= Solution
The elements $1,-1$ commute with the whole <quaternion group>, so they form singleton classes. Each of the three remaining pairs is a <conjugacy class>: for example conjugation by $j$ sends $i$ to $-i$, and likewise cyclically. No such pair can mix with a different pair, because the quotient by $\{\pm1\}$ is abelian and its <cosets> $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. Hence
$$
\boxed{\{1\},\quad\{-1\},\quad\{i,-i\},\quad\{j,-j\},\quad\{k,-k\}.}
$$
For a finite group over $\mathbb C$, the number of inequivalent <irreducible representations> equals the number of <conjugacy classes>, so there are five. The <sum of squares of irreducible representation dimensions> is the <group order>, eight. With five positive integer dimensions, the only possibility is
$$
\boxed{1,1,1,1,2.}
$$
Indeed five one-dimensional contributions give five, and replacing one by a two-dimensional contribution adds exactly three; any further enlargement would exceed eight. These are the dimensions used in the <character table of the quaternion group>.
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