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

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:
 * $\DD$ has $19 \, 422$ elements.

Proof
Let $n$ denote the cardinality of $\DD$.

Equivalently, $n$ equals the number of operations $\circ$ on $S$ on which the only automorphism is $I_S$.

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:

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

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


 * $N = \card \AA + \card \BB + \card {\CC_1} + \card {\CC_2} + \card {\CC_3} + \card {\DD}$

From Count of Binary Operations on Set:


 * $N = 3^{3^2} = 3^9 = 19 \, 683$

Then we have:
 * From Automorphism Group of $\AA$: $\card \AA = 3$
 * From Automorphism Group of $\BB$: $\card \BB = 3^3 - 3 = 24$
 * From Automorphism Group of $\CC_n$: for $n = 1, 2, 3: \card {\CC_n} = 3^4 - 3 = 78$

Hence we have: