# Schröder T-quasigroups

V.A. Shcherbacov

###### Abstract

We prolong research of Schröder quasigroups and Schröder T-quasigroups.

2000 Mathematics Subject Classification: 20N05

Key words and phrases: quasigroup, loop, groupoid, Schröder quasigroups,
Schröder identity.

##  1 Introduction

This paper is connected in main with the following Belousov problem 1 [2] :
Find a complete characterization of groups isotopic to quasigroups which
satisfy one of the identities: $x(y\cdot yx)=y$, $xy\cdot yx=y$ (Stein 3rd law
[15]), $xy\cdot yx=x$ (Schröder 2nd law [15]).

Some of these identities guarantee that a quasigroup is orthogonal to its
parastrophe [2, 16].

Here we find conditions imposed not only on a group $G$ over which various
Schröder quasigroups are defined, but on the group of automorphisms $Aut\,G$,
over which various T-objects (quasigroups and groupoids) are defined.

Necessary definitions can be found in [1, 4, 13, 16].

######  Definition 1.

Binary groupoid $(Q,\circ)$ is called a left quasigroup if for any ordered
pair $(a,b)\in Q^{2}$ there exist the unique solution $x\in Q$ to the equation
$a\circ x=b$ [1].

######  Definition 2.

Binary groupoid $(Q,\circ)$ is called a right quasigroup if for any ordered
pair $(a,b)\in Q^{2}$ there exist the unique solution $y\in Q$ to the equation
$y\circ a=b$ [1].

######  Definition 3.

A quasigroup $(Q,\cdot)$ with an element $1\in Q$, such that $1\cdot x=x\cdot
1=x$ for all $x\in Q$, is called a loop.

######  Definition 4.

Binary groupoid $(Q,\cdot)$ is called medial if this groupoid satisfies the
following medial identity:

$xy\cdot uv=xu\cdot yv$ (1)

for all $x,y,u,v\in Q$ [1].

We recall

######  Definition 5.

Quasigroup $(Q,\cdot)$ is a T-quasigroup if and only if there exists an
abelian group $(Q,+)$, its automorphisms $\varphi$ and $\psi$ and a fixed
element $a\in Q$ such that $x\cdot y=\varphi x+\psi y+a$ for all $x,y\in Q$
[9].

A T-quasigroup with the additional condition $\varphi\psi=\psi\varphi$ is
medial.

######  Definition 6.

Garrett Birkhoff [3] has defined an equational quasigroup as an algebra with
three binary operations $(Q,\cdot,/,\backslash)$ that satisfies the following
six identities:

$x\cdot(x\backslash y)=y,$ (2)

$(y/x)\cdot x=y,$ (3)

$x\backslash(x\cdot y)=y,$ (4)

$(y\cdot x)/x=y,$ (5)

$x/(y\backslash x)=y,$ (6)

$(x/y)\backslash x=y.$ (7)

Ernst Schröder (a German mathematician mainly known for his work on algebraic
logic) introduced and studied the following identity of generalized
associativity [6]:

$(y\cdot z)\backslash x=z(x\cdot y).$ (8)

See also [7, 8, 10] for detail. Notice, article [10] is well written and very
detailed.

In the quasigroup case the identity (8) is equivalent to the following
identity [14, 16]:

$(y\cdot z)\cdot(z\cdot(x\cdot y))=x$ (9)

If in idempotent quasigroup $(Q;\cdot)$ the identity (9) we put $x=y$, then we
obtain the following standard Schröder identity:

$(x\cdot y)\cdot(y\cdot x)=x.$ (10)

######  Definition 7.

Any quasigroup with the identity (10) is called a Schröder quasigroup.

So we have different objects that have name Schröder. Namely,

(i) the following identity of generalized associativity [6]:

$(y\cdot z)\backslash x=z(x\cdot y)$ (8);

(ii) the Schröder identity of generalized associativity in quasigroups: (9);

(iii) Schröder identity (Schröder 2-nd identity [15]) $(x\cdot y)\cdot(y\cdot
x)=x$ (10);

Identity $(x\cdot y)\cdot(y\cdot x)=y$ is named by Albert Sade [15] as Stein
3-rd identity.

Each of these identities deserves a separate study in the class of groupoids,
left (right) quasigroups, in the classes of quasigroups and of T-quasigroups.

