Group of Order 3 is Unique

Theorem
There exists exactly $1$ group of order $3$, up to isomorphism:


 * $C_3$, the cyclic group of order $3$.

Proof
From Existence of Cyclic Group of Order n we have that one such group of order $3$ is the cyclic group of order $3$.

This is exemplified by the additive group of integers modulo $3$, whose Cayley table can be presented as:

Consider an arbitrary group $\struct {G, \circ}$ whose identity element is $e$.

Let the underlying set of $G$ be:
 * $G = \set {e, a, b}$

where $a, b \in G$ are arbitrary.

Since $e$ is the identity, we can start off the Cayley table for $G$ as:


 * $\begin{array}{c|ccc}

\circ & e & a & b \\ \hline e & e & a & b \\ a & a &  &   \\ b & b &  &   \\ \end{array}$

Consider the element at $a \circ a$.

We have that $a \circ a$ must be either $e$ or $b$, as from Group has Latin Square Property it cannot be $a$.

If $a \circ a = e$, then that leaves only $b$ to complete the middle row.

But $b$ already exists in the final column.

So $a \circ a$ cannot be $e$, so must be $b$.

Hence the Cayley table for $G$ so far is:


 * $\begin{array}{c|ccc}

\circ & e & a & b \\ \hline e & e & a & b \\ a & a & b &  \\ b & b &  &   \\ \end{array}$

The rest of the table is completed by following the result that Group has Latin Square Property:


 * $\begin{array}{c|ccc}

\circ & e & a & b \\ \hline e & e & a & b \\ a & a & b & e \\ b & b & e & a \\ \end{array}$

and it is seen by inspection that $G$ is indeed the cyclic group of order $3$.