Sets of Operations on Set of 3 Elements/Automorphism Group of A/Cayley Table

Theorem

Let $S = \set {a, b, c}$ be a set with $3$ elements.

Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.

From Automorphism Group of $\AA$, there are $3$ such operations $\circ$ on $S$.

One is the right operation, one is the left operation, and the third is neither.

The Cayley table of the operation $\circ$ on $S$ such that:

every permutation of $S$ is an automorphism on $\struct {S, \circ}$

$\circ$ is neither the right operation nor the left operation

can be presented as follows:

$\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$