###  1.1  Schröder identity of generalized associativity

It is convenient to call this identity the Schröder identity of generalized
associativity. Often various variants of associative identity, which are true
in a quasigroup, guarantee that this quasigroup is a loop.

It is not so in the case with the identity. We give an example of quasigroup
which is not a loop with the identity (9) [14]. See also [16]. A quasigroup
from this example does not have left and right identity element.

Quasigroups with Schröder identity of generalized associativity are not
necessary idempotent and associative. See the following example.

Table 1

$\begin{array}[]{l|llllllll}\cdot&0&1&2&3&4&5&6&7\\\ \hline\cr
0&1&4&7&0&6&5&2&3\\\ 1&5&2&3&6&0&1&4&7\\\ 2&0&7&4&1&5&6&3&2\\\
3&6&3&2&5&1&0&7&4\\\ 4&4&1&0&7&3&2&5&6\\\ 5&3&6&5&2&4&7&0&1\\\
6&7&0&1&4&2&3&6&5\\\ 7&2&5&6&3&7&4&1&0\\\ \end{array}$

The left cancellation (left division) groupoid with the identity (9) and with
the identity $(x/x=y/y)$ (in a quasigroup this identity guarantees existence
of the left identity element) is a commutative group of exponent two [14].

The similar results are true for the right case [14]. In this case we use the
identity $(x\backslash x=y\backslash y)$.

It is clear that this result is true for any quasigroup with the left or right
identity element.

Notice, any 2-group $(G,+)$ (in such group $x+x=0$ for any $x\in G$) satisfies
Schröder identity of generalized associativity.

###  1.2  Schröder identity $(x\cdot y)\cdot(y\cdot x)=x$

We recall information from [11].

######  Lemma 1.

In any Schröder quasigroup $(Q,\cdot)$ the equality $x\cdot x=y\cdot y$
implies $x=y$, and the equality $x\cdot y=y\cdot x$ implies $x=y$.

###### Proof.

Suppose $(x\cdot x)=(y\cdot y)$. Then from the identity (10) we have
$x=(x\cdot x)\cdot(x\cdot x)=(y\cdot y)\cdot(y\cdot y)=y.$

Suppose $x\cdot y=y\cdot x$. Then we have $y=(x\cdot y)\cdot(y\cdot x)=(y\cdot
x)\cdot(x\cdot y)=x$. ∎

######  Theorem 1.

Necessary condition for the existence of an idempotent Schröder quasigroup
$(Q,\cdot)$ of order v is that v = 0 or = 1 pmod 4 [11].

In the proofs of Theorems 1 are used sufficiently results from [17].

######  Example 1.

Define groupoid $GF(2^{r},\ast)$ over the Galois field $GF(2^{r})$ in the
following way:

$x\ast y=a\cdot x+(a+1)y,$ (11)

where $x,y\in GF(2^{r})$, the element $a$ is a fixed element of the set
$(GF(2^{r}))$, the operations $+$ and $\cdot$ are binary operations of this
field.

The groupoid $(GF(2^{r}),\ast)$ is an idempotent medial Schröder quasigroup
[11].

######  Example 2.

In [11] it is mentioned that do not exist Schröder quasigroups $(Q,\cdot)$ of
order 5. Using Mace 5 we construct quasigroups with Stein 3-rd law of order 5
(left loop and idempotent quasigroup).

$\star$ | 0 | 1 | 2 | 3 | 4
---|---|---|---|---|---
0 | 0 | 1 | 2 | 3 | 4
1 | 2 | 3 | 1 | 4 | 0
2 | 3 | 0 | 4 | 2 | 1
3 | 4 | 2 | 0 | 1 | 3
4 | 1 | 4 | 3 | 0 | 2
$\cdot$ | 0 | 1 | 2 | 3 | 4
---|---|---|---|---|---
0 | 0 | 2 | 3 | 4 | 1
1 | 3 | 1 | 4 | 2 | 0
2 | 4 | 0 | 2 | 1 | 3
3 | 1 | 4 | 0 | 3 | 2
4 | 2 | 3 | 1 | 0 | 4

We remark, in quasigroups from Schröder identity does not follow Stein 3-rd
(quasigroup $(Q,\ast)$) and vice.

In quasigroup $(Q,\ast)$ is true identity (10) and is not true identity (7),
in quasigroup $(Q,\circ)$ is true identity (7) and is not true identity (10).

