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.

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" |


 * align="left"  |
 * $\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a & a &  &   \\ b &  & b &   \\ c &  &   & c \\ \end {array}$
 * align="left" |
 * 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" |


 * 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" |


 * align="left"  |
 * $\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a &  & a &   \\ b &  &   & b \\ c & c &  &   \\ \end {array}$
 * align="left" |
 * 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" |


 * 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" |


 * align="left"  |
 * $\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a &  &   & a \\ b & b &  &   \\ c &  & c &   \\ \end {array}$
 * align="left" |
 * 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" |


 * 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.