= Solution
Use the identity relation as printed in the PDF, $XYZ=I$. Since $Z^2=I$, multiplying on the right by $Z$ gives $XY=Z$. Also
$$
YX=(XY)^{-1}=Z^{-1}=Z=XY,
$$
because $X,Y,Z$ are involutions. Thus $X$ and $Y$ commute, and so does their product $Z$ with either of them. Therefore \b[the group is Abelian], and is the <Klein four-group>.
Each <conjugacy class> is a singleton. Over the complex numbers every <irreducible representation of a finite abelian group> is one-dimensional. Choose independent signs $\varepsilon=\chi(X)$ and $\eta=\chi(Y)$; the relations force $\varepsilon^2=\eta^2=1$ and $\chi(Z)=\varepsilon\eta$. Hence the complete <character table> is
$$
\boxed{\begin{array}{c|rrrr}
&I&X&Y&Z\\\hline
\chi_{++}&1&1&1&1\\
\chi_{+-}&1&1&-1&-1\\
\chi_{-+}&1&-1&1&-1\\
\chi_{--}&1&-1&-1&1
\end{array}.}
$$
These four distinct <one-dimensional representations> exhaust the irreducible possibilities: there are four <conjugacy classes>, and their squared dimensions already sum to $|G|=4$. The rows also satisfy <character orthogonality>. The local transcription damages the final symbol in the defining relation; the PDF makes the identity $I$ unambiguous.
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