$\ast$ | 0 | 1 | 2 | 3
---|---|---|---|---
0 | 0 | 1 | 3 | 2
1 | 2 | 3 | 1 | 0
2 | 1 | 0 | 2 | 3
3 | 3 | 2 | 0 | 1
$\circ$ | 0 | 1 | 2 | 3
---|---|---|---|---
0 | 0 | 2 | 3 | 1
1 | 3 | 1 | 0 | 2
2 | 1 | 3 | 2 | 0
3 | 2 | 0 | 1 | 3

###  1.3 Quasigroups with identity $x(y\cdot yx)=y$

Next Cayley tables of quasigroups with identity $x(y\cdot yx)=y$ were obtained
using Mace [12].

*: | 0 | 1 | 2
---|---|---|---
0 | 0 | 1 | 2
1 | 2 | 0 | 1
2 | 1 | 2 | 0
*: | 0 | 1 | 2 | 3
---|---|---|---|---
0 | 0 | 2 | 3 | 1
1 | 1 | 3 | 2 | 0
2 | 2 | 0 | 1 | 3
3 | 3 | 1 | 0 | 2
*: | 0 | 1 | 2 | 3 | 4
---|---|---|---|---|---
0 | 0 | 2 | 4 | 1 | 3
1 | 2 | 1 | 3 | 4 | 0
2 | 4 | 3 | 2 | 0 | 1
3 | 1 | 4 | 0 | 3 | 2
4 | 3 | 0 | 1 | 2 | 4
*: | 0 | 1 | 2 | 3 | 4 | 5 | 6
---|---|---|---|---|---|---|---
0 | 0 | 2 | 3 | 1 | 5 | 6 | 4
1 | 4 | 1 | 6 | 0 | 3 | 2 | 5
2 | 5 | 0 | 2 | 4 | 6 | 1 | 3
3 | 6 | 5 | 0 | 3 | 1 | 4 | 2
4 | 1 | 6 | 5 | 2 | 4 | 3 | 0
5 | 2 | 3 | 4 | 6 | 0 | 5 | 1
6 | 3 | 4 | 1 | 5 | 2 | 0 | 6

*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
0 | 0 | 2 | 4 | 1 | 6 | 3 | 7 | 5
1 | 6 | 1 | 5 | 2 | 0 | 7 | 3 | 4
2 | 7 | 4 | 2 | 5 | 3 | 6 | 0 | 1
3 | 4 | 7 | 0 | 3 | 5 | 1 | 2 | 6
4 | 5 | 3 | 6 | 7 | 4 | 2 | 1 | 0
5 | 2 | 0 | 7 | 6 | 1 | 5 | 4 | 3
6 | 3 | 5 | 1 | 4 | 7 | 0 | 6 | 2
7 | 1 | 6 | 3 | 0 | 2 | 4 | 5 | 7
*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
---|---|---|---|---|---|---|---|---|---
0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
1 | 2 | 0 | 1 | 8 | 6 | 4 | 5 | 3 | 7
2 | 1 | 2 | 0 | 7 | 5 | 6 | 4 | 8 | 3
3 | 4 | 5 | 6 | 0 | 3 | 7 | 8 | 1 | 2
4 | 3 | 7 | 8 | 4 | 0 | 1 | 2 | 5 | 6
5 | 8 | 3 | 7 | 6 | 2 | 0 | 1 | 4 | 5
6 | 7 | 8 | 3 | 5 | 1 | 2 | 0 | 6 | 4
7 | 6 | 4 | 5 | 2 | 8 | 3 | 7 | 0 | 1
8 | 5 | 6 | 4 | 1 | 7 | 8 | 3 | 2 | 0

