 Solution

The subgroups $⟨A,B⟩$ and $⟨C,D⟩$ each have order eight, their elements commute across the two subgroups, and their intersection is exactly $⟨Z⟩$. ThereforeSince $G^{\prime }=⟨Z⟩$, the abelianization $G/G^{\prime }$ 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 $\pm 1$ 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.