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 $Q_8 / Q_8' \simeq \mathbb{Z}^2 \times \mathbb{Z}^2$


Two hints:

  • 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.

The latter hint does lead to a construction of the missing character, but I'm not sure how representation theoretical that construction is :-/

  • 1
    $\begingroup$ You may prefer to use the orthogonality relations to find the 2D character as opposed to the representation of quaternions. $\endgroup$ – Jyrki Lahtonen Feb 16 '15 at 19:06

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.