*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---|---|---
0 | 0 | 2 | 4 | 1 | 5 | 3 | 7 | 9 | 6 | 10 | 8
1 | 8 | 1 | 7 | 10 | 0 | 2 | 9 | 6 | 3 | 5 | 4
2 | 6 | 8 | 2 | 4 | 9 | 0 | 1 | 10 | 5 | 7 | 3
3 | 10 | 6 | 0 | 3 | 1 | 9 | 8 | 4 | 7 | 2 | 5
4 | 7 | 5 | 6 | 0 | 4 | 10 | 3 | 2 | 1 | 8 | 9
5 | 9 | 0 | 3 | 8 | 7 | 5 | 2 | 1 | 10 | 4 | 6
6 | 4 | 10 | 9 | 5 | 8 | 1 | 6 | 3 | 2 | 0 | 7
7 | 5 | 3 | 8 | 2 | 10 | 6 | 4 | 7 | 9 | 1 | 0
8 | 2 | 7 | 10 | 9 | 3 | 4 | 5 | 0 | 8 | 6 | 1
9 | 3 | 4 | 1 | 7 | 6 | 8 | 10 | 5 | 0 | 9 | 2
10 | 1 | 9 | 5 | 6 | 2 | 7 | 0 | 8 | 4 | 3 | 10

*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
---|---|---|---|---|---|---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 7 | 6 | 5 | 8 | 11 | 10 | 9
1 | 0 | 1 | 3 | 2 | 5 | 4 | 7 | 6 | 9 | 8 | 11 | 10
2 | 2 | 3 | 0 | 1 | 6 | 5 | 4 | 7 | 10 | 9 | 8 | 11
3 | 3 | 2 | 1 | 0 | 7 | 6 | 5 | 4 | 11 | 10 | 9 | 8
4 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 0 | 1 | 2 | 3
5 | 7 | 4 | 5 | 6 | 10 | 8 | 11 | 9 | 2 | 0 | 3 | 1
6 | 6 | 7 | 4 | 5 | 9 | 11 | 8 | 10 | 1 | 3 | 0 | 2
7 | 5 | 6 | 7 | 4 | 11 | 10 | 9 | 8 | 3 | 2 | 1 | 0
8 | 8 | 9 | 10 | 11 | 0 | 2 | 1 | 3 | 4 | 5 | 6 | 7
9 | 11 | 8 | 9 | 10 | 1 | 0 | 3 | 2 | 5 | 6 | 7 | 4
10 | 10 | 11 | 8 | 9 | 2 | 3 | 0 | 1 | 6 | 7 | 4 | 5
11 | 9 | 10 | 11 | 8 | 3 | 1 | 2 | 0 | 7 | 4 | 5 | 6

We recall,

######  Lemma 2.

If two algebraic systems, say $A$ and $B$ satisfy an identity, then direct
product of these systems $A\times B$ satisfy this identity [5, 16].

###### Proof.

It follows from Definition 6. ∎

######  Lemma 3.

Quasigroups with identity $x(y\cdot yx)=y$ there exists for any $n$, where

$n\in\\{3^{n_{1}}\cdot 4^{n_{2}}\cdot 5^{n_{3}}\cdot 7^{n_{4}}\cdot
8^{n_{5}}\cdot 11^{n_{6}}\cdot 23^{n_{7}}\\},$ (12)

$n_{i}\in\mathbb{N}$, $i\in\overline{\\{1,7\\}}$.

###### Proof.

This follows from Lemma 2, given in this section examples and Example 5. ∎

###  1.4 Stein 3-rd identity

Next Cayley tables of quasigroups with identity $xy\cdot yx=y$ were obtained
using Mace [12].

*: | 0 | 1 | 2 | 3
---|---|---|---|---
0 | 0 | 1 | 3 | 2
1 | 2 | 3 | 1 | 0
2 | 1 | 0 | 2 | 3
3 | 3 | 2 | 0 | 1
*: | 0 | 1 | 2 | 3 | 4
---|---|---|---|---|---
0 | 0 | 1 | 2 | 3 | 4
1 | 2 | 3 | 1 | 4 | 0
2 | 3 | 0 | 4 | 2 | 1
3 | 4 | 2 | 0 | 1 | 3
4 | 1 | 4 | 3 | 0 | 2
*: | 0 | 1 | 2 | 3 | 4 | 5
---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 5
1 | 0 | 1 | 3 | 2 | 5 | 4
2 | 2 | 3 | 4 | 5 | 0 | 1
3 | 3 | 2 | 5 | 4 | 1 | 0
4 | 4 | 5 | 0 | 1 | 2 | 3
5 | 5 | 4 | 1 | 0 | 3 | 2

