Algebra Loop/Examples/Order 4

From ProofWiki
Jump to navigation Jump to search

Example of Algebra Loop

The following are the Cayley tables of the operations $\circ$ on $S = \set {e, a, b, c}$ such that $\struct {S, \circ}$ is an algebra loop whose identity is $e$:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & b & c & e \\ b & b & c & e & a \\ c & c & e & a & b \\ \end{array} \qquad \begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array} \qquad \begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & c & e & b \\ b & b & e & c & a \\ c & c & b & a & e \\ \end{array} \qquad \begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & a & e \\ c & c & b & e & a \\ \end{array}$


The first two of these are the Cayley tables of:

the cyclic group of order $4$
the Klein $4$-group

while the $3$rd and $4$th are also isomorphic to the cyclic group of order $4$.


Proof

The initial specification allows us to populate the first few elements of the Cayley table:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & & & \\ b & b & & & \\ c & c & & & \\ \end{array}$


Let us consider $a \circ a$.

This cannot be $a$ as there is already an $a$ in the row and column.


Let $a \circ a = e$
$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & & \\ b & b & & & \\ c & c & & & \\ \end{array}$

This immediately forces:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & & \\ c & c & b & & \\ \end{array}$

There are two ways to complete this.

Either $b \circ b = e$ which gives us the Klein $4$-group:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$

or $b \circ b = a$, which gives us:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & a & e \\ c & c & b & e & a \\ \end{array}$


Let $a \circ a = b$
$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & b & & \\ b & b & & & \\ c & c & & & \\ \end{array}$

We note that $a \circ b = e$ will not work, because that forces $a \circ c = c$ which is not allowed because there is already a $c$ in the $c$ column.

Hence we have $a \circ b = c$, which forces the completion of the cyclic group of order $4$:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & b & c & e \\ b & b & c & e & a \\ c & c & e & a & b \\ \end{array}$


Let $a \circ a = c$
$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & c & & \\ b & b & & & \\ c & c & & & \\ \end{array}$

This forces the completion of the following:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & c & e & b \\ b & b & e & c & a \\ c & c & b & a & e \\ \end{array}$


We then note that we can rearrange the order of the rows and columns of the remaining two tables to demonstrate that they are the cyclic group of order $4$:


Let us take:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & c & e & b \\ b & b & e & c & a \\ c & c & b & a & e \\ \end{array}$

Rearranging the order of the rows and columns as follows:

$\begin{array}{r|rrr} \circ & e & b & c & a \\ \hline e & e & b & c & a \\ b & b & c & a & e \\ c & c & a & e & b \\ a & a & e & b & c \\ \end{array}$

from which the cyclic group of order $4$ is verified by inspection.


Similarly:

$\begin{array}{r|rrr} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & a & e \\ c & c & b & e & a \\ \end{array}$

Rearranging the order of the rows and columns as follows:

$\begin{array}{r|rrr} \circ & e & b & a & c \\ \hline e & e & b & a & c \\ b & b & a & c & e \\ a & a & c & e & b \\ c & c & e & b & a \\ \end{array}$

from which the cyclic group of order $4$ is also verified by inspection.

$\blacksquare$


Sources