= Solution
The <conjugacy class> of $g\in G$ is $\{hgh^{-1}:h\in G\}$. An <irreducible representation> acts on a nonzero vector space with no nonzero proper subspace invariant under every group element.
In the <quaternion group>, $1$ and $-1$ commute with all elements, so each forms its own conjugacy class. The element $i$ commutes with $\pm1,\pm i$, while conjugation by $j$ or $k$ changes it to $-i$. Thus its class is $\{i,-i\}$. The same argument for $j,k$ gives all five classes:
$$
\boxed{\{1\},\qquad\{-1\},\qquad\{i,-i\},\qquad\{j,-j\},\qquad\{k,-k\}.}
$$
Their sizes sum to eight. Equivalently each noncentral element has a four-element centralizer and hence a two-element class.
Use the following finite-group representation theorems over $\mathbb C$: the number of inequivalent irreducible representations equals the number of conjugacy classes, and their dimensions obey the <sum of squares of irreducible representation dimensions>, $\sum d_j^2=|G|$. Thus there are five irreducibles, with $\sum_{j=1}^5d_j^2=8$. Starting with five dimensions equal to one uses five of those eight units; increasing one dimension from one to two uses exactly the remaining three, while any larger increase would exceed eight. Therefore
$$
\boxed{\text{five complex irreducible representations, with dimensions }1,1,1,1,2.}
$$
These are the dimensions underlying the <character table of the quaternion group>. The scalar field matters: the quaternionic irreducible becomes four-dimensional over the reals. The class-count and dimension-sum statements used here have their standard complex meaning.
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