Algebra Loop/Examples/Order 5
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Example of Algebra Loop
The following is the Cayley table of an operation $\circ$ on $S = \set {e, a, b, c, d}$ such that $\struct {S, \circ}$ is an algebra loop whose identity is $e$, but which is not actually a group:
- $\begin{array}{r|rrr}
\circ & e & a & b & c & d \\ \hline e & e & a & b & c & d \\ a & a & e & d & b & c \\ b & b & c & e & d & a \\ c & c & d & a & e & b \\ d & d & b & c & a & e \\ \end{array}$
Proof
We note that:
| \(\ds \paren {a \circ a} \circ b\) | \(=\) | \(\ds e \circ b\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
| \(\ds a \circ \paren {a \circ b}\) | \(=\) | \(\ds a \circ d\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c\) |
demonstrating that $\circ$ is not associative.
Hence $\struct {S, \circ}$ is not a group.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.8 \ \text {(c)}$