Algebra Loop/Examples/Order 3

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Example of Algebra Loop

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

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

Hence, up to isomorphism, there is only one algebra loop with $3$ elements.

This is isomorphic to the additive group of integers modulo $3$.


Proof

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

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


Then we note that:

\(\ds a \circ b\) \(=\) \(\ds e\)
\(\ds b \circ a\) \(=\) \(\ds e\)

because both $a$ and $b$ already appear in the row and column of those cells.


Hence we have:

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

and the remaining cells are likewise forced.


The isomorphism $\phi$ to the additive group of integers modulo $3$ can be established as:

\(\ds \map \phi e\) \(=\) \(\ds \eqclass 0 3\)
\(\ds \map \phi a\) \(=\) \(\ds \eqclass 1 3\)
\(\ds \map \phi b\) \(=\) \(\ds \eqclass 2 3\)

The fact that this is indeed an isomorphism follows by inspection.

$\blacksquare$


Sources

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