*: | 0 | 1 | 2 | 3 | 4 | 5 | 6
---|---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 6 | 5
1 | 0 | 1 | 3 | 2 | 5 | 4 | 6
2 | 2 | 3 | 0 | 6 | 1 | 5 | 4
3 | 3 | 2 | 4 | 5 | 6 | 0 | 1
4 | 5 | 6 | 1 | 4 | 0 | 2 | 3
5 | 4 | 5 | 6 | 1 | 2 | 3 | 0
6 | 6 | 4 | 5 | 0 | 3 | 1 | 2
*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 7 | 6 | 5
1 | 0 | 1 | 3 | 2 | 5 | 4 | 7 | 6
2 | 2 | 3 | 0 | 1 | 6 | 5 | 4 | 7
3 | 3 | 2 | 1 | 0 | 7 | 6 | 5 | 4
4 | 4 | 5 | 6 | 7 | 0 | 1 | 2 | 3
5 | 7 | 4 | 5 | 6 | 1 | 2 | 3 | 0
6 | 6 | 7 | 4 | 5 | 2 | 3 | 0 | 1
7 | 5 | 6 | 7 | 4 | 3 | 0 | 1 | 2

*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
---|---|---|---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 8 | 7 | 6 | 5
1 | 0 | 1 | 3 | 2 | 5 | 4 | 8 | 7 | 6
2 | 2 | 3 | 0 | 1 | 6 | 5 | 4 | 8 | 7
3 | 3 | 2 | 1 | 0 | 7 | 6 | 5 | 4 | 8
4 | 4 | 5 | 6 | 7 | 8 | 1 | 0 | 3 | 2
5 | 8 | 4 | 5 | 6 | 0 | 7 | 3 | 2 | 1
6 | 7 | 8 | 4 | 5 | 1 | 2 | 6 | 0 | 3
7 | 6 | 7 | 8 | 4 | 2 | 3 | 1 | 5 | 0
8 | 5 | 6 | 7 | 8 | 3 | 0 | 2 | 1 | 4
*: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---|---
0 | 1 | 0 | 2 | 3 | 4 | 7 | 8 | 5 | 6 | 9
1 | 0 | 1 | 3 | 2 | 5 | 4 | 9 | 8 | 7 | 6
2 | 2 | 3 | 0 | 1 | 6 | 5 | 4 | 9 | 8 | 7
3 | 3 | 2 | 1 | 0 | 7 | 6 | 5 | 4 | 9 | 8
4 | 4 | 5 | 6 | 7 | 8 | 9 | 0 | 1 | 2 | 3
5 | 8 | 7 | 9 | 6 | 0 | 1 | 2 | 3 | 4 | 5
6 | 9 | 8 | 7 | 5 | 1 | 2 | 3 | 6 | 0 | 4
7 | 5 | 4 | 8 | 9 | 2 | 3 | 6 | 7 | 1 | 0
8 | 6 | 9 | 4 | 8 | 3 | 0 | 7 | 2 | 5 | 1
9 | 7 | 6 | 5 | 4 | 9 | 8 | 1 | 0 | 3 | 2

######  Lemma 4.

Quasigroups with identity $xy\cdot yx=y$ there exists for any $n$, where

$n\in\\{4^{n_{1}}\cdot 5^{n_{2}}\cdot 6^{n_{3}}\cdot 7^{n_{4}}\cdot
8^{n_{5}}\cdot 9^{n_{6}}\cdot 10^{n_{7}}\cdot 13^{n_{8}}\cdot 17^{n_{9}}\cdot
29^{n_{10}}\\},$ (13)

$n_{i}\in\mathbb{N}$, $i\in\overline{\\{1,10\\}}$.

###### Proof.

This follows from Lemma 2, given in this paragraph examples and Examples 9,
10, 11. ∎

##  2 Schröder identities in T-quasigroups

###  2.1 Schröder identity of generalized associativity in T-quasigroups

######  Theorem 2.

In T-quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$ Schröder
identity of generalized associativity is true if and only if $\varphi
x=\psi^{-2}x$, $\varepsilon=\varphi^{7}$, $\varepsilon=\psi^{14}$,
$\varphi\psi z+\psi\varphi z=0$.

