 Solution

The conjugacy class of $g\in G$ is $\{hg{h}^{-1}:h\in G\}$. 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 $\pm 1,\pm i$, 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: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 ${\sum }_{j=1}^5{d}_j^2=8$. 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. ThereforeThese 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.