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

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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:

\(\ds \CC_1\) \(:\) \(\ds \set {I_S, \tuple {a, b} }\)
\(\ds \CC_2\) \(:\) \(\ds \set {I_S, \tuple {a, c} }\)
\(\ds \CC_3\) \(:\) \(\ds \set {I_S, \tuple {b, c} }\)

where $I_S$ is the identity mapping on $S$.

Then:

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


Isomorphism Classes

Let $\CC = \CC_1 \cup \CC_2 \cup \CC_3$.

Let $\oplus \in \CC$.

Then the isomorphism class of $\oplus$ consists of $3$ elements: exactly one operation from each of $\CC_1$, $\CC_2$ and $\CC_3$.


Operations with Identity

$9$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ has an identity element.


Commutative Operations

$8$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ is commutative.


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


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 $S$.


Without loss of generality, 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 1

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

$c \circ c = c$

$\Box$


Lemma 2

\(\ds a \circ a = a\) \(\iff\) \(\ds b \circ b = b\)
\(\ds a \circ a = b\) \(\iff\) \(\ds b \circ b = a\)
\(\ds a \circ a = c\) \(\iff\) \(\ds b \circ b = c\)
\(\ds a \circ b = a\) \(\iff\) \(\ds b \circ a = b\)
\(\ds a \circ b = b\) \(\iff\) \(\ds b \circ a = a\)
\(\ds a \circ b = c\) \(\iff\) \(\ds b \circ a = c\)
\(\ds a \circ c = a\) \(\iff\) \(\ds b \circ c = b\)
\(\ds a \circ c = b\) \(\iff\) \(\ds b \circ c = a\)
\(\ds a \circ c = c\) \(\iff\) \(\ds b \circ c = c\)
\(\ds c \circ a = a\) \(\iff\) \(\ds c \circ b = b\)
\(\ds c \circ a = b\) \(\iff\) \(\ds c \circ b = a\)
\(\ds c \circ a = c\) \(\iff\) \(\ds c \circ b = c\)

$\Box$


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.

$\blacksquare$


Sources