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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.

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 $\{\pm 1\}$ is abelian and its cosets $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. HenceFor 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 isIndeed 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.

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.