Phy5646:Problem 7.1 Solution

Question

Show that there is only one group of order 3.

Solution

Let $G$ be our group and $e,a,b\in G$. Such that $a$, $b$, $e$ are distinct. Therefore $a\cdot a=e$ or $a\cdot a=b$. Well, if $a\cdot a=e$ then $a\cdot b=b$, however, if $a\cdot b=b$ then $a=a\cdot b\cdot b^{-1}=b\cdot b^{-1}=e$ and the group is not of order 3. Therefore, $a\cdot a=b$. Similarly, $b\cdot b=a$.

Our group multiplication table is then unique:

${\begin{matrix}e&a&b\\a&b&e\\b&e&a\end{matrix}}$