Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 10 / c / 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 $\{\pm 1\}$ 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.