# Abstract Nonsense

## Character Table for the Quaternions

Point of post: In this post we use the techniques we’ve devoloped to construct the character table for the quaternions.

Motivation

We continue our effort of constructing character tables by focusing now on the Quaternions, $Q$. We will once again show how beautiful the theory we’ve devoloped can be by constructing the entire table without actually finding a non-trivial irreducible character.

Character Table for $Q$

We begin by finding the number of irreducible characters of $Q$ and their degree. To do this, we note that if $d_\alpha$ is the degree of any irreducible character that $d_\alpha\mid |Q|=8$ so that $d_\alpha=1,2,4,8$. Moreover, since $d_\alpha^2\leqslant 8$ we may conclude that $d_\alpha=1,2$. Thus, we are looking for solutions to $m+4n=8$ where we know that $m\geqslant 1$ since the inclusion of the trivial character and $n\geqslant 1$ since otherwise all the irreps of $Q$ would be degree one and thus $Q$ would be abelian. But, a quick check shows that the only possible solution to this is $m=4$ and $n=1$. Thus, we may conclude that $\widehat{Q}=\left\{\alpha_{\text{triv}},\alpha_1,\alpha_2,\alpha_3,\alpha_4\right\}$ where $d_{\alpha_{\text{triv}}}=d_{\alpha_1}=d_{\alpha_2}=d_{\alpha_3}=1$ and $d_{\alpha_4}=2$. Thus, we may begin to construct our character table (using the ideas we used before  to start the table)

$\begin{array}{c|ccccc} Q & e & \mathcal{C}_1 & \mathcal{C}_2 & \mathcal{C}_3 & \mathcal{C}_4\\ \hline \chi^{\text{triv}} & 1 & 1 & 1 & 1 & 1\\ \chi^{(\alpha_1)} & 1 & a_1 & b_1 & c_1 & d_1\\ \chi^{(\alpha_2)} & 1 & a_2 & b_2 & c_2 & d_2\\ \chi^{(\alpha_3)} & 1 & a_3 & b_3 & c_3 & d_3\\ \chi^{(\alpha_4)} & 2 & a_4 & b_4 & c_4 & d_4\end{array}$

We next wish to ascertain the true identities of $\mathcal{C}_1,\mathcal{C}_2,\mathcal{C}_3$ and $\mathcal{C}_4$. One of these, say $\mathcal{C}_1$, is obvious. Namely, it is clear by inspection that $-1\in\mathcal{Z}\left(Q\right)$ and so $-1$ is in a class by itself–i.e. $\mathcal{C}_1=\{-1\}$. Thus, by first principles we know then that $\#\left(\mathcal{C}_2\right)+\#\left(\mathcal{C}_3\right)+\#\left(\mathcal{C}_4\right)=6$. Now, since $\#\left(\mathcal{C}_m\right)\mid 8\;\; m=2,3,4$ and $\#\left(\mathcal{C}_m\right)\ne 1\;\; m=2,3,4$  (since clearly none of $Q-\{-1,1\}$ are in the center of $Q$) we may conclude by a simple case analysis to conclude that $\#\left(\mathcal{C}_m\right)=2\;\; m=2,3,4$. Noting that $i$ is conjugate to $-i$, $j$ to $-j$, and $k$ to $-k$ we may conclude that $\mathcal{C}_2=\{i,-i\},\mathcal{C}_3=\{j,-j\}$ and $\mathcal{C}_4=\{k,-k\}$. Noting though that since $\chi^{(\alpha_m)}(e)=1\;\; m=2,3,4$ we may conclude that $|b_m|=|c_m|=|d_m|=1\;\; m=1,2,3$. But, by CT 3 we know that

$\displaystyle 1+|b_1|^2+|b_2|^2+|b_3|^2+|b_4|^2=\frac{8}{2}=4$

and from CT 9 we may conclude that $|b_4|^2=0$ and so $b_4=0$. Similarly, $c_4=0$ and $d_4=0$.  But, by CT 2  we know that

$0=1\cdot(1\cdot 2)+1\cdot(1\cdot a_4)+2\cdot(1\cdot b_4)+2\cdot(1\cdot c_4)+2\cdot(1\cdot d_4)=2+a_4$

Therefore, $a_4=-2$. Thus, our character table begins to take shape as

$\displaystyle \begin{array}{c|ccccc} Q & e & -1 & i & j & k\\ \hline \chi^{\text{triv}} & 1 & 1 & 1 & 1 & 1\\ \chi^{(\alpha_1)} & 1 & a_1 & b_1 & c_1 & d_1\\ \chi^{(\alpha_2)} & 1 & a_2 & b_2 & c_2 & d_2\\ \chi^{(\alpha_3)} & 1 & a_3 & b_3 & c_3 & d_3\\ \chi^{(\alpha_4)} & 2 & -2 & 0 & 0 & 0\end{array}$

Note though that by CT 2 we know by comparing rows two, three, and four with row five respectively gives us

$\displaystyle 0=1\cdot(2\cdot 1)+1\cdot(-2\cdot a_m)+2\cdot(b_m\cdot 0)+2\cdot(c_m\cdot 0)+2\cdot(d_m\cdot 0)=2-2a_m$

so that $a_m=1$ for $m=1,2,3$. Thus, our character table can now be rewritten

$\begin{array}{c|ccccc} Q & e & -1 & i & j & k\\ \hline \chi^{\text{triv}} & 1 & 1 & 1 & 1 & 1\\ \chi^{(\alpha_1)} & 1 & 1 & b_1 & c_1 & d_1\\ \chi^{(\alpha_2)} & 1 & 1 & b_2 & c_2 & d_2\\ \chi^{(\alpha_3)} & 1 & 1 & b_3 & c_3 & d_3\\ \chi^{(\alpha_4)} & 2 & -2 & 0 & 0 & 0\end{array}$

Note next that by our previous remarks we have that $|b_m|=|c_m|=|d_m|=1$ for $m=1,2,3$. But, by CT 7 we know that $b_m,c_m,d_m$ for $m=1,2,3$ are real and thus are all either plus or minus one. But, by considering CT 2 for rows two, three, and four with row one we may easily conclude that two of $b_m,c_m,d_m$ are $-1$ and the other $1$ for $m=1,2,3$. But, it’s easy to see from CT 2 that two rows cannot be exactly the same. And, since the first two columns of each of these rows agree it follows that it is at least one of the values in which we are interested which differs. But, since there are precisely $\displaystyle 3$ ways to consider the aforementioned configurations of $1$‘s and $-1$‘s it follows that all three of these configurations occur. Thus, since the characters are interchangable we may finally fill in the character table as

$\begin{array}{c|ccccc} Q & 1 & -1 & i & j & k\\ \hline \chi^{\text{triv}} & 1 & 1 & 1 & 1 & 1\\ \chi^{(\alpha_1)} & 1 & 1 & 1 & -1 & -1\\ \chi^{(\alpha_2)} & 1 & 1 & -1 & -1 & 1\\ \chi^{(\alpha_3)} & 1 & 1 & -1 & 1 & -1\\ \chi^{(\alpha_4)} & 2 & -2 & 0 & 0 & 0\end{array}$

Consequences

Using CT 5 and CT6 we may conclude from the above that

$\left\{\text{Normal Subgroups}\right\}=\bigg\{\{1\},\{1,-1\},\{1,-1,i,-i\},\{1,-1,j,-j\},\{1,-1,k,-k\},Q\bigg\}$

and $\mathcal{Z}\left(Q\right)=\{-1,1\}$.

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.