# Finding the character table of $Q_8$

Assume $$G$$ is a finite group.

I am trying to construct the character table of $$Q_8$$, which is defined by

$$Q_8=\{\pm 1,\pm i, \pm j,\pm k \}, \ i^2=j^2=k^2=-1, \ ij=k,jk=i,ki=j$$

By considering the $$G'$$, I can prove that $$|G|=8=1^2+1^2+1^2+1^2+2^2$$. Therefore $$G$$ has four $$1$$-dimensional representations and one $$2$$-dimensional representation. So

$$\begin{array}{|c|c|c|c|} \hline &1& -1 & \{\pm i \} & \{\pm j \} & \{\pm k \} \\ \hline \chi_0 & 1 & 1 & 1 & 1 & 1 \\ \hline \chi_1 & 1 & ? & ?&? & ? \\ \hline \chi_2 & 1 & ? &? & ?& ? \\ \hline \chi_3 & 1 & ? & ?&? & ? \\ \hline \chi_4 & 2 & ? & ? & ? & ? \\ \hline \end{array}$$

I guessing the first vertical column is a 2 at the bottom because $$tr(I_2)=2$$. The first row is $$1$$'s because it is the 1-d trivial representation.

I am unsure of how to proceed from here. Should I be able to identity the $$\chi_i$$'s?

Please do not give full solutions as otherwise I will not learn.

If it helps its fairly easy to show that $$Q_8 / Q_8' \simeq \mathbb{Z}_2 \times \mathbb{Z}_2$$

If you just try to guess them it works out just fine. You know that the commutator subgroup is $\{\pm 1 \}$, so $\chi_i$ are trivial on $-1$, so all the values of $\chi_i$ are in fact $\pm 1$ since every element has order dividing $4$.
Since there are $3$ characters it makes sense that they should be "symmetric" in $i, j, k$. So let $\chi_1(i) = 1$, $\chi_2(i)=\chi_3(i)=-1$, and define the other two analogously. You can check these extend to characters, and then you can use column orthogonality to fill in the table.
• The 1-dimensional representations of a finite group $G$, have the commutator subgroup $G'=[G,G]$ in their kernel. Indeed, their number is equal to the index $[G:G']$. Can you find four 1-dimensional representations of the quotient group $Q_8/Q_8'$?
• The ring of Hamiltonian quaternions can be identified as the subring of complex $2\times2$-matrices consisting of matrices of the form $$\left(\begin{array}{rr}z_1&z_2\\-z_2^*&z_1^*\end{array}\right)$$ with $z_1,z_2$ arbitrary complex numbers.