= Solution
The <matrices> obey $Q^5=P^2=I$ and $PQP=Q^{-1}$. Moving every $P$ to the right therefore reduces every product to $Q^k$ or $Q^kP$, with $0\leq k<5$. The five diagonal <matrices> $Q^k$ are distinct because $\alpha$ is a primitive fifth <root of unity>; the five $Q^kP$ are distinct and off-diagonal. Thus
$$
\boxed{G=\{Q^k,Q^kP:0\leq k<5\},\qquad |G|=10.}
$$
This is the <dihedral group> of the regular pentagon, with $Q$ a rotation and $P$ a reflection.
Conjugation by $P$ interchanges $Q^k$ and $Q^{-k}$, while conjugation by a power of $Q$ fixes rotations. Also
$$
Q^j(Q^kP)Q^{-j}=Q^{k+2j}P.
$$
Since multiplication by $2$ permutes the residues modulo $5$, all five reflections form one <conjugacy class>. The full list of <conjugacy classes> is
$$
\boxed{\{I\},\quad\{Q,Q^4\},\quad\{Q^2,Q^3\},\quad\{P,QP,Q^2P,Q^3P,Q^4P\}.}
$$
To prove the <representation> is <irreducible> over $\mathbb C$, any proper nonzero <invariant subspace> would be a line invariant under $Q$. Its two distinct <eigenvalues> force such a line to be one of its coordinate <eigenspaces>. But $P$ exchanges those two lines, so neither is $G$-invariant. Hence the given <complex representation> is <irreducible>.
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