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..