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

Theorem
Let $S = \set {a, b, c}$ be a set with $3$ elements.

Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.

Then:
 * $\AA$ has $3$ 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 $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$

Hence $\AA$ can be defined as the set of operations $\circ$ on $S$ such that every permutation on $S$ is an automorphism of $\struct {S, \circ}$.

The set $\map \Gamma S$ of all permutations on $S = \set {a, b, c}$ has $6$ elements.

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 the various elements of $S$.

Let us denote each of the remaining permutations on $S$ as follows:

So, let $\circ \in \AA$.

From Permutation of Set is Automorphism of Set under Left Operation and Permutation of Set is Automorphism of Set under Right Operation:
 * ${\to} \in \AA$
 * ${\gets} \in \AA$

where $\to$ and $\gets$ are the right operation and left operation respectively.

Next we note that from Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent:
 * $\forall x \in S: x \circ x = x$

Let us explore various options for $a \circ b$.

Let $a \circ b = a$.

Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:

This is none other than the left operation, which has already been counted.

Let $a \circ b = b$. Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:

This is none other than the right operation, which has already been counted.

Let $a \circ b = c$. Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:

Thus we have identified another operation on $S$ for which all permutations on $S$ are automorphisms of $\struct {S, \circ}$.

We now note that once the product element $a \circ b$ has been selected to be either $a$, $b$ or $c$, the full structure of $\struct {S, \circ}$ is forced.

Hence there are no other operations $\circ$ on $S$ but these ones we have counted.

That is, there is a total of $3$ such operations.