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

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

Let $\DD$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S}$, where $I_S$ is the identity mapping on $S$.

Then:
 * $696$ of the operations of $\DD$ is commutative.

Proof
Let $n$ denote the number of commutative operations of $\DD$.

Recall these definitions:

Let $\AA$, $\BB$, $\CC_1$, $\CC_2$ 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:

Let $N$ be the total number of commutative operations on $S$.

Let:
 * $A$ denote the number of commutative operations in $\AA$
 * $B$ denote the number of commutative operations in $\BB$
 * $C$ denote the total number of commutative operations in $\CC_1$, $\CC_2$ and $\CC_3$.

From the lemma, and from the Fundamental Principle of Counting:


 * $N = A + B + C + D$

From Count of Commutative Binary Operations on Set:


 * $N = 3^6 = 729$

Then we have:
 * From Automorphism Group of $\AA$: Operations with Identity: $A = 1$
 * From Automorphism Group of $\BB$: Operations with Identity: $B = 8$
 * From Automorphism Group of $\CC_n$: Operations with Identity: $C = 3 \times 8$

Hence we have: