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

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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:

\(\ds \CC_1\) \(:\) \(\ds \set {I_S, \tuple {a, b} }\)
\(\ds \CC_2\) \(:\) \(\ds \set {I_S, \tuple {a, c} }\)
\(\ds \CC_3\) \(:\) \(\ds \set {I_S, \tuple {b, c} }\)

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

Without loss of generality, we will analyse the nature of $\CC_1$.

Recall this lemma:

Lemma

\(\ds a \circ a = a\) \(\iff\) \(\ds b \circ b = b\)
\(\ds a \circ a = b\) \(\iff\) \(\ds b \circ b = a\)
\(\ds a \circ a = c\) \(\iff\) \(\ds b \circ b = c\)
\(\ds a \circ b = a\) \(\iff\) \(\ds b \circ a = b\)
\(\ds a \circ b = b\) \(\iff\) \(\ds b \circ a = a\)
\(\ds a \circ b = c\) \(\iff\) \(\ds b \circ a = c\)
\(\ds a \circ c = a\) \(\iff\) \(\ds b \circ c = b\)
\(\ds a \circ c = b\) \(\iff\) \(\ds b \circ c = a\)
\(\ds a \circ c = c\) \(\iff\) \(\ds b \circ c = c\)
\(\ds c \circ a = a\) \(\iff\) \(\ds c \circ b = b\)
\(\ds c \circ a = b\) \(\iff\) \(\ds c \circ b = a\)
\(\ds c \circ a = c\) \(\iff\) \(\ds c \circ b = c\)

$\Box$


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

\(\ds a \circ b = b\) \(\iff\) \(\ds b \circ a = a\)

However, we note that:

\(\ds a \circ c = a\) \(\iff\) \(\ds b \circ c = b\)
\(\ds c \circ a = a\) \(\iff\) \(\ds c \circ b = b\)

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 $\CC_3$.

$\blacksquare$


Sources

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