= Solution
Observe that $A=-I$, $B^2=I$ and $C=-B$. Thus $A^2=B^2=C^2=I$, while $AB=BA=C$, $AC=CA=B$ and $BC=CB=A$. The <multiplication table>, with the left factor labeling the row, is
$$
\boxed{\begin{array}{c|cccc}
&I&A&B&C\\\hline
I&I&A&B&C\\
A&A&I&C&B\\
B&B&C&I&A\\
C&C&B&A&I
\end{array}.}
$$
It proves <closure> of the set. <Associativity> is inherited from <matrix multiplication>, $I$ is the <identity element>, and every element is its own <inverse element>. Therefore this set is a <group>. The table is symmetric, so it is an <Abelian group>; its three nonidentity elements have order two, identifying it as the <Klein four-group> $C_2\times C_2$.
Back to article page
