime $p\geq 5$ such that $p|n$, then

        1. A.

if $\alpha_{i}\neq 0$ for some $i\geq 2$, we know $G$ is $3$ magic or else

        2. B.

$G$ is $3$-magic if and only if $9|n$.

Proof. Let $G$ be a finitely generated abelian group. If $G$ is infinite then
$G$ is $3$-magic by Proposition 2, so let $G$ be an abelian group with
$|G|=n$. Then we have:

  1. 1.

If $n$ is odd, then $G$ is $3$-magic if and only if $n\geq 9$ by Proposition 2

  2. 2.

If $n$ is even, write its Sylow-2-subgroup in the form
$C_{2}^{\alpha_{1}}\times C_{2^{2}}^{\alpha_{2}}\times
C_{2^{3}}^{\alpha_{3}}\times\ldots\times C_{2^{l}}^{\alpha_{l}}$.

    1. (a)

If $G$ is a $2$-group, then

      1. i.

if $\alpha_{i_{0}}\neq 0$ for some $i_{0}\geq 4$, then $C_{2^{i_{0}}}\leq G$,
which is $3$-magic by Proposition 2.

      2. ii.

if $\alpha_{i}=0$ for all $i\geq 4$, then we know that if $\alpha_{2}\geq 2$
or $\alpha_{3}\geq 2$, then $C_{4}^{2}\leq G$ or $C_{8}^{2}\leq G$, both of
which are $3$-magic by Proposition 2. Suppose then that $\alpha_{i}=0$ for all
$i\geq 4$, $\alpha_{2}\leq 1$, and $\alpha_{3}\leq 1$.

      3. iii.

Otherwise, we have the following:

        1. A.

if $\alpha_{1}\neq 0$ and $\alpha_{3}=1$, we have $C_{2}\times C_{8}\leq G$,
which is $3$-magic by Proposition 2.

        2. B.

if $\alpha_{2}=1$ and $\alpha_{3}=1$, then $C_{4}\times C_{8}\leq G$, which
was shown to be $3$-magic in Proposition 2.

        3. C.

if $\alpha_{3}=0$, then $G$ is not $3$-magic by Propositions 2 and 2.

    2. (b)

If $G$ is not a $2$-group, then we can say that:

      1. i.

if $p|n$, where $p\geq 5$, then $C_{2p}\leq G$, which is $3$-magic by
Proposition 2.

      2. ii.

if there is no prime $p\geq 5$ such that $p|n$, then $3|n$ in order for $G$ to
not be a $2$-group. It follows that:

        1. A.

if $\alpha_{i_{0}}\neq 0$ for some $i_{0}\geq 2$, then $C_{2^{i_{0}}(3)}\leq
G$, which is $3$ magic by Proposition 2.

        2. B.

if $\alpha_{i}=0$ for all $i\geq 2$, then if $9|n$, we have $C_{9}\leq G$ or
$C_{3}^{2}\leq G$, both of which are $3$-magic by Proposition 2 and 2,
respectively, and if $9\nmid n$, then $G$ is not $3$-magic by Proposition 2.

$\square$

##  4 Nonabelian Groups

We now briefly turn our attention to nonabelian groups and first present some
sufficient conditions for $3$-magic groups.

Proposition 4.1. Let $G$ be a group with $|G|=n$. If any of the following are
true, then $G$ is $3$-magic.

  1. 1.

There is a prime $p\geq 11$ such that $p|n$.

  2. 2.

There is a prime $p\neq 2$ such that $p^{2}|n$.

Proof.

  1. 1.

By Cauchy’s Theorem, $C_{p}\leq G$, which is $3$-magic by Propositon 2.

  2. 2.

As a consequence of the Sylow Theorems, $H\leq G$, where $|H|=p^{2}$. This
means that $H\cong C_{p^{2}}$ or $H\cong C_{p}^{2}$, both of which are
$3$-magic by Proposition 2.

$\square$

Next we present a nonabelian group that is $3$-magic but not covered by the
previous proposition.

Example 4.2. When the homomorphism is acting faithfully, $C_{7}\rtimes C_{3}$
is $3$-magic.

Proof. When the homomorphism is acting faithfully, we can write $C_{7}\rtimes
C_{3}=\left<a,b\mid a^{7}=b^{3}=1,bab^{-1}=a^{4}\right>$. Then

$\displaystyle\begin{bmatrix}a&ab&a^{3}b^{2}\\\ a^{2}b^{2}&1&a^{6}b\\\
a^{2}b&a^{5}b^{2}&a^{6}\end{bmatrix}$

is a magic square with magic product $1$. $\square$

##  5 Summary

In this paper, we defined a magic group by likening it to a magic square using
the language of groups. We were then able to characterize the finitely
generated abelian $3$-magic groups. Beyond this, we also discussed whether
some nonabelian groups are $3$-magic.

##  6 Future Endeavors

There are multiple topics of study for future investigations. For example, we
could explore “magic cubes” or “magic tesseracts” in groups, potentially as an
extension of the concept briefly touched on in [4]. Another idea is to
investigate whether various structures (e.g. groups, graphs, etc.) have
3-magic automorphism groups. Finally, in our future study, we would like to
find groups that are $n$-magic for $n>3$.

## References

  * [1] Onno Cain  “Gaussian Integers, Rings, Finite Fields, and the Magic Square of Squares”, 2019  arXiv:1908.03236 [math.RA]
  * [2] “Magic Square of Squares”  URL: https://www.mathpages.com/home/kmath417/kmath417.htm
  * [3] Numberphile  “The Parker Square - Numberphile”  YouTube  URL: https://www.youtube.com/watch?v=aOT_bG-vWyg
  * [4] Wen Yihui; Hugo Sun  “Note on Magic Squares and Magic Cubes on Abelian Groups”  In _Journal of Mathematical Research & Exposition_ 17.2, 1997, pp. 176–178