Congruence lattice of direct power of Steiner seven element quasigroup

Question

The Steiner seven element quasigroup is the algebra $\mathbf{S}_7 = \langle S_7, \cdot \rangle$, where $S_7 = \{ 1,2,3,4,5,6,7 \}$, and, up to isomorphism, its multiplication table is the one given here.

The defining identities for the variety of Steiner quasigroups are $$x \cdot x \approx x, \quad x \cdot y \approx y \cdot x, \quad x \cdot (x \cdot y) \approx y.$$ From these, it is possible to deduce that members of this variety have uniform congruences, that is, for each congruence relation, all the classes have the same cardinality.
One consequence of the uniformity of congruences is that, if a finite algebra $\mathbf{A}$ has a prime number of elements, then that algebra is simple (that is, its only congruences are $\Delta_A = \{ \langle a,a \rangle : a \in A \}$ and $\nabla_A = A^2$).
So in particular, $\mathbf{S}_7$ is simple.
Now, I' like to prove that the congruence lattice of $\mathbf{S}_7^2$ has precisely four elements (and I know, indirectly, that this is so; I could say why, if someone finds it can be helpful).
The elements would be $$\Delta_{S_7^2}, \ker\pi_1, \ker\pi_2, \nabla_{S_7^2},$$ where $\pi_i$ is the $i$th projection of $\mathbf{S}_7^2$ onto $\mathbf{S}_7$.
These four elements are certainly members of the congruence lattice, and if $\theta$ is another congruence, given its uniformity and the fact that the divisors of $|S_7^2| = 49$ are $1$, $7$ and $49$, then $\theta$ ought to have seven congruence classes, each with seven elements;
from this, it follows that $\theta \cap \ker\pi_i = \Delta_{S_7^2}$ and $\theta \vee \ker\pi_i = \nabla_{S_7^2}$.

A consequence of these later identities and the uniformity of congruences is that, if we lay the elements of $S_7^2$ in a table with seven rows and seven columns, such that, in each each row the first coordinate is constant and in each column the second coordinate is constant, then each congruence class of $\theta$ picks precisely a member from each row and each column (thus, no two members of the same row or column).
Then I tried to derive some contradiction from this (using the product to conclude that eventually, we would get two elements of the same row or column in the same congruence class), and that might be possible, but after a few pages of calculations I thought there must be a better way, a more clever way.

Any hints? Thanks in advance.

Answer 1

Background comments. The $7$-element Steiner quasigroup $\mathbb S$ is built out of the Fano plane, $\textrm{PG}(2,2)$. This is the projective plane over the $2$-element field, as such it has seven points and seven $3$-element lines. The multiplication of the Steiner quasigroup $\mathbb S = \langle \textrm{PG}(2,2); \cdot\rangle$ is defined so that $p\cdot p = p$ for every point $p$ (which says that $\mathbb S$ is idempotent), and if $p\neq q$ the product $p\cdot q$ = is the third point on the line through $p$ and $q$. That is, $\{p, q, p\cdot q\}$ is either a singleton or a line.

$\textrm{PG}(2,2)$ and $\mathbb S$ have the same automorphism group: both automorphism groups are equal to the group of permutations of $\textrm{PG}(2,2)$ which map points to points, lines to lines and preserve incidence. The equality of these groups derives from the fact that the lines are exactly the subsets of the form $\{p, q, p\cdot q\}$, $p\neq q$. Thus $\textrm{Aut}(\mathbb S)= \textrm{Aut}(\textrm{PG}(2,2))= \textrm{PGL}(3,2)$ = the nonabelian simple group of order $168$. End of background comments.

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The fact that $\mathbb S\times \mathbb S$ has no skew congruence follows from:

Observation. If $\mathbb T$ is a finite simple algebra in an idempotent, congruence uniform variety, and $\mathbb T\times \mathbb T$ has a skew congruence, then $\textrm{Aut}(\mathbb T)$ has a regular normal subgroup.