###### Proof.

We rewrite identity (9) in the following form:

$\varphi^{2}y+\psi^{3}y+\varphi\psi z+\psi\varphi z+\psi^{2}\varphi x=x.$ (14)

If we substitute in equality (14) $y=z=0$ then we have

$\varphi x=\psi^{-2}x.$ (15)

If we substitute in equality (14) $x=z=0$ then we have

$\varphi^{2}y+\psi^{3}y=0.$ (16)

Taking into consideration equality (15), we can re-write equality (16) in the
form

$\psi^{-4}y+\psi^{3}y=0,$ (17)

or in the form

$\psi^{3}=I\psi^{-4},$ (18)

where $Ix=-x$ for all $x\in Q$. Notice, the permutation $I$ is an automorphism
of the group $(Q,+)$ here. Therefore, we can rewrite previous equalities in
the form

$\varepsilon=I\psi^{-7},I=\psi^{-7},\varepsilon=\psi^{-14},\varepsilon=\psi^{14},\varepsilon=\varphi^{7}.$
(19)

If we substitute in equality (14) $x=y=0$ then we have

$\varphi\psi z+\psi\varphi z=0.$ (20)

Converse. If we substitute in identity (9) the expression $x\cdot y=\varphi
x+\psi y$, then we obtain equality (14), which is true taking into
consideration equalities (15), (16), (20). Then we obtain, that identity (9)
is true in this case. ∎

######  Corollary 1.

In medial quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$
Schröder identity of generalized associativity is true if and only if the
group $(Q,+)$ is an abelian 2-group (i.e. $x+x=0$ for any $x\in Q$), $\varphi
x=\psi^{-2}x$, $\varepsilon=\varphi^{7}$, $\varepsilon=\psi^{14}$.

###### Proof.

From the identity of mediality it follows that $\varphi\psi z+\psi\varphi
z=2\cdot\varphi\psi z=0$ for all $z\in Q$, i.e., the group $(Q,+)$ is an
abelian 2-group. ∎

######  Example 3.

We start from the group $GL(3,2)\cong PSL(2,7)$. This is the group of non-
degenerate matrices of size $3\times 3$ over the field of order 2. Or the
group of non-degenerate matrices of size $2\times 2$ over the field of order
7. The order of this group is equal to 8, $|Aut(GL(3,2)|=168=3\times 7\times
8$.

We can present elements of the group $(Z_{2}^{3},+)$ in the following form :
$1=(000),2=(001),3=(010),4=(011),5=(100),6=(101),7=(110),8=(111).$

$\begin{array}[]{l|llllllll}+&1&2&3&4&5&6&7&8\\\ \hline\cr
1&1&2&3&4&5&6&7&8\\\ 2&2&1&4&3&6&5&8&7\\\ 3&3&4&1&2&7&8&5&6\\\
4&4&3&1&1&8&7&6&5\\\ 5&5&6&7&8&1&2&3&4\\\ 6&6&5&8&7&2&1&4&3\\\
7&7&8&5&6&3&4&1&2\\\ 8&8&7&6&5&4&3&2&1\\\ \end{array}$

We have the following automorphisms of the group $(Z_{2}^{3},+)$:

$\varphi=\begin{pmatrix}1&1&0\\\ 1&0&1\\\ 0&1&0\\\ \end{pmatrix}$,
$\psi=\begin{pmatrix}0&1&0\\\ 1&1&1\\\ 0&1&1\\\ \end{pmatrix}$.

Notice that $\varphi^{7}=\psi^{7}=\varepsilon$, $\varphi=\psi^{-2}$,
$\varphi\psi=\psi\varphi$.

Therefore Schröder medial quasigroup $(Q,\circ)$ of generalized associativity
can have the form $x\circ y=\varphi x+\psi y$:

$\begin{array}[]{l|llllllll}\circ&1&2&3&4&5&6&7&8\\\ \hline\cr
1&1&4&8&5&3&2&6&7\\\ 2&3&2&6&7&1&4&8&5\\\ 3&6&7&3&2&8&5&1&4\\\
4&8&5&1&4&6&7&3&2\\\ 5&7&6&2&3&5&8&4&1\\\ 6&5&8&4&1&7&6&2&3\\\
7&4&1&5&8&2&3&7&6\\\ 8&2&3&7&6&4&1&5&8\\\ \end{array}$

###  2.2 Schröder identity

######  Theorem 3.

In T-quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$ Schröder
identity is true if and only if $\varphi^{2}+\psi^{2}=\varepsilon$,
$\varphi\psi y+\psi\varphi y=0$.

