= Solution
Let $B=\left(\begin{smallmatrix}0&b\\c&0\end{smallmatrix}\right)$ be the lower block. Then
$$
B^2=bcI_2,\qquad T(a)^2=\operatorname{diag}(1,bc,bc),\qquad T(a)^4=\operatorname{diag}(1,(bc)^2,(bc)^2).
$$
The defining relation $a^4=I$ requires $T(a)^4=I_3$, so for real constants $bc=\pm1$. Both possibilities ensure invertibility.
If $bc=-1$, the generator matrix has order four, because its square is not the identity while its fourth power is. Its four powers are distinct, giving a <faithful representation>. If $bc=1$, its square is the identity but it is not itself the identity; its lower diagonal entries are zero. The image then has order two and the <kernel of a group homomorphism> is $\{I,a^2\}$. Thus the <faithfulness criterion for a block representation of the cyclic group of order four> gives
$$
\boxed{\text{faithful exactly when }bc=-1;\qquad\text{unfaithful representation exactly when }bc=1.}
$$
Other parameter values give no representation of $Z_4$, not additional unfaithful cases; $bc=0$ also makes the assigned matrix singular. No orthogonality of these matrices was required.
For the final construction take
$$
\boxed{D(a)=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad D(a^r)=\begin{pmatrix}1&0\\0&(-1)^r\end{pmatrix}.}
$$
Multiplication adds exponents, so $D(a^r)D(a^s)=D(a^{r+s})$ and $D(a)^4=I_2$. The representing matrix is the identity exactly for even exponents. Therefore this is a two-dimensional representation with kernel \b[$\{I,a^2\}$], as required. It factors through the quotient cyclic group of order two.
Back to article page
