Sets of Operations on Set of 3 Elements/Commutative Operations
Theorem
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\PP$ be the set of all commutative operations $\circ$ on $S$.
Then the elements of $\PP$ are divided in $129$ isomorphism classes.
That is, up to isomorphism, there are $129$ commutative operations on $S$ which have an identity element.
Proof
From Automorphism Group of $\AA$: Commutative Operations:
- there is exactly $1$ commutative operation in $\AA$.
From Automorphism Group of $\BB$: Commutative Operations:
- there are $8$ commutative operations in $\BB$.
From Automorphism Group of $\CC_n$: Commutative Operations:
- there are $3 \times 8$ commutative operations in $\CC$.
From Automorphism Group of $\DD$: Commutative Operations:
- there are $696$ commutative operations in $\DD$.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
- the elements of $\BB$ form isomorphism classes in pairs.
From Automorphism Group of $\CC_n$: Isomorphism Classes:
- the elements of $\CC$ form isomorphism classes in threes.
From Automorphism Group of $\DD$: Isomorphism Classes:
- the elements of $\DD$ form isomorphism classes in sixes.
Hence there are:
- $\dfrac 8 2 = 4$ isomorphism classes of commutative operations in $\BB$.
- $\dfrac {3 \times 8} 3 = 8$ isomorphism classes of commutative operations in $\CC$.
- $\dfrac {696} 6 = 116$ isomorphism classes of commutative operations in $\DD$.
Thus there are $1 + 4 + 8 + 116 = 129$ isomorphism classes of operations $\circ$ on $S$ which have an identity element..
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$