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



$\{ x^2,x^6 \}$





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?


  • $\begingroup$ Your latter $H$ is not a subgroup. A subgroup containing $x$ surely also contains $x^2$. But the subgroup $K$ of even powers of $x$ seems to be normal. Looks like $G/K$ is isomorphic to the Klein four group $C_2\times C_2$. You can lift all four 1-dimensional reps of this to a rep of $G$ (you have already identified 3 of the 4). $G/\langle x^4\rangle$ looks like a dihedral group, so... $\endgroup$ Commented Aug 11, 2011 at 15:41
  • $\begingroup$ Oh no, you are right! Ok i'll change it. $\endgroup$
    – Otis
    Commented Aug 11, 2011 at 15:46
  • $\begingroup$ But what if the quotient by the center wasn't the dihedral group? I see how to do this in this exact case, but what I'd like is if someone could show me how to find a $2$-dim rep without using the fact that the quotient is something we can handle. For example, sometimes the questions give e.g. $\langle x^7=x^4=1, yxy^{-1} = x^{-1}\rangle$, for which I don't have any ideas for how to find a higher degree representation $\endgroup$
    – Otis
    Commented Aug 11, 2011 at 15:50
  • $\begingroup$ sorry that should be $y^4$ in my presentation $\endgroup$
    – Otis
    Commented Aug 11, 2011 at 15:50
  • $\begingroup$ I don't think that you will get all the irreducible reps from lifting those of $G/\langle x^4 \rangle$. After all, all those will have $x^4$ acting as identity. Tensoring with a 1-d character doesn't change that fact. You need to find something else! I would induce such a representation of $\langle x \rangle$, where $x^4$ is mapped to $-1$ as opposed to $+1$. $\endgroup$ Commented Aug 11, 2011 at 15:55

3 Answers 3


Short answer:

Multiplication by one-dimensionals will not be sufficient. Looking at the central quotient will not be sufficient.

Finding one dimensional representations of a finitely presented group is easy: every such representation is a representation of the abelianization. In you case: $$\begin{align} G/[G,G] &\cong \langle x, y : x^8 = y^2 = 1, x = yxy^{-1} = x^3 \rangle \\ &= \langle x, y : x^2 = y^2 = 1, xy = yx \rangle \\ &\cong C_2 \times C_2\end{align}$$

Hence it has 4 one-dimensional representations which take x to ±1 and y to ±1.

Since we happen to know G has order 16, this leaves 12 to be written as a sum of squares of divisors of 16, so there are three two-dimensional representations.

One optimistic way to proceed is to think $H=\langle x \rangle$ is abelian, normal, and the conjugation action (and transversal) are written right in the presentation. This means (1) I know all of the representations of H and (2) I can easily induce a representation of H to G. Luckily all irreducible representations of H are one-dimensional, and inducing to G will double their dimension.

In particular, take the representation of H which sends x to ζ, a primitive eighth root of unity. The induced representation sends x to $\left(\begin{smallmatrix} \zeta & 0 \\ 0 & \zeta^3 \end{smallmatrix}\right)$ and y to $\left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right)$, which is clearly irreducible (the only eigenspaces of x are not eigenspaces of y).

One can multiply this representation by $x \mapsto -1$, $y \mapsto \pm1$ to get a second irreducible representation of degree 2.

Neither of these two representations are representations of $G/\langle x^4\rangle$, because they are faithful.

However, the third irreducible twoo-dimensional representation could be found from the quotient group. I would just induce it from $x \mapsto \zeta^2 = i$ to get $x \mapsto \left(\begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix}\right)$ and $y\mapsto \left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$. This irreducible representation is unchanged (up to isomorphism) by multiplication by any of the one-dimensional representations.

You asked about generality: the group you are given is incredibly special. It is called a quasi-dihedral group. Having a prime index cyclic normal subgroup is very, very special, and I suspect you are expected to recognize this property and use it.

If your presentation exhibits a chief series with abelian factors, you can use similar techniques to find the character table and representations. In general though, a presentation provides almost no algorithmic information. I recommend Pahlings–Lux's textbook if you are interested in methodically finding character tables in realistic and difficult situations.

  • $\begingroup$ Thanks! This is great! What does "and transversal" mean in the paragraph beginning "one optimistic way.." $\endgroup$
    – Otis
    Commented Aug 11, 2011 at 16:16
  • 1
    $\begingroup$ @Otis: The transversal in question is {1,y} so that every element of G can be written as an element of 1H or yH. It is part of the definition of induced representation. Index 2 makes it easy to find transversals, since one of them is 1 and the other is something not in H. $\endgroup$ Commented Aug 11, 2011 at 16:22
  • 1
    $\begingroup$ I agree with Jack that this group is too special. I recommend working out character tables for $S_4, S_5, S_6$ (and maybe $A_4, A_5, A_6$ as well) if you haven't; they're good exercises in induction. $\endgroup$ Commented Aug 11, 2011 at 16:35

If you're given a presentation, you can compute the abelianization of $G$. Every $1$-dimensional representation factors through the abelianization, and conversely every irreducible representation of the abelianization gives an irreducible representation of $G$, so you get precisely all of the $1$-dimensional characters this way. Here the abelianization is

$$G/[G, G] \cong \langle x, y | x^8 = y^2 = 1, x = x^3 \rangle \cong C_2 \times C_2$$

so the four $1$-dimensional characters take value $\pm 1$ on $x$ and $\pm 1$ on $y$. Note that this requires no guesswork, and in particular doesn't require you to fiddle around with normal subgroups; it's much cleaner to fiddle around with quotients.

The remaining three characters are, as you say, necessarily $2$-dimensional. (In general when doing this step, keep in mind that dimensions divide $|G|$.) Probably the best general procedure for finding irreducible characters of dimension greater than $1$ is induction. It's a good idea to induce from subgroups of small index, and also probably a good idea to induce from subgroups whose irreducible characters have low dimension (e.g. abelian subgroups). As others have noted, here you have an abelian subgroup of index $2$...!

Other general techniques to keep in mind:

  • Tensoring is good, but so is taking the symmetric or exterior square.
  • A Galois conjugate of a character is still a character (exercise; or see this math.SE question).
  • The number of self-inverse conjugacy classes is equal to the number of self-dual irreducible characters (exercise).
  • If $g \in G$ has order $d$ and is conjugate to $g^k \in G$, then $\chi(g)$ always lies in the subfield of $\mathbb{Q}(\zeta_d)$ ($\zeta_d$ a primitive $d^{th}$ root of unity) fixed by $\zeta_d \mapsto \zeta_d^k$ (exercise). In particular, if $g$ is conjugate to $g^k$ for all $(d, k) = 1$, then $\chi(d)$ is always (rational, hence) an integer.
  • $\begingroup$ Thank you!! This helps a ton $\endgroup$
    – Otis
    Commented Aug 11, 2011 at 16:14

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 $$|G| = \sum_{g \in G}|\chi(g)^2| \geq \sum_{z \in Z(G)} |\chi(z)|^2 = |Z(G)|\chi(1)^2).$$


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