Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Commutative Operations

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

Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:

where $I_S$ is the identity mapping on $S$.

Then: $8$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ is commutative.

Proof
, we will analyse the nature of $\CC_1$.

Recall this lemma:

Lemma
The operations brought about by $a \circ a$, $b \circ b$ and $c \circ c$ have no effect on whether those operations are commutative.

The operations in which $a \circ b = a$ and $a \circ b = b$ are definitely not commutative, as these are such that $a \circ b \ne b \circ a$.

Hence all commutative operations of $\CC_1$ have $a \circ b = c = b \circ a$.

Hence, of the above, we have:
 * $3$ options of $a \circ a$ each give rise to a partial operation.


 * Only $1$ option of $a \circ b$ gives rise to a partial operation, that is $a \circ b$.


 * The $3$ options of $a \circ c$ each give rise to a partial operation, but they force the corresponding options of $c \circ a$, $b \circ c$ and $c \circ b$.

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.