1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
|
\import{algebra/magma.tex}
\section{Quasigroups}
\begin{struct}\label{quasigroup}
A quasigroup $A$ is a magma equipped with
\begin{enumerate}
\item $\ldiv$
\item $\rdiv$
\end{enumerate}
such that
\begin{enumerate}
\item\label{quasigroup_ldiv_type} for all $a,b\in\carrier[A]$ we have $\ldiv[A](a,b)\in\carrier[A]$.
\item\label{quasigroup_rdiv_type} for all $a,b\in\carrier[A]$ we have $\rdiv[A](a,b)\in\carrier[A]$.
\item\label{quasigroup_mul_ldiv} for all $a,b\in\carrier[A]$ we have $b = \mul[A](a,\ldiv[A](a,b))$.
\item\label{quasigroup_ldiv_mul} for all $a,b\in\carrier[A]$ we have $b = \ldiv[A](a,\mul[A](a,b))$.
\item\label{quasigroup_rdiv_mul} for all $a,b\in\carrier[A]$ we have $b = \mul[A](\rdiv[A](b,a),a)$.
\item\label{quasigroup_mul_rdiv} for all $a,b\in\carrier[A]$ we have $b = \rdiv[A](\mul[A](b,a),a)$.
\end{enumerate}
\end{struct}
% Cancelling an element on the left.
\begin{lemma}\label{quasigroup_cancel_left}
Let $A$ be a quasigroup.
Let $a,b,c\in\carrier[A]$.
Suppose $\mul[A](a,b) = \mul[A](a,c)$.
Then $b = c$.
\end{lemma}
\begin{proof}
We have $b=\ldiv[A](a,\mul[A](a,b))$ by \cref{quasigroup_ldiv_mul}.
We have $c=\ldiv[A](a,\mul[A](a,c))$ by \cref{quasigroup_ldiv_mul}.
Follows by assumption.
\end{proof}
% Cancelling an element on the right.
\begin{lemma}\label{quasigroup_cancel_right}
Let $A$ be a quasigroup.
Let $a,b,c\in\carrier[A]$.
Suppose $\mul[A](a,c) = \mul[A](b,c)$.
Then $a = b$.
\end{lemma}
\begin{proof}
We have $a=\rdiv[A](\mul[A](a,c),c)$ by \cref{quasigroup_mul_rdiv}.
We have $b=\rdiv[A](\mul[A](b,c),c)$ by \cref{quasigroup_mul_rdiv}.
Follows by assumption.
\end{proof}
|
