= Solution
The subgroups $\langle A,B\rangle$ and $\langle C,D\rangle$ each have order eight, their elements commute across the two subgroups, and their intersection is exactly $\langle Z\rangle$. Therefore
$$
|G|=\frac{8\cdot8}{2}=32.
$$
Since $G'=\langle Z\rangle$, the <abelianization> $G/G'$ has order $16$ and is generated by the four commuting involutions represented by $A,B,C,D$. It is isomorphic to $(C_2)^4$, so it has sixteen one-dimensional complex representations, obtained by independently sending $A,B,C,D$ to $\pm1$ and sending $Z$ to $1$.
These characters contribute $16$ to the <sum of squares of irreducible degrees>. The given irreducible representation has degree four and contributes another $16$. Part (a) and $|G|=32$ show that no other irreducible representations exist. Thus the complete list is the sixteen linear characters and the given four-dimensional representation.
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