Sets of Operations on Set of 3 Elements/Automorphism Group of C n/Operations with Identity

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

Let $\CC_1$, $\CC_2$ and $\CC_3$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:

where $I_S$ is the identity mapping on $S$.

Then:
 * $9$ of the operations of each of $\CC_1$, $\CC_2$ and $\CC_3$ has an identity element.

Proof
, we will analyse the nature of $\CC_1$.

Recall this lemma:

Lemma
We observe that neither $a$ nor $b$ can be an identity, because:

However, we note that:

and so from here it may still be the case that $c$ may be an identity for at least one element of $\CC_1$.

There are still $3$ options for $a \circ a$ and $a \circ b$ which do not affect the behaviour of $c \circ x$ or $x \circ c$.

From the Product Rule for Counting it follows that there are $3 \times 3 = 9$ operations of $\CC_1$ which have an identity element.

As $\CC_2$ and $\CC_3$ can be obtained from $\CC_1$ just by renaming elements of $S$, the same applies to both $\CC_2$ and $\CC_3$.