Group of Order 3 is Unique

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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:

$\begin{array}{r|rrr} \struct {\Z_3, +_3} & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \hline \eqclass 0 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \eqclass 1 3 & \eqclass 1 3 & \eqclass 2 3 & \eqclass 0 0 \\ \eqclass 2 3 & \eqclass 2 3 & \eqclass 0 3 & \eqclass 1 3 \\ \end{array}$

$\Box$


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 $3$.

$\blacksquare$


Sources

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