Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Commutative Operations
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:
$8$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ is commutative.
Proof
Without loss of generality, we will analyse the nature of $\CC_1$.
Recall this lemma:
Lemma
\(\ds a \circ a = a\) | \(\iff\) | \(\ds b \circ b = b\) | ||||||||||||
\(\ds a \circ a = b\) | \(\iff\) | \(\ds b \circ b = a\) | ||||||||||||
\(\ds a \circ a = c\) | \(\iff\) | \(\ds b \circ b = c\) | ||||||||||||
\(\ds a \circ b = a\) | \(\iff\) | \(\ds b \circ a = b\) | ||||||||||||
\(\ds a \circ b = b\) | \(\iff\) | \(\ds b \circ a = a\) | ||||||||||||
\(\ds a \circ b = c\) | \(\iff\) | \(\ds b \circ a = c\) | ||||||||||||
\(\ds a \circ c = a\) | \(\iff\) | \(\ds b \circ c = b\) | ||||||||||||
\(\ds a \circ c = b\) | \(\iff\) | \(\ds b \circ c = a\) | ||||||||||||
\(\ds a \circ c = c\) | \(\iff\) | \(\ds b \circ c = c\) | ||||||||||||
\(\ds c \circ a = a\) | \(\iff\) | \(\ds c \circ b = b\) | ||||||||||||
\(\ds c \circ a = b\) | \(\iff\) | \(\ds c \circ b = a\) | ||||||||||||
\(\ds c \circ a = c\) | \(\iff\) | \(\ds c \circ b = c\) |
$\Box$
The operations brought about by $a \circ a$, $b \circ b$ and $c \circ c$ have no effect on whether those operations are commutative.
The operations in which $a \circ b = a$ and $a \circ b = b$ are definitely not commutative, as these are such that $a \circ b \ne b \circ a$.
Hence all commutative operations of $\CC_1$ have $a \circ b = c = b \circ a$.
Hence, of the above, we have:
- $3$ options of $a \circ a$ each give rise to a partial operation.
- Only $1$ option of $a \circ b$ gives rise to a partial operation, that is $a \circ b$.
- The $3$ options of $a \circ c$ each give rise to a partial operation, but they force the corresponding options of $c \circ a$, $b \circ c$ and $c \circ b$.
Hence by the Product Rule for Counting there are $3 \times 3 = 9$ commutative operations which can be constructed thus.
However, note that one of these:
- $\begin {array} {c|ccc}
\circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$
has already been accounted for in Automorphism Group of $\AA$: Commutative Operations.
This commutative operations is such that the group of automorphisms of $\struct {S, \circ}$ forms the complete symmetric group on $S$, not just the given permutations.
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 $8.14 \ \text{(d)}$