# What is this subgroup of $\mathfrak S_{12}$?

On some occasion I was gifted a calendar. It displays a math quizz every day of the year. Not really exciting in general, but at least one of them let me raise a group-theoretic question.

The quizz: consider an hexagon where the vertices and the middle points of the edges are marked, as in the figure

$\hskip11em$

One is asked to place the numbers $1,2,3,4,5,6,8,9,10,11,12,13$ (mind that $7$ is omitted) on points $a,\ldots,\ell$, in such a way that the sum on each edge equals $21$. If you like, you may search a solution, but this is not my question.

Of course the solution is non unique. You may apply any element of the isometry group of the hexagon. A little subtler is the fact that the permutation $(bc)(ef)(hi)(kl)(dj)$ preserves the set of solutions (check this).

Question. What is the invariance group of the solutions set ? Presumably, it is generated by the elements described above. What is its order ? Because it is not too big, it must be isomorphic to a known group. Which one ?

• What do you mean by edge? – Mariano Suárez-Álvarez Nov 12 '10 at 16:50
• I assume edges are subsets of the vertices, and that you want to distribute the numbers among the vertices so that the sum of the numbers corresponding to the vertices of each edge sum up to 14. But then 13 can only be accompanied in an edge by 1, so it belongs to exactly one edge... – Mariano Suárez-Álvarez Nov 12 '10 at 17:04
• Grumblh... my English is good enough for PDEs and for matrices, but not for geometry. By "edge", I meant a segment joining two consecutive vertices of the hexagone. There is one point at each vertex, and one point between two consecutive vertices. The sum of the numbers allocated to two consecutive vertices and to their middle point must be $21$. – Denis Serre Nov 12 '10 at 17:20
• You wrote $14$ in the question, not $21$ :) – Mariano Suárez-Álvarez Nov 12 '10 at 17:22
• Well, I has assume that since that was impossible you had something stranger in mind for "edge"... I wonder what the people who voted the question understood! :) – Mariano Suárez-Álvarez Nov 12 '10 at 17:27

The invariance group of the solutions set can be given a geometric interpretation as follows. Note that $\mathfrak{S}_4 \times \frac{\mathbf{Z}}{2\mathbf{Z}}$ is none other than the group of isometries of the cube.

It is known that if one cuts a cube by the bisecting plane of a space diagonal, the cross-section is a regular hexagon (see the picture at the middle of this page). The vertices of this hexagon are midpoints of (some) edges of the cube. Let $X$ be the set of corners and middles of this hexagon (it has cardinality $12$). Let us consider the following bijection between $X$ and the set $E$ of edges of the cube : if $[AB]$ is a side of the hexagon, with midpoint $M$, we map $A$ (resp. $B$) to the unique edge $e_A$ (resp. $e_B$) in $E$ containing it, and we map $M$ to the unique edge $e_M \in E$ such that $e_A$, $e_B$ and $e_M$ meet at a common vertex of the cube.

Given any solution of the initial problem, we can label the edges of the cube using the above bijection. This labelling has the following nice property : the sum of three edges meeting at a common vertex is always 21. Proof : by construction, six of these eight summing conditions are satisfied. The remaining two conditions read $b+f+j=d+h+\ell=21$ using Denis' notations, and are implied by the first six conditions.

So we found an equivalent ($3$-dimensional) formulation of the problem, namely labelling the edges of a cube. It is now clear that the symmetry group of the cube acts on the set of solutions. It remains to prove that the solution is unique up to isometry, which can be done by hand, here is how I did it : note that only two possible sums involve $1$ (resp. $13$), namely $1+8+12$ and $1+9+11$ (resp. $2+6+13$ and $3+5+13$). Therefore $1$ and $13$ must sit on opposite edges. Then $4$ and $10$ must sit on the unique edges which are parallel to $1$ and $13$. It is the easy to complete the cube.

The resulting labelling has some amusing properties For example, the sum of edges of a given face is always $28$. The sum of two opposite edges is always $14$. Finally, the sum of edges along a cyclohexane-like circuit is always $42$.

• +1. This is awesomely beautiful! – Alex B. Nov 17 '10 at 13:49
• Et +1 pour la communication entre membres d'un même laboratoire sur MO ! – Maxime Bourrigan Nov 17 '10 at 17:57
• @Maxime. Indeed. Discussing this problem at lunch helped much finding the main idea. This shows to me the importance of discussions with colleagues. – François Brunault Nov 17 '10 at 22:22
• Your construction tells that the graph of the benzene ($C_6H_6$) can be folded over the cube. To do so, label the hydrogens $1,\ldots,6$ clockwise. Then pull upward the vertices $1,3,5$ and merge them. Likewise, pull downward the vertices $2,4,6$ and merge them. You obtain a (combinatorial) cube. Of course, this does not have a chemical counterpart. On the one hand, benzene is a rigid molecule. On the other hand, the bonds $C\sim C$ are not equivalent to the bonds $C-H$. For instance, their lengths are different. – Denis Serre Nov 20 '10 at 8:30
• @Denis. This is a nice way to visualize the construction. Regarding benzene, we could also say that it's a planar molecule (according to Wikipedia). The shape of cyclohexane ($C_6 H_{12}$) looks more like a cube, but it has too many hydrogen atoms, and the $C-C-C$ angles are 109°, not 90°. Note also that cyclohexane has two conformations, namely "chair" (more stable) and "boat". But the sum of edges along a "boat-like" circuit on the cube is not constant. – François Brunault Nov 20 '10 at 9:50

