View source for Phy5646:Problem 7.1 Solution

=Question=
Show that there is only one group of order 3.

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

Our group multiplication table is then unique:

<math>
\begin{matrix}
e & a & b \\
a & b & e \\
b & e & a 
\end{matrix}
</math>