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

From ProofWiki
Jump to navigation Jump to search

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:

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


Proof

Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:

$\begin {array} {c|ccc}

\to & a & b & c \\ \hline a & a & b & c \\ b & a & b & c \\ c & a & b & c \\ \end {array} \qquad \begin {array} {c|ccc} \gets & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$


We have from Algebraic Structures formed by Left and Right Operations are not Isomorphic for Cardinality Greater than 1 that $\struct {S, \to}$ and $\struct {S, \gets}$ are not isomorphic.


Observe from the Cayley table $\circ$ that:

$(1): \quad \forall x, y \in S: x \ne y \implies x \circ y \ne x \land x \circ y \ne y$


Aiming for a contradiction, suppose there exists an isomorphism $\phi$ from $\struct {S, \to}$ to $\struct {S, \circ}$.

We have:

Then:

\(\ds \map \phi {a \to b}\) \(=\) \(\ds \map \phi a \circ \map \phi b\) Definition of Isomorphism
\(\ds \) \(=\) \(\ds \map \phi c\) from $(1)$
\(\ds \) \(=\) \(\ds \map \phi b\) Definition of Right Operation

Hence:

$\map \phi c = \map \phi b$

which means $\phi$ is not an injection.

That is, $\phi$ is not a bijection and hence not an isomorphism.

Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \to}$ to $\struct {S, \circ}$.


In a similar way:

Aiming for a contradiction, suppose there exists an isomorphism $\phi$ from $\struct {S, \gets}$ to $\struct {S, \circ}$.

We have:

Then:

\(\ds \map \phi {a \gets b}\) \(=\) \(\ds \map \phi a \circ \map \phi b\) Definition of Isomorphism
\(\ds \) \(=\) \(\ds \map \phi c\) from $(1)$
\(\ds \) \(=\) \(\ds \map \phi a\) Definition of Left Operation

Hence:

$\map \phi c = \map \phi a$

which means $\phi$ is not an injection.

That is, $\phi$ is not a bijection and hence not an isomorphism.

Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \gets}$ to $\struct {S, \circ}$.


It has been shown that none of the elements of $\AA$ is isomorphism to any of the others.

The result follows by definition of isomorphism class.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sets_of_Operations_on_Set_of_3_Elements/Automorphism_Group_of_A/Isomorphism_Classes&oldid=566974"