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.