= Solution
Direct <matrix multiplication> gives $P^2=P-I$ and hence $P^3=-I$. It follows that
$$
\begin{aligned}
P^0&=\begin{pmatrix}1&0\\0&1\end{pmatrix},&P&=\begin{pmatrix}1&1\\-1&0\end{pmatrix},&P^2&=\begin{pmatrix}0&1\\-1&-1\end{pmatrix},\\
P^3&=\begin{pmatrix}-1&0\\0&-1\end{pmatrix},&P^4&=\begin{pmatrix}-1&-1\\1&0\end{pmatrix},&P^5&=\begin{pmatrix}0&-1\\1&1\end{pmatrix},
\end{aligned}
$$
and $P^6=I$. These six matrices are distinct; consequently no smaller positive power is the <identity element>. Thus
$$
\boxed{\langle P\rangle=\{I,P,P^2,P^3,P^4,P^5\}\cong C_6,\qquad\operatorname{ord}(P)=6.}
$$
Products add exponents modulo six and $P^{-1}=P^5$. This verifies the <cyclic group> structure explicitly.
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