[This observation establishes that $\mathbb S\times \mathbb S$ has no skew congruence, since a regular normal subgroup of $\textrm{Aut}(\mathbb S)=\textrm{PGL}(3,2)$ would be of order $|S|=7$, and a simple group of order $168$ cannot have a normal subgroup of order $7$.]

Proof of the Observation. Let $\Theta$ be a skew congruence on $\mathbb T\times \mathbb T$. Let the partition of $T\times T$ into $\Theta$-classes be $\Pi_{\Theta}=\{\alpha,\beta,\gamma,\ldots\}$. As noted in the original question, any class of $\Theta$ is a subset of $T\times T$ that contains one element of every row and column, that is, each $\alpha\in\Pi_{\Theta}$ is the graph of a permutation of $T$. Because $\mathbb T\times \mathbb T$ is idempotent, congruence classes are subalgebras, so each $\alpha\in\Pi_{\Theta}$ is in fact the graph of an automorphism of $\mathbb T$. (Here I am using that an automorphism of $\mathbb T$ is nothing other than a permutation of $T$ whose graph is a subalgebra of $\mathbb T\times \mathbb T$.)

If $\lambda,\mu\in\textrm{Aut}(\mathbb T)$, then $\lambda\times \mu:(x,y)\mapsto (\lambda(x),\mu(y))$ is an automorphism of $\mathbb T\times \mathbb T$, and $(\lambda\times\mu)(\Theta)$ is another skew congruence of $\mathbb T\times \mathbb T$. Its associated partition is $$ \Pi_{(\lambda\times\mu)(\Theta)} = \{\mu\alpha\lambda^{-1},\mu\beta\lambda^{-1},\ldots\}. $$ In particular, $\Theta'=(\alpha\times\textrm{id})(\Theta)$ is another skew congruence of $\mathbb T\times \mathbb T$, and its partition is $\Pi_{\Theta'} =\{\textrm{id},\beta\alpha^{-1},\ldots\}$. This partition contains a class that is (the graph of) $\textrm{id}$, hence $\Theta'$ is another skew congruence of $\mathbb T\times \mathbb T$, but it is nicer in the sense that it has the diagonal as a class. Let's rewrite its partition as $\Pi_{\Theta'}=\{\textrm{id}, \alpha',\beta',\ldots\}$. I complete the argument by showing that the set of automorphisms $\Pi_{\Theta'}=\{\textrm{id}, \alpha',\beta',\ldots\}$ is a regular normal subgroup of $\textrm{Aut}(\mathbb T)$.

The partition associated to $\Theta'$ is $\Pi_{\Theta'}=\{\textrm{id}, \alpha',\beta',\ldots\}$, so the partition associated to the skew congruence $(\textrm{id}\times\alpha')(\Theta')$ is $\{\alpha',\alpha'\cdot\alpha', \alpha'\cdot\beta',\ldots\}$, which shares a class $\alpha'$ with $\Pi_{\Theta'}=\{\textrm{id}, \alpha',\beta',\ldots\}$, so by congruence uniformity they are equal. This shows that $\alpha',\beta'\in \Pi_{\Theta'}$ implies $\alpha'\cdot\beta'\in\Pi_{\Theta'}$. Similarly, if $\lambda\in\textrm{Aut}(\mathbb T)$, then the partition associated to the skew congruence $(\lambda\times \lambda)(\Theta')$ is $\{\textrm{id}, \lambda\alpha'\lambda^{-1},\lambda\beta'\lambda^{-1},\ldots\}$, which shares a class $\textrm{id}$ with $\Pi_{\Theta'}$, hence we conclude the partitions are equal. Altogether this shows that the set $\Pi_{\Theta'}=\{\textrm{id}, \alpha',\beta',\ldots\}$ is closed under multiplication and conjugation, so it is a normal subgroup of $\textrm{Aut}(\mathbb T)$. The regularity of this subgroup is equivalent to the fact that the set $\Pi_{\Theta'}= \{\textrm{id}, \alpha',\beta',\ldots\}$ partitions $T\times T$. \\\

Answer 2

This answer uses exercises 1-3 of Chapter 8 of Commutator Theory for Congruence Modular Varieties by Freese and McKenzie. I use the notation $0_A$ for the bottom congruence of an algebra $\mathbf{A}$ and $1_A$ for the top.

