Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Lemma 2
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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:
| \(\ds a \circ a = a\) | \(\iff\) | \(\ds b \circ b = b\) | ||||||||||||
| \(\ds a \circ a = b\) | \(\iff\) | \(\ds b \circ b = a\) | ||||||||||||
| \(\ds a \circ a = c\) | \(\iff\) | \(\ds b \circ b = c\) | ||||||||||||
| \(\ds a \circ b = a\) | \(\iff\) | \(\ds b \circ a = b\) | ||||||||||||
| \(\ds a \circ b = b\) | \(\iff\) | \(\ds b \circ a = a\) | ||||||||||||
| \(\ds a \circ b = c\) | \(\iff\) | \(\ds b \circ a = c\) | ||||||||||||
| \(\ds a \circ c = a\) | \(\iff\) | \(\ds b \circ c = b\) | ||||||||||||
| \(\ds a \circ c = b\) | \(\iff\) | \(\ds b \circ c = a\) | ||||||||||||
| \(\ds a \circ c = c\) | \(\iff\) | \(\ds b \circ c = c\) | ||||||||||||
| \(\ds c \circ a = a\) | \(\iff\) | \(\ds c \circ b = b\) | ||||||||||||
| \(\ds c \circ a = b\) | \(\iff\) | \(\ds c \circ b = a\) | ||||||||||||
| \(\ds c \circ a = c\) | \(\iff\) | \(\ds c \circ b = c\) |
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 $r: S \to S$:
- $r := \map r a = b, \map r b = a, \map r c = c$
In Lemma 1 it has been established that:
| \(\ds a \circ a = a\) | \(\iff\) | \(\ds b \circ b = b\) | ||||||||||||
| \(\ds a \circ a = b\) | \(\iff\) | \(\ds b \circ b = a\) | ||||||||||||
| \(\ds a \circ a = c\) | \(\iff\) | \(\ds b \circ b = c\) |
We select values for $x$ in the expression $a \circ b = x$ and determine how $r$ constrains other product elements.
Thus:
| \(\ds a \circ x\) | \(=\) | \(\ds a\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map r a \circ \map r x\) | \(=\) | \(\ds \map r a\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds b \circ \map r x\) | \(=\) | \(\ds b\) |
Hence:
| \(\ds a \circ b = a\) | \(\iff\) | \(\ds b \circ a = b\) | ||||||||||||
| \(\ds a \circ b = b\) | \(\iff\) | \(\ds b \circ a = a\) | ||||||||||||
| \(\ds a \circ b = c\) | \(\iff\) | \(\ds b \circ a = c\) |
Similarly:
| \(\ds x \circ a\) | \(=\) | \(\ds a\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map r x \circ \map r a\) | \(=\) | \(\ds \map r a\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map r x \circ b\) | \(=\) | \(\ds b\) |
Hence:
| \(\ds c \circ a = a\) | \(\iff\) | \(\ds c \circ b = b\) | ||||||||||||
| \(\ds c \circ a = b\) | \(\iff\) | \(\ds c \circ b = a\) | ||||||||||||
| \(\ds c \circ a = c\) | \(\iff\) | \(\ds c \circ b = c\) |
$\blacksquare$