Sets of Operations on Set of 3 Elements/Automorphism Group of B/Commutative Operations
Theorem
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\BB$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ forms the set $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$, where $I_S$ is the identity mapping on $S$.
Then:
- Exactly $8$ of the operations of $\BB$ is commutative.
Proof
Recall Automorphism Group of $\BB$.
Consider each of the categories of $\BB$ induced by each of $a \circ a$, $a \circ b$ and $a \circ c$, illustrated by the partially-filled Cayley tables to which they give rise:
- $(1)
- \quad a \circ a$
- $\begin {array} {c|ccc}
\circ & a & b & c \\ \hline a & a & & \\ b & & b & \\ c & & & c \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & b & & \\ b & & c & \\ c & & & a \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & c & & \\ b & & a & \\ c & & & b \\ \end {array}$
- $(2)
- \quad a \circ b$
- $\begin {array} {c|ccc}
\circ & a & b & c \\ \hline a & & a & \\ b & & & b \\ c & c & & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & b & \\ b & & & c \\ c & a & & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & c & \\ b & & & a \\ c & b & & \\ \end {array}$
- $(3)
- \quad a \circ c$
- $\begin {array} {c|ccc}
\circ & a & b & c \\ \hline a & & & c \\ b & a & & \\ c & & b & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & & b \\ b & c & & \\ c & & a & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & & c \\ b & a & & \\ c & & b & \\ \end {array}$
With a view to Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, we inspect the various combinations of these partial Cayley tables.
It is apparent that the result of $x \circ x$ makes no difference to whether an operation is commutative.
However, note that each of the partial operations in $(2)$ can be conjoined with exactly $1$ of the partial operations in $(3)$ to make a commutative operation:
- $\begin {array} {c|ccc}
\circ & a & b & c \\ \hline a & & a & c \\ b & a & & b \\ c & c & b & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & b & a \\ b & b & & c \\ c & a & c & \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & & c & b \\ b & c & & a \\ c & b & a & \\ \end {array}$
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 $8.14 \ \text{(d)}$