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 $Q_8 / Q_8' \simeq \mathbb{Z}_2 \times \mathbb{Z}_2$
Best Answer
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.