= Solution
The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. <Associativity> is inherited from <matrix multiplication>, $W$ is the <identity element>, and the inverses from (a), with $( -A)^{-1}=-A^{-1}$, remain in the set. Thus the <group axioms> hold. This is the <quaternion group> $Q_8$.
In the element order specified in the question, the <multiplication table> is
$$
\boxed{\begin{array}{c|cccccccc}
&W&X&Y&Z&\widehat W&\widehat X&\widehat Y&\widehat Z\\\hline
W&W&X&Y&Z&\widehat W&\widehat X&\widehat Y&\widehat Z\\
X&X&\widehat W&Z&\widehat Y&\widehat X&W&\widehat Z&Y\\
Y&Y&\widehat Z&\widehat W&X&\widehat Y&Z&W&\widehat X\\
Z&Z&Y&\widehat X&\widehat W&\widehat Z&\widehat Y&X&W\\
\widehat W&\widehat W&\widehat X&\widehat Y&\widehat Z&W&X&Y&Z\\
\widehat X&\widehat X&W&\widehat Z&Y&X&\widehat W&Z&\widehat Y\\
\widehat Y&\widehat Y&Z&W&\widehat X&Y&\widehat Z&\widehat W&X\\
\widehat Z&\widehat Z&\widehat Y&X&W&Z&Y&\widehat X&\widehat W
\end{array}}
$$
Interpreting “character of the <group>” as the <character of a representation> specified by these <matrices>, take the <trace> of each <matrix>:
$$
\boxed{\chi(W)=2,\qquad\chi(\widehat W)=-2,\qquad
\chi(X)=\chi(Y)=\chi(Z)=\chi(\widehat X)=\chi(\widehat Y)=\chi(\widehat Z)=0.}
$$
A character belongs to a <representation>; the underlying abstract <group> can have other characters.
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