###### Proof.

From identity

$xy\cdot yx=x$ (21)

we have

$\varphi(\varphi x+\psi y)+\psi(\varphi y+\psi x)=x.$ (22)

If in (22) $y=0$, then

$\varphi^{2}+\psi^{2}=\varepsilon.$ (23)

If in (22) $x=0$, then

$\varphi\psi y+\psi\varphi y=0.$ (24)

Converse. If we substitute in identity (10) the expression $x\cdot y=\varphi
x+\psi y$, then we obtain equality (22), which is true taking into
consideration equalities (23), (24). Then we obtain, that identity (10) is
true in this case. ∎

######  Corollary 2.

In medial quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$
Schröder identity is true if and only if $\varphi^{2}+\psi^{2}=\varepsilon$,
the group $(Q,+)$ is an abelian 2-group (i.e., $x+x=0$ for any $x\in Q$).

###### Proof.

This follows from mediality of quasigroup $(Q,\cdot)$. ∎

###  2.3 T-quasigroup with identity $x(y\cdot yx)=y$

######  Theorem 4.

In T-quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$ identity
$x(y\cdot yx)=y$ is true if and only if $\varphi=I\psi^{3}$,
$\psi^{4}+\psi^{5}=I$.

###### Proof.

From identity

$x(y\cdot yx)=y$ (25)

we have

$\varphi x+\psi(\varphi y+\psi(\varphi y+\psi x)=y.$ (26) $\varphi
x+\psi\varphi y+\psi^{2}\varphi y+\psi^{3}x=y.$ (27)

If in (27) $y=0$, then

$\varphi+\psi^{3}=0,$ (28)

$\varphi=I\psi^{3}.$ (29)

If in (27) $x=0$, then

$\psi\varphi+\psi^{2}\varphi=\varepsilon.$ (30)

Taking in account (29) we have

$\psi^{4}+\psi^{5}=I.$ (31)

Converse. If we substitute in identity (25) the expression $x\cdot y=\varphi
x+\psi y$, then we obtain equality (26), which is true taking into
consideration equalities (35), (30). Then we obtain, that identity (25) is
true in this case. ∎

######  Corollary 3.

In medial quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$
identity $x(y\cdot yx)=y$ is true if and only if $\varphi=I\psi^{3}$,
$\psi^{4}+\psi^{5}=I$.

###### Proof.

This follows from mediality of quasigroup $(Q,\cdot)$. ∎

######  Example 4.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\psi=-2$, then
$\varphi=I\psi^{3}=8$, $\psi^{4}+\psi^{5}=16-32=-16=-1$. The last is true if
$-16\equiv-1\pmod{n}$, $15\equiv 0\pmod{n}$.

Then quasigroup $(Z_{15},\circ)$ of the form $x\circ y=8\cdot x-2\cdot
y\pmod{15}$ is medial quasigroup with identity $x(y\cdot yx)=y$.

Check. $8x-2(8y-2(8y-2x))=y\pmod{15}$, $8x-16y+32y-8x=y\pmod{15}$,
$16y=y\pmod{15}$, $y=y\pmod{15}$.

######  Example 5.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\psi=-3$, then
$\varphi=I\psi^{3}=27$, $\psi^{4}+\psi^{5}=81-243=-162=-1$. The last is true
if $-162\equiv-1\pmod{n}$, $161\equiv 0\pmod{n}$.

Quasigroup $(Z_{23},\circ)$ of the form $x\circ y=4\cdot x-3\cdot y\pmod{23}$
is medial quasigroup with identity $x(y\cdot yx)=y$.

Check. $4x-3(4y-3(4y-3x))=y\pmod{23}$, $4x-12y+36y-27x=y\pmod{23}$,
$24y=y\pmod{23}$, $y=y\pmod{23}$.

######  Example 6.

Quasigroup $(Z_{161},\circ)$ of the form $x\circ y=27\cdot x-3\cdot
y\pmod{161}$ is medial quasigroup with identity $x(y\cdot yx)=y$.

Check. $27x-3(27y-3(27y-3x))=y\pmod{161}$, $27x-81y+243y-27x=y\pmod{161}$,
$162y=y\pmod{161}$, $y=y\pmod{161}$.

###  2.4 T-quasigroups with Stein 3-rd identity

######  Theorem 5.

In T-quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$ Stein 3-rd
identity is true if and only if $\varphi^{2}+\psi^{2}=0$, $\varphi\psi
y+\psi\varphi y=\varepsilon$.

