What is the number of irreducible characters of $G$ and their dimensions when $|G|=p^3$ is non abelian?

Question

I know $G'=Z(G)$ and $|Z(g)|=p$. $|G/G'|=p^2$ is the number of irreducible characters of dimension $1$. If $\chi $ is the characteristic of the regular representation

$$p^3=|G|=\langle\chi,\chi \rangle =\sum_{i=1}^{k} n_i^2= p^2+ \sum_{i=p^2+1}^kn_i^2$$

Where $k$ is the number of conjugacy classes (and the number of irreducible non isomorphic representations). We also know by the orbit stabilizer theorem that $|g^G|=|G*x|=|G|/|\text{stab}(G)|$. So the order of each conjugacy class divides $|G|=p^3$. $G=\dot{\cup}_{g_i}g_i^G$ and $|Z(g)|=p$, so there are $p$ elements whose conjugacy class are singletons.

$$p^3=\sum_i |g_i^G|=\sum_j p^2+\sum_k p+1+1+...+1 =jp^2 + p(p^2-jp-1)+p $$ It is impossible to have a conjugacy class with $p^3$ elements (because that would mean a positive number being equal to a negative). Also, we know $j+p^2-jp-1+p=k> p^2$, because of the first equation we have writen. Therefore, $j(1-p)>(p-1)\Rightarrow j<1 $, so $j=0$ and $k=p^2+p-1$ and we have:

$$p^3=p^2+\sum_{i=p^2+1}^{p^2+p-1}n_i^2$$

So we have the number of conjugacy classes (the same numebr of irreducible representations!). But I still need to compute their dimension and it is not clear how I would solve these $n_i$...

Answer 1

One appraoch is to use a theorem of Ito which states that if $A$ is an abelian normal subgroup of $G$, then the degree of any irreducible complex representation divides the index $|G:A|$. So, since every $p$-group has a normal $p$-subgroup of every possible order, and groups of order $p^2$ are abelian, we have that the degree of any irr. complex representation of $G$ divides $p$ when $|G| = p^3$. In other words, the only nonlinear characters of $G$ will have degree $p$.

Now, you've correctly observed that in this case, $G$ has $p^2$ linear characters, and we may use the formula to see that $p^3 = p^2 + np^2$ for some integer $n$. Solving for $n$ in terms of $p$ gives us $n = p -1$. There are $p - 1$ nonlinear characters for this group.

Note: There's probably a more elementary way to conclude that every nonlinear character has degree $p$ which does not rely on Ito's theorem, but this is a valid approach I think.

Edit: It is actually easy to see that a nonabelian group of order $p^3$ cannot have an irreducible rpesentation of degree $p^2$. If there were such a representation than since $|G| = \sum_{i=0}^k n_i^2$, and one of the $n_i$'s is $p^2$, the sum would have to be at least $p^4$ which is impossible.

Answer 2

An easier way is as follows. You stated you know that $G'=Z(G)$ and $|G:Z(G)|=|G:G'|=p^2$. Hence there are $p^2$ linear characters. Now use the basic fact that for any $\chi \in Irr(G)$, we have $\chi(1)^2 \leq |G:Z(G)|$ (see for example (2.30)Corollary in M.I. Isaacs, Character Theory of Finite Groups). Hence all the non-linear characters must have degree $p$. Using the Frobenius-Molien formula ($|G|$ is the sum of squares of all degrees), we see that the number of non-linear must be $\frac{p^3-p^2}{p^2}=p-1$. So $\#Irr(G)= p^2+p-1$. No need for Itô's Theorem!