Character Tables of $D_{4}$ and $Q_{8}$

Question

Is there an intuitive reason that the Quaternion group and the Dihedral group on four vertices have the same character table? Does this indicate something special about the two groups? Or is it more of a coincidence the tables are the same?

Answer 1

Suppose $G$ is a non-abelian group of order $8$. If its irreps. are of dimensions $n_i$, then from the formula $\sum n_i^2 = 8$ (and remembering that we have at least one $n_i = 1$, since there is the trivial representation, but not all $n_i =1$, since $G$ is non-abelian), we find that the $n_i$ are equal to $1,1,1,1,2$.

The abelianization of $G$ is necessarily the Klein $4$-group (since if the abelianization of $G$ were cyclic, then $G$ would be generated by $[G,G]$ --- or order $2$ and normal --- and a lift of the generator of $G/[G,G]$, and hence would be abelian, contradicting the assumption that it is non-abelian).

This already forces a fair bit of information about the $4$ one-dimensional characters.

Since $G$ has a unique $2$-dimensional representation, it is isomorphic to its twist by any of the one-dimensional characters. This forces more information about the character table.

In fact, if my back-of-the-envelope computation is correct, this completely determines the character table.

So, essentially, for a non-abelian group of order $8$ the constraints are just too tight for there to be more than one possible character table.