Sets of Operations on Set of 3 Elements/Automorphism Group of D/Lemma
Lemma for Sets of Operations on Set of 3 Elements
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$, $\CC_3$ and $\DD$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
\(\ds \AA\) | \(:\) | \(\ds \map \Gamma S\) | where $\map \Gamma S$ is the symmetric group on $S$ | |||||||||||
\(\ds \BB\) | \(:\) | \(\ds \set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }\) | where $I_S$ is the identity mapping on $S$ | |||||||||||
\(\ds \CC_1\) | \(:\) | \(\ds \set {I_S, \tuple {a, b} }\) | ||||||||||||
\(\ds \CC_2\) | \(:\) | \(\ds \set {I_S, \tuple {a, c} }\) | ||||||||||||
\(\ds \CC_3\) | \(:\) | \(\ds \set {I_S, \tuple {b, c} }\) | ||||||||||||
\(\ds \DD\) | \(:\) | \(\ds \set {I_S}\) |
Then:
- $\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$ forms a partition of the set of all operations on $S$.
Proof
Let $\OO$ denote the set of all operations on $S$.
Let $\SS := \set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.
First we note that from Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Thus every operation on $S$ belongs to at least one of $\set {\AA, \BB, \CC_1, \CC_2, \CC_3, \DD}$.
That is:
- $\OO = \ds \bigcup \SS$
Then we note that from:
each of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$ is non-empty.
Consider for example the constant operation on $S$:
- $\forall x, y \in S: x \circ y = a$
From the above cited results, $\circ$ as described does not belong to any of $\AA$, $\BB$, $\CC_1$, $\CC_2$ and $\CC_3$.
Hence it must belong to $\DD$.
Thus $\DD$ is also non-empty.
It remains to be shown that $\SS$ is a pairwise disjoint set of sets.
By definition, $\DD$ is disjoint with all other elements of $\SS$.
By construction, $\AA$ is also disjoint with all other elements of $\SS$.
This needs considerable tedious hard slog to complete it. In particular: Need to establish that $\BB \cap \CC_1 = \O$ etc. and that $\CC_x \cap \CC_y = \O$ To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |