Sets of Operations on Set of 3 Elements/Automorphism Group of A/Operations with Identity

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, $\map \Gamma S$.

Then:

None of the operations of $\AA$ has an identity element.

Proof

Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:

$\begin{array}{c|ccc}

\to & a & b & c \\ \hline a & a & b & c \\ b & a & b & c \\ c & a & b & c \\ \end{array} \qquad \begin{array}{c|ccc} \gets & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ \end{array} \qquad \begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end{array}$

The result is apparent by inspection.

$\blacksquare$

Sources

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