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

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:
 * Each of $\CC_1$, $\CC_2$ and $\CC_3$ has $3^4 - 3$ elements.

Proof
Recall the definition of (group) automorphism:


 * $\phi$ is an automorphism on $\struct {S, \circ}$ :
 * $\phi$ is a permutation of $S$
 * $\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$

From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.

Hence it is not necessary to analyse the effect of $I_S$ on $S$.

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

Let $n$ be the number of operations $\circ$ on $S$ such that $\tuple {a, b}$ is an automorphism of $\struct {S, \circ}$.

Let us denote the permutation $\tuple {a, b}$ as $r: S \to S$, defined as:


 * $r = \map r a = b, \map r b = a, \map r c = c$

We select various product elements $x \circ y \in S$ and determine how $r$ constrains other product elements.

Lemma 2
Hence, selecting each of the $3$ options for:
 * $a \circ a$
 * $a \circ b$
 * $a \circ c$
 * $c \circ a$

generates a unique operation on $S$.

From the Product Rule for Counting it follows that:
 * $n = 3 \times 3 \times 3 \times 3 = 3^4$

From Automorphism Group of $\AA$, $3$ of those operations are elements of the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ are $\map \Gamma S$.

Hence those $3$ elements are not in $\CC_1$, and are removed from the overall count.

Hence the result.