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Since $a^4=I$, multiplication adds exponents modulo four: $a^j{a}^k=a^{j+k\mathrm{mod}4}$. Hence the Cayley table of the cyclic group of order four, with the left factor indexing rows, isThe table is symmetric because the group is Abelian. Its row for each element permutes all four members, as cancellation requires.

The three order-four subgroups areIn each case the generator has powers $A,A^2=\hat{W},A^3=\hat{A},A^4=W$, so all three are cyclic and mutually isomorphic to $C_4$, a cyclic group of order four. For $G_1$ the table isThe maps $X^j\mapsto Y^j$ and $X^j\mapsto Z^j$ are explicit group isomorphisms. These are all order-four subgroups: any such subgroup contains an element other than $W,\hat{W}$, and that element has order four and already generates the subgroup. They share the order-two subgroup $\{W,\hat{W}\}$ and are each normal, being of index two.

The complete subgroup list isThus there are ten subgroups: one trivial, five of order two, three of order four, and the whole group. The five order-two subgroups are mutually isomorphic. The two noncyclic order-four subgroups are mutually isomorphic Klein four-groups; neither is isomorphic to the cyclic group of order four.

For completeness, Lagrange's theorem permits only orders $1,2,4,8$. An order-four subgroup containing $U$ or $U^3$ must be $⟨U⟩$. Otherwise it can contain only $I,U^2$ and elements $U^{j}V$. The product of two distinct such elements is $U^{j-k}$. To avoid generating $U$ or $U^3$, their exponents must differ by two, giving exactly the two listed Klein four-groups. The order-two subgroups come from the five involutions $U^2$ and $U^{j}V$. This proves completeness.

Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a cyclic group of order four. Its multiplication table isWrite the specified image of $a$ as $A$. Direct matrix multiplication givesA group representation is a group homomorphism to a general linear group, so the images of all powers of $a$ are forced:All four images are invertible. Defining $\Phi (a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell }=A^j{A}^{\ell }$ verifies every product in the table. These observations prove both existence and uniqueness of this representation with the prescribed image.

The requested two-dimensional irreducible representation does not exist for the Vierergruppe. The Klein four-group $V_4$ is an abelian group in which every nonidentity element has order two. Its complex irreducible representations are all one-dimensional. The same holds over the real numbers here: the commuting representing involutions are simultaneously diagonalizable with real eigenvalues $\pm 1$, so every two-dimensional representation has an invariant line. Each genuine irreducible character satisfies $\chi (g^2)=\chi (1)=1$, and hence has Frobenius-Schur indicator $\nu =1$.

The printed notation $C_4$ suggests a different possible interpretation, which must not be conflated with the Klein four-group. For the cyclic group of order four, a real two-dimensional irreducible representation is generated byIts character on $1,J,-I,-J$ is $(2,0,-2,0)$, giving $\nu =(2-2+2-2)/4=0$. Over the complex numbers this representation splits into two conjugate one-dimensional characters, so it is not a two-dimensional complex irreducible representation. These alternatives explain what can be computed without silently changing the printed group or the ground field.

For the symmetric group representation in part (a), the identity and three transpositions contribute $4I$ to $A={\sum }_{g}D(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. ThusMore 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 $A=(4\cdot 1/1)I$.

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 $A=0=(4\cdot 0/2)I$ in that example. This equality is a direct calculation; the complex version of Schur's lemma cannot be applied to its reducible complexification.