= Solution
A <cyclic group> of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from <Lagrange's theorem>; for even $p$, the unique such element is $c=g_1^{p/2}$. Hence
$$
\boxed{p\text{ is even},\qquad C=\{I,g_1^{p/2}\}.}
$$
Because $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta=2\pi q/p$, with $\gcd(q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:
$$
\boxed{D(c)=-I_2,\qquad D(g_i)=\begin{pmatrix}\cos(i\vartheta)&-\sin(i\vartheta)\\\sin(i\vartheta)&\cos(i\vartheta)\end{pmatrix},\quad\vartheta=\frac{2\pi q}{p},\quad\gcd(q,p)=1.}
$$
The <character> is the trace:
$$
\boxed{\chi(g_i)=2\cos\frac{2\pi qi}{p},\qquad\chi(I)=2.}
$$
Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the <irreducible real representations of a finite cyclic group>.
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