Sets of Operations on Set of 3 Elements/Automorphism Group of B/Commutative Operations

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\BB$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
:Exactly $8$ of the operations of $\BB$ is commutative.
== Proof ==
Recall Automorphism Group of $\BB$.
Consider each of the categories of $\BB$ induced by each of $a \circ a$, $a \circ b$ and $a \circ c$, illustrated by the partially-filled Cayley tables to which they give rise:
;$(1): \quad a \circ a$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & & \\
b & & b & \\
c & & & c \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b & & \\
b & & c & \\
c & & & a \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c & & \\
b & & a & \\
c & & & b \\
\end {array}$
;$(2): \quad a \circ b$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & \\
b & & & b \\
c & c & & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & \\
b & & & c \\
c & a & & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & \\
b & & & a \\
c & b & & \\
\end {array}$
;$(3): \quad a \circ c$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & b \\
b & c & & \\
c & & a & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array}$
With a view to Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, we inspect the various combinations of these partial Cayley tables.
It is apparent that the result of $x \circ x$ makes no difference to whether an operation is commutative.
However, note that each of the partial operations in $(2)$ can be conjoined with exactly $1$ of the partial operations in $(3)$ to make a commutative operation:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & c \\
b & a & & b \\
c & c & b & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & a \\
b & b & & c \\
c & a & c & \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & b \\
b & c & & a \\
c & b & a & \\
\end {array}$
Hence by the Product Rule for Counting there are $3 \times 3 = 9$ commutative operations which can be constructed thus.
However, note that one of these:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
has already been accounted for in Automorphism Group of $\AA$: Commutative Operations.
This commutative operations is such that the group of automorphisms of $\struct {S, \circ}$ forms the complete symmetric group on $S$, not just the given permutations.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$
Category:Sets of Operations on Set of 3 Elements