= Solution
For the <symmetric group> representation in part (a), the identity and three transpositions contribute $4I$ to $A=\sum_gD(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. Thus
$$
\boxed{A=3I=\frac{6\cdot1}{2}I.}
$$
More generally, the <central average of representation matrices at squares> commutes with the representation, because conjugation permutes the summands. For a complex <irreducible representation>, <Schur's lemma> therefore gives $A=cI$. Taking the <trace> yields $c\dim D=|G|\nu$, proving the claimed formula under its correct hypotheses.
For the <Klein four-group> there is no two-dimensional irreducible representation, as established in part (b). Its genuine one-dimensional irreducibles have $A=4$ and $\nu=1$, so $A=(4\cdot1/1)I$.
If the intended alternative is the real quarter-turn representation of the <cyclic group of order four>, its squared matrices sum to $I-I+I-I=0$. Thus $A=0=(4\cdot0/2)I$ in that example. This equality is a direct calculation; the complex version of <Schur's lemma> cannot be applied to its reducible complexification.
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