It is quite easy to prove that the group is exactly what you wrote. It is enough to show that, up to the action of an element of that group, the unique solution is $13.3.5.4.12.8.1.11.9.10.2.6$. First, one may assume of course $a=13$. Since the only way to decompose $8$ is $2+6$ and $3+5$, one might also assume, up to elements in the group, that $b=3$ and $c=5$. Clearly $\{l,k\}=\{2,6\}$.

An elementary calculation shows that the sums $b+f+j$ and $d+h+l$ equal $21$, so actually each of the $12$ vertices belongs to exactly two sets of $3$ vertices with sum $21$. Now, let's see where $12$ may be. Its complement to $21$ is $9$, which has $3$ writings : $1+8$, $4+5$ and $3+6$. The last one can not occur since $3$ and $6$ are already placed. It is then clear that one must have $d=4$ and $e=12$. From here it is easy to fill out the hexagone: Since $l+d+h=21$, $l$ can not be $2$, so $l=6$ and $h=11$. Further, $\{f,g\}=\{1,8\}$. If $f=1$, the sum $f+b+j$ can not be $21$. So $f=8$, $g=1$, and finally $i=9$.

It's not very elegant, but it works...

• It looks like you are assuming 13 and 12 lie on corners of the hexagon. – S. Carnahan Nov 12 '10 at 18:07
• Yes, for 13 is clear, since some elements of Denis' group exchange corners and middles. For 12 it is a consequence of my argument: if it were in the middle, i.e. d=12, then either l=6, so h=5, contradiction, or l=2, so h=7, contradiction. – Andrei Moroianu Nov 12 '10 at 18:10
• I agree that my argument is not explained in detail, I am new on this site and I have no experience in writing down in real time... – Andrei Moroianu Nov 12 '10 at 18:15
• Oh, I had not noticed the exchange. +1. – S. Carnahan Nov 12 '10 at 18:24
• @Andrei. Your argument is OK. I had to check the impossibility of placing $12$ on $d$, but this is OK on MO. – Denis Serre Nov 13 '10 at 9:12

Here is the MAGMA code to generate your group:

G:=sub<Sym(12)|(1,3,5,7,9,11)*(2,4,6,8,10,12), (2,12)*(3,11)*(4,10)*(5,9)*(6,8), (2,3)*(5,6)*(8,9)*(11,12)*(4,10)>;


I have a small function written by Tim Dokchitser that recognises direct and semi-direct products of standard groups like cyclic, symmetric, dihedral, etc. The group you have described is isomorphic to $C_2\times {\mathfrak S}_4$. The unique normal subgroup of order 2 is generated by a reflection in the centre of the hexagon, so its non-trivial element is given by (a,g)(b,h)(c,i)(d,j)(e,k)(f,l).

The 4-cycles are somewhat harder to visualise. One of them is (a,d,g,j)(b,c,e,f)(h,i,k,l). The copy of ${\mathfrak S}_4$ that this is contained in (there are three normal subgroups isomorphic to ${\mathfrak S}_4$) is generated by the following elements:

• (b, l)(c, k)(d, j)(e, i)(f,h) (reflection in the axis through a and g)
• (a, i, e)(b, j, f)(c, k, g)(d, l, h) (counter-clockwise rotation by $2\pi/3$)
• (a, g)(b, e)(c, f)(d, j)(h, k)(i, l) (square of the above 4-cycle)
• (a, j)(b, k)(c, i)(d, g)(e, h)(f, l) ( a weird thing somewhat similar to the previous one)

• Even though I have never seen a line of MAGMA code, I can understand this one. However, I am not sure to understand the result $C_2\times S_4$. Do you mean $\mathfrak S_4$ ? If so, how do you visualize an action of $\mathfrak S_4$ here ? What do you mean by "generated by a reflection in the central point of the hexagon" ? Finally, what kind of output message do you get from MAGMA ? – Denis Serre Nov 12 '10 at 16:58
• @Andrei and Alex. You both deserve $+1$ for your answers. But I am embarrassed to choose which one to accept as the answer to my question. Alex identified completely the group from the generators I gave, but Andrei's analysis shows that this is the invariance group of the solution set (every solution is, up to the action of the group, that one he calculated). – Denis Serre Nov 13 '10 at 9:25