= Solution
Write the two <Pauli matrices> as
$$
A=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
B=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad
C=AB=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
They satisfy $A^2=B^2=I$, $BA=-AB$ and $C^2=-I$. Moving every $A$ to the left of every $B$ reduces each word to one of eight distinct <matrices>. Consequently
$$
\boxed{G=\{I,-I,A,-A,B,-B,C,-C\},\qquad |G|=8.}
$$
Taking $r=C$, $s=A$ gives $r^4=s^2=I$ and $srs=r^{-1}$, identifying the <group> with the <dihedral group> of a square.
The elements $\pm I$ are central. Conjugating $A$ by $B$ changes its sign, and similarly for $B$ and $C$. No conjugation mixes these pairs: modulo the central <subgroup> $\{\pm I\}$ the <quotient group> is the <Klein four-group> and hence <Abelian>. Thus the <conjugacy classes> are precisely
$$
\boxed{\{I\},\quad\{-I\},\quad\{A,-A\},\quad\{B,-B\},\quad\{C,-C\}.}
$$
There are five complex <irreducible representations>, one for each <conjugacy class>. In a <one-dimensional representation>, the images of $A,B$ are signs $s,t\in\{1,-1\}$. Since they commute, $-I$ maps to $1$, and all four choices of $s,t$ satisfy the relations. These are four inequivalent <one-dimensional representations>. The <sum of squares of irreducible representation dimensions> leaves $8-4=4$ for the fifth, so the <dimensions> are
$$
\boxed{1,1,1,1,2.}
$$
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