First note that $[1_{S_7},1_{S_7}]=1_{S_7}$ since $\mathbf{S}_7$ is not abelian. Thus $\mathrm{Con}(\mathbf{S}_7)$ satisfies $[\theta,\phi]=\theta\wedge\phi$ for all $\theta,\phi\in\mathrm{Con}(\mathbf{S}_7)$. Such algebras are called neutral.

By exercise 2, the product $\mathbf{S}_7\times\mathbf{S}_7$ is neutral as well.

By exercise 1, the congruence lattice $\mathrm{Con}(\mathbf{S}_7\times\mathbf{S}_7)$ is distributive.

Now, if $\theta\in\mathrm{Con}(\mathbf{S}_7\times\mathbf{S}_7)$ and $\eta_1,\eta_2$ are the two projection kernels, then $\theta=\theta\vee(\eta_1\wedge\eta_2)=(\theta\vee\eta_1)\wedge(\theta\vee\eta_2)$. Thus $\theta$ is of the form $\phi\times\psi$ where $\phi,\psi\in\mathrm{Con}(\mathbf{S}_7)$. (This idea is used in the solution to exercise 3.)

Answer 3

Let me start by giving the multiplication table for $\mathbf{S}_7$ for future reference.

\begin{array}{c|ccccccc} \cdot &1 &2 &3 &4 &5 &6 &7\\ \hline 1 &1 &3 &2 &7 &6 &5 &4\\ 2 &3 &2 &1 &6 &7 &4 &5\\ 3 &2 &1 &3 &5 &4 &7 &6\\ 4 &7 &6 &5 &4 &3 &2 &1\\ 5 &6 &7 &4 &3 &5 &1 &2\\ 6 &5 &4 &7 &2 &1 &6 &3\\ 7 &4 &5 &6 &1 &2 &3 &7 \end{array}

The following might not be the ideal solution, but for want of a more elegant one, here it goes...

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Following the reasoning advanced in the post, it is enough to show that if $\theta$ is a congruence of $\mathbf{S}_7^2$ such that $\langle a,b \rangle \theta \langle c,d \rangle$, with $a \neq c$ and $b \neq d$, then $\theta = \nabla_{S_7^2}$.
Since such congruence has to have, in each of its classes, one element from each row and column (thinking of the elements of $\mathbf{S}_7^2$ in a matrix), it is enough to show that $$\Theta(\langle 1,1 \rangle, \langle 2,b \rangle) = \nabla_{S_7^2},$$ whenever $b \neq 1$ (if $b = 1$ this would be $\ker\pi_2$).
So let us see that for each such $b$, we still have two elements of the same column in the same class, thus obtaining a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,2 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,4 \rangle \;\theta\; \langle 2,2 \rangle \cdot \langle 1,4 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,5 \rangle \;\theta\; \langle 2,2 \rangle \cdot \langle 2,5 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,7 \rangle \;\theta\; \langle 3,6 \rangle\\ \langle 3,6 \rangle \;\theta\; \langle 2,7 \rangle \end{cases} \Rightarrow \langle 1,7 \rangle \;\theta\; \langle 2,7 \rangle, $$ a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,3 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,4 \rangle \;\theta\; \langle 2,3 \rangle \cdot \langle 1,4 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,6 \rangle \;\theta\; \langle 2,3 \rangle \cdot \langle 2,6 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,7 \rangle \;\theta\; \langle 3,5 \rangle\\ \langle 3,5 \rangle \;\theta\; \langle 2,7 \rangle \end{cases} \Rightarrow \langle 1,7 \rangle \;\theta\; \langle 2,7 \rangle, $$ a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,4 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,2 \rangle \;\theta\; \langle 2,4 \rangle \cdot \langle 1,2 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,5 \rangle \;\theta\; \langle 2,4 \rangle \cdot \langle 2,5 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,3 \rangle \;\theta\; \langle 3,6 \rangle\\ \langle 3,6 \rangle \;\theta\; \langle 2,3 \rangle \end{cases} \Rightarrow \langle 1,3 \rangle \;\theta\; \langle 2,3 \rangle, $$ a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,5 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,2 \rangle \;\theta\; \langle 2,5 \rangle \cdot \langle 1,2 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,4 \rangle \;\theta\; \langle 2,5 \rangle \cdot \langle 2,4 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,3 \rangle \;\theta\; \langle 3,7 \rangle\\ \langle 3,7 \rangle \;\theta\; \langle 2,3 \rangle \end{cases} \Rightarrow \langle 1,3 \rangle \;\theta\; \langle 2,3 \rangle, $$ a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,6 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,2 \rangle \;\theta\; \langle 2,6 \rangle \cdot \langle 1,2 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,7 \rangle \;\theta\; \langle 2,6 \rangle \cdot \langle 2,7 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,3 \rangle \;\theta\; \langle 3,4 \rangle\\ \langle 3,4 \rangle \;\theta\; \langle 2,3 \rangle \end{cases} \Rightarrow \langle 1,3 \rangle \;\theta\; \langle 2,3 \rangle, $$ a contradiction.

