Sets of Operations on Set of 3 Elements/Automorphism Group of B

== 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:
:$\BB$ has $3^3 - 3$ elements.
=== Isomorphism Classes ===
=== Operations with Identity ===
=== Commutative Operations ===
== Proof ==
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ :
::$\phi$ 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 $S$.
Let us denote each of the remaining elements of $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$ as follows:
We select various product elements $x \circ y \in S$ and determine how $p$ and $q$ constrain other product elements as follows:
Then by definition of the above mappings $p$ and $q$, and the definition of a homomorphism, we obtain as follows:
;$(1): \quad a \circ a$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & & \\
b & & b & \\
c & & & c \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b & & \\
b & & c & \\
c & & & a \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c & & \\
b & & a & \\
c & & & b \\
\end {array}$
So selecting $a \circ a$ fixes $b \circ b$ and $c \circ c$.
;$(2): \quad a \circ b$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & a & \\
b & & & b \\
c & c & & \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & b & \\
b & & & c \\
c & a & & \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & c & \\
b & & & a \\
c & b & & \\
\end {array}$
So selecting $a \circ b$ fixes $b \circ c$ and $c \circ a$.
;$(3): \quad a \circ c$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & a \\
b & b & & \\
c & & c & \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & b \\
b & c & & \\
c & & a & \\
\end {array}$
|-
| align="left" |
| align="left" |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & & & c \\
b & a & & \\
c & & b & \\
\end {array}$
So selecting $a \circ c$ fixes $b \circ a$ and $c \circ b$.
There are $3$ elements $x$ of $S$ with which $a \circ x$ can be made.
For each of these $3$, there are $3$ different elements $y$ such that $a \circ x = y$.
These collectively fix all the possible values of $b \circ x$ and $c \circ x$.
Hence they exhaust all possible operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Thus there are independently:
:$3$ different options for $a \circ a$
:$3$ different options for $a \circ b$
:$3$ different options for $a \circ c$
and therefore $3 \times 3 \times 3 = 3^3$ operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Note that from Automorphism Group of $\AA$, $3$ of these are also such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$.
So these are excluded from our count.
The result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$
Category:Sets of Operations on Set of 3 Elements