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

Lemma for Sets of Operations on Set of 3 Elements
Let $S = \set {a, b, c}$ be a set with $3$ elements.

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

Then:
 * $\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$ forms a partition of the set of all operations on $S$.

Proof
Let $\OO$ denote the set of all operations on $S$.

Let $\SS := \set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.

First we note that from Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.

Thus every operation on $S$ belongs to at least one of $\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.

That is:
 * $\OO = \ds \bigcup \SS$

Then we note that from:


 * Automorphism Group of $\AA$
 * Automorphism Group of $\BB$
 * Automorphism Group of $\CC_n$

each of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ is non-empty.

Consider for example the constant operation on $S$:
 * $\forall x, y \in S: x \circ y = a$

From the above cited results, $\circ$ as described does not belong to any of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$.

Hence it must belong to $\DD$.

Thus $\DD$ is also non-empty.

It remains to be shown that $\SS$ is a pairwise disjoint set of sets.

By definition, $\DD$ is disjoint with all other elements of $\SS$.

By construction, $\AA$ is also disjoint with all other elements of $\SS$.