Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Lemma 2

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

Let $\CC_1$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b} }$, where $I_S$ is the identity mapping on $S$.

Then:

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 $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$

Let us denote $\tuple {a, b}$ as the mapping $r: S \to S$:
 * $r := \map r a = b, \map r b = a, \map r c = c$

In Lemma 1 it has been established that:

We select values for $x$ in the expression $a \circ b = x$ and determine how $r$ constrains other product elements.

Thus:

Hence:

Similarly:

Hence: