TITLE: Character Table From Presentation
QUESTION [12 upvotes]: I've recently learned about character tables, and some of the tricks for computing them for finite groups (quals...) but I've been having problems actually doing it. Thus, my question is (A) how to finish the following question (I am ok with general techniques, I can work out the particulars of the calculations) and (B) any tricks that are helpful to do the following types of questions:
The question is 
Let $G$ be the group of order 16 with the presenation
$$
\langle x,y | x^8=y^2 = 1, yxy^{-1} = x^3 \rangle
$$
Compute the conjugacy classes and construct the character table.

I compute the conjugacy classes to be 
$\{id\}$
$\{x,x^3\}$ 
$\{ x^2,x^6 \}$
$\{x^4\}$
$\{x^5,x^7\}$
$\{xy,x^3y,x^7y,x^5y\}$
$\{y,x^2y,x^6y,x^4y\}$

So, I calculate that there are 7 conjugacy classes, so there are 7 representations. The trivial represenation is one, and there is another one given by the nontrivial representation of $\mathbb{Z}/2\mathbb{Z} \simeq G /\langle x\rangle$, i.e. $-1$ on the last two conjugacy classes and $1$ otherwise. 
So, now I need 5 squares to sum to 14, which must be $1^2+1^2+2^2+2^2+2^2$. So, there are still two more $1$-dimensional representations, so there should be another normal subgroup, so some playing around (is there a fast way to see this?) I the even powers of $x$ to be a normal subgroup, with quotient the Klein 4 group, so I can fill in the other 1 dim reps.
Now, here is where I get stuck. How do I find a 2 dimensional representation? I assume that once I find one, I can just tensor the 1-dim reps to finish the character table? Is this the best way? I don't see an obvious way to write down a representation of $G$, so perhaps I should use induction? From what subgroup?

Alternatively, I could use the quotient $G/\langle x^4\rangle$, which is a nonabelian group of order 8, so without much work I should be able to write down its character table, but do other groups with similar presentations that I will see still have this nice property that the center quotients to something simpler - and even if it did, how do I construct a higher dim representation in any other case if I dont recognize the quotient group as something I can explicitly write as a symmetry group of some shape?
Thanks!

REPLY [7 votes]: You could also proceed like this, once you have (correctly) established that $G$ has four irreducible characters of degree 1 (linear characters). Note that $x^{2}$ is in $G^{\prime}$, the derived group of $G$, since $x^{-1}y^{-1}xy = x^{2}.$ Hence $x^{2}$ is in the kernel of each linear character of $G$, so $x^{4}$ is certainly in all these kernels. But there must be some irreducible character $\chi$ of $G$ whose kernel does not contain $x^{4}.$ Let $\sigma$ be a representation affording $\chi$, so you have established already that $\sigma$ must be two-dimensional. Now $\sigma(x)$ must have order $8$, since $\sigma(x^{4}) \neq I_{2 \times 2}$, by the way we chose $\chi$.
What can the eigenvalues of $\sigma(x)$ be? At least one, say $\omega$, is a primitive
$8$-th root of unity.  But then $\omega^{3}$ is also an eigenvalue of $\sigma(x)$, since
if $\sigma(x).v = \omega v$, then $\sigma(y^{-1}xy).\sigma(y^{-1}).v = \omega^{3}\sigma(y^{-1})v$.
This tells us that $\chi(x) = \omega + \omega^{3}$ (and, more generally, tells us that 
$\chi(x^{r}) = \omega^{r} + \omega^{3r}$ for $ 0 \leq r \leq 7).$ You could finish by noting that $\sum_{j=0}^{7} |\chi(x^j)|^{2} = 16,$ so that $\chi$ must vanish outside $\langle x \rangle$, but there are several other ways to use the orthogonality relations
to deduce this. Note that this really gives us two different irreducible characters, since we could have used ${\bar \omega}$ instead of $\omega$. However, this procedure 
is rather specific to this particular group.
Here are a couple of well-known useful general facts. If $G$ is a finite non-Abelian $p$-group, then the number of distinct linear characters of $G$ is divisible by $p^{2}.$ This is because $p^{2}$ divides $|G|$, and divides $\chi(1)^{2}$ for each non-linear irreducible character of $G$. Also, if $G$ is a finite $p$-group, and $\chi$ is an irreducible character of $G$, then $\chi(1)^2$ divides $[G:Z(G)].$ (Since both of these are powers of $p$, it suffices to prove that $\chi(1)^2 \leq [G:Z(G)].$ But $$|G| = \sum_{g \in G}|\chi(g)^2|
\geq \sum_{z \in Z(G)} |\chi(z)|^2 = |Z(G)|\chi(1)^2).$$