Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Lemma 1

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:

$c$ is an idempotent element under $\circ$, that is:

$c \circ c = c$

Proof

Recall the definition of (group) automorphism:

$\phi$ is an automorphism on $\struct {S, \circ}$ if and only if:

$\phi$ 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 $r: S \to S$:

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

Aiming for a contradiction, suppose $c$ is not idempotent.

Then $c \circ c = x$, where $x = a$ or $x = b$.

\(\ds c \circ c\)

\(=\)

\(\ds x\)

\(\ds \leadsto \ \ \)

\(\ds \map r c \circ \map r c\)

\(=\)

\(\ds \map r x\)

\(\ds \)

\(\ne\)

\(\ds x\)

So it cannot be the case that $c \circ c = a$ or $c \circ c = b$.

Hence $c \circ c = c$, that is, $c$ is idempotent.

$\blacksquare$