If $\langle 1,1 \rangle \;\theta\; \langle 2,7 \rangle$, then $$ \begin{cases} \langle 1,1 \rangle \cdot \langle 1,2 \rangle \;\theta\; \langle 2,7 \rangle \cdot \langle 1,2 \rangle\\ \langle 1,1 \rangle \cdot \langle 2,6 \rangle \;\theta\; \langle 2,7 \rangle \cdot \langle 2,6 \rangle \end{cases} \Rightarrow \begin{cases} \langle 1,3 \rangle \;\theta\; \langle 3,5 \rangle\\ \langle 3,5 \rangle \;\theta\; \langle 2,3 \rangle \end{cases} \Rightarrow \langle 1,3 \rangle \;\theta\; \langle 2,3 \rangle, $$ a contradiction.

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Update (Trying to follow the comment of Keith Kearnes.)

Let $\mathbf{G}$ be the group of automorphisms of $\mathbf{S}_7$ which fix $1$ and $2$ (and also $3$, since $1 \cdot 2 = 3$).
Then $\mathbf{G}$ acts transitively on $\{4,5,6,7\}$ (we may consider the automorphisms given by $(45)(67)$, $(46)(57)$ and $(47)(56)$).
Now we can reduce the last four cases to a single one, like this:
If $\langle 1,1 \rangle \,\theta\, \langle 2, \phi(4) \rangle$, where $\phi \in G$, then $$\langle 1,1 \rangle \cdot \langle 1,2 \rangle \;\theta\; \langle 2, \phi(4) \rangle \cdot \langle 1,2 \rangle,$$ that is, $\langle 1,3 \rangle \,\theta\, \langle 3, 2 \cdot \phi(4) \rangle$.
On the other hand, $$\langle 1,1 \rangle \cdot \langle 2, 1 \cdot (2 \cdot \phi(4)) \rangle \;\theta\; \langle 2, \phi(4) \rangle \cdot \langle 2, 1 \cdot (2 \cdot \phi(4)) \rangle,$$ that is, $\langle 3, 2 \cdot \phi(4) \rangle \,\theta\, \langle 2, \phi(4) \cdot (1 \cdot (2 \cdot \phi(4))) \rangle$.

Now it is an easily verifiable fact of the Fano plane (see connection with $\mathbf{S}_7$ here; also in the beginning of Keith Kearnes' answer), that if $\{u,v,uv\}$ is a line, and $x \notin \{v,uv\}$, then $u(vx) = (uv)x$, whence $$x(u(vx))=uv.$$ In particular, since $1 \cdot 2 = 3$, we have $x(1(2x))=3$, whenever $x \in S_7 \setminus \{2,3\}$, yielding $\phi(4) \cdot (1 \cdot (2 \cdot \phi(4))) = 3$, and therefore, $\langle 1,3 \rangle \,\theta\, \langle 2,3 \rangle$, a contradiction.