= Solution
The given <permutations> satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight elements $r^j,r^js$ for $0\leq j<4$ are distinct: $s\notin\langle r\rangle$, whose only element of order two is $r^2=(13)(24)$. Thus $H$ is the order-eight <dihedral group>, and we may use the <matrices>
$$
R=\sigma(r)=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
S=\sigma(s)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$
They obey the same relations and describe a quarter-turn and a <reflection> of a <square>.
The <eigenvalues> of $R$ over the <complex numbers> are $i$ and $-i$, with respective <eigenlines> spanned by $(1,-i)$ and $(1,i)$. Since $S$ interchanges these <eigenlines>, neither is an <invariant subspace> for $H$. Any proper nonzero <subrepresentation> of this two-dimensional <complex representation> would be a line invariant under both $R$ and $S$, and an $R$-invariant line must be one of those <eigenlines>. None qualifies, so \b[$W$ is irreducible].
For the calculation in part (e), its <character> is
$$
\chi_W(1)=2,\qquad \chi_W(r^2)=-2,\qquad
\chi_W(r)=\chi_W(r^3)=\chi_W(r^js)=0\quad(0\leq j<4).
$$
Indeed $R^2=-I$, while a quarter-turn or a <reflection> has <trace> zero. The <character inner product> $(2^2+(-2)^2)/8=1$ provides a second irreducibility check.
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