###### Proof.

From identity

$xy\cdot yx=y$ (32)

we have

$\varphi(\varphi x+\psi y)+\psi(\varphi y+\psi x)=y.$ (33)

If in (33) $y=0$, then

$\varphi^{2}+\psi^{2}=0.$ (34)

$\varphi^{2}=I\psi^{2}.$ (35)

If in (33) $x=0$, then

$\varphi\psi+\psi\varphi=\varepsilon.$ (36)

Converse. If we substitute in identity (32) the expression $x\cdot y=\varphi
x+\psi y$, then we obtain equality (33), which is true taking into
consideration equalities (34), (36). Then we obtain, that identity (32) is
true in this case. ∎

######  Corollary 4.

In medial quasigroup $(Q,\cdot)$ of the form $x\cdot y=\varphi x+\psi y$ Stein
3-rd identity is true if and only if $\varphi^{2}+\psi^{2}=0$,
$2\varphi\psi=\varepsilon$.

###### Proof.

This follows from mediality of quasigroup $(Q,\cdot)$. ∎

######  Example 7.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=1$,
$\psi=3$ then $\varphi^{2}+\psi^{2}=1+9=0\pmod{5},n=5$. Further
$2\varphi\psi=2\cdot 1\cdot 3=6=\varepsilon=1\pmod{5}$, $x\cdot
y=x+3y\pmod{5}$.

Check. $(x+3y)+3(y+3x)=y\pmod{5}$, $x+3y+3y+9x=y\pmod{5}$, $y=y\pmod{5}$.

######  Example 8.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=4$,
$\psi=2$ then $\varphi^{2}+\psi^{2}=16+4=0\pmod{5},n=5$. Further
$2\varphi\psi=2\cdot 4\cdot 2=16=\varepsilon=1\pmod{5}$, $x\cdot
y=4x+2y\pmod{5}$.

Check. $4(4x+2y)+2(4y+2x)=y\pmod{5}$, $16x+8y+8y+4x=y\pmod{5}$, $y=y\pmod{5}$.

######  Example 9.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=2$,
$\psi=10$ then $\varphi^{2}+\psi^{2}=4+100=104=0\pmod{13},n=13$. Further
$2\varphi\psi=2\cdot 2\cdot 10=40=\varepsilon=1\pmod{13}$, $x\cdot
y=2x+10y\pmod{13}$.

Check. $2(2x+10y)+10(2y+10x)=y\pmod{13}$, $4x+20y+20y+100x=y\pmod{13}$,
$y=y\pmod{13}$.

######  Example 10.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=6$,
$\psi=10$ then $\varphi^{2}+\psi^{2}=36+100=136=0\pmod{17},n=17$. Further
$2\varphi\psi=2\cdot 6\cdot 10=120=\varepsilon=1\pmod{17}$, $x\cdot
y=6x+10y\pmod{17}$.

Check. $6(6x+10y)+10(6y+10x)=y\pmod{17}$, $36x+60y+60y+100x=y\pmod{17}$,
$y=y\pmod{17}$.

######  Example 11.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=8$,
$\psi=20$ then $\varphi^{2}+\psi^{2}=64+400=464=0\pmod{29},n=29$. Further
$2\varphi\psi=2\cdot 8\cdot 20=320=\varepsilon=1\pmod{29}$, $x\cdot
y=8x+20y\pmod{29}$.

Check. $8(8x+20y)+20(8y+20x)=y\pmod{29}$, $64x+160y+160y+400x=y\pmod{29}$,
$y=y\pmod{65}$.

######  Example 12.

Suppose we have the group $Z_{n}$ of residues modulo $n$. If $\varphi=3$,
$\psi=11$ then $\varphi^{2}+\psi^{2}=9+121=130=0\pmod{65},n=65$. Further
$2\varphi\psi=2\cdot 3\cdot 11=66=\varepsilon=1\pmod{65}$, $x\cdot
y=3x+11y\pmod{65}$.

Check. $3(3x+11y)+11(3y+11x)=y\pmod{65}$, $9x+33y+33y+121x=y\pmod{65}$,
$y=y\pmod{65}$.

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Victor Shcherbacov
“Vladimir Andrunachievici”
Institute of Mathematics and Computer Science
5 Academiei str., Chişinău MD-2028
Moldova
E-mail<EMAIL_ADDRESS>