Sets of Operations on Set of 3 Elements/Operations with Identity

Theorem

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

Let $\NN$ be the set of all operations $\circ$ on $S$ which have an identity element.

Then the elements of $\NN$ are divided in $45$ isomorphism classes.

That is, up to isomorphism, there are $45$ operations on $S$ which have an identity element.

Proof

From Automorphism Group of $\AA$: Operations with Identity:

there are no elements of $\AA$ which have an identity element.

From Automorphism Group of $\BB$: Operations with Identity:

there are no elements of $\BB$ which have an identity element.

From Automorphism Group of $\CC_n$: Operations with Identity:

there are $3 \times 9$ elements of $\CC$ which have an identity element.

From Automorphism Group of $\DD$: Operations with Identity:

there are $216$ elements of $\DD$ which have an identity element.

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:

$\dfrac {3 \times 9} 3 = 9$ isomorphism classes of elements of $\CC$ which have an identity element.

$\dfrac {216} 6 = 36$ isomorphism classes of elements of $\DD$ which have an identity element.

Thus there are $3 + 36 = 45$ 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 $8.14 \ \text{(c)}$