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

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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.


Cayley Table

The Cayley table of the operation $\circ$ on $S$ such that:

every permutation of $S$ is an automorphism on $\struct {S, \circ}$
$\circ$ is neither the right operation nor the left operation

can be presented as follows:

$\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$


Isomorphism Classes

The elements of $\AA$ are each in its own isomorphism class.


Operations with Identity

None of the operations of $\AA$ has an identity element.


Commutative Operations

Exactly $1$ of the operations of $\AA$ is commutative.


Proof

Recall the definition of (group) automorphism:

$\phi$ is an automorphism on $\struct {S, \circ}$ if and only if:
$\phi$ 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:

\(\ds p\) \(:\) \(\ds \map p a = b, \map p b = c, \map p c = a\)
\(\ds q\) \(:\) \(\ds \map q a = c, \map q b = a, \map q c = b\)
\(\ds r\) \(:\) \(\ds \map r a = a, \map r b = c, \map r c = b\)
\(\ds s\) \(:\) \(\ds \map s a = c, \map s b = b, \map s c = a\)
\(\ds t\) \(:\) \(\ds \map t a = b, \map t b = a, \map t c = c\)


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:

\(\ds \map p a \circ \map p b\) \(=\) \(\ds \map p a\)
\(\ds \leadsto \ \ \) \(\ds b \circ c\) \(=\) \(\ds b\)
\(\ds \map q a \circ \map q b\) \(=\) \(\ds \map q a\)
\(\ds \leadsto \ \ \) \(\ds c \circ a\) \(=\) \(\ds c\)
\(\ds \map r a \circ \map r b\) \(=\) \(\ds \map r a\)
\(\ds \leadsto \ \ \) \(\ds a \circ c\) \(=\) \(\ds a\)
\(\ds \map s a \circ \map s b\) \(=\) \(\ds \map s a\)
\(\ds \leadsto \ \ \) \(\ds c \circ b\) \(=\) \(\ds c\)
\(\ds \map t a \circ \map t b\) \(=\) \(\ds \map t a\)
\(\ds \leadsto \ \ \) \(\ds b \circ a\) \(=\) \(\ds b\)

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

$\Box$


Let $a \circ b = b$.

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

\(\ds \map p a \circ \map p b\) \(=\) \(\ds \map p b\)
\(\ds \leadsto \ \ \) \(\ds b \circ c\) \(=\) \(\ds c\)
\(\ds \map q a \circ \map q b\) \(=\) \(\ds \map q b\)
\(\ds \leadsto \ \ \) \(\ds c \circ a\) \(=\) \(\ds a\)
\(\ds \map r a \circ \map r b\) \(=\) \(\ds \map r b\)
\(\ds \leadsto \ \ \) \(\ds a \circ c\) \(=\) \(\ds c\)
\(\ds \map s a \circ \map s b\) \(=\) \(\ds \map s b\)
\(\ds \leadsto \ \ \) \(\ds c \circ b\) \(=\) \(\ds b\)
\(\ds \map t a \circ \map t b\) \(=\) \(\ds \map t b\)
\(\ds \leadsto \ \ \) \(\ds b \circ a\) \(=\) \(\ds a\)

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

$\Box$


Let $a \circ b = c$.

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

\(\ds \map p a \circ \map p b\) \(=\) \(\ds \map p c\)
\(\ds \leadsto \ \ \) \(\ds b \circ c\) \(=\) \(\ds a\)
\(\ds \map q a \circ \map q b\) \(=\) \(\ds \map q c\)
\(\ds \leadsto \ \ \) \(\ds c \circ a\) \(=\) \(\ds b\)
\(\ds \map r a \circ \map r b\) \(=\) \(\ds \map r c\)
\(\ds \leadsto \ \ \) \(\ds a \circ c\) \(=\) \(\ds b\)
\(\ds \map s a \circ \map s b\) \(=\) \(\ds \map s c\)
\(\ds \leadsto \ \ \) \(\ds c \circ b\) \(=\) \(\ds a\)
\(\ds \map t a \circ \map t b\) \(=\) \(\ds \map t c\)
\(\ds \leadsto \ \ \) \(\ds b \circ a\) \(=\) \(\ds c\)

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

$\Box$


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.

$\blacksquare$


Sources

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