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:
\(\ds \CC_1\) | \(:\) | \(\ds \set {I_S, \tuple {a, b} }\) | ||||||||||||
\(\ds \CC_2\) | \(:\) | \(\ds \set {I_S, \tuple {a, c} }\) | ||||||||||||
\(\ds \CC_3\) | \(:\) | \(\ds \set {I_S, \tuple {b, c} }\) |
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$