Irreducible reps of $U(3,p)$ I am trying to determine the irreducible reps of $U(3,p)$, or the Heisenberg group of order p^3.
I see that determining the one dimensional reps is equivalent to determining $\hom(\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}, \mathbb{C}^\times)$, but I don't see how to determine the reps with dimension greater than 1. 
 A: I will simply note $G = U(3,p)$. 
Besides the abelianisation quotient $\pi : G \to G^{\mathrm{ab}}$ you mentioned, whose image $G^{\mathrm{ab}}$ is an elementary abelian group of order $p^2$, I will take advantage of the center of $G$, which is $ZG = \left\{ \begin{pmatrix} 1 & 0 & * \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \right\}$, a cyclic group of order $p$. It is actually the kernel of $\pi$.


*

*Conjugacy classes of $G$.
As $\pi : G \to G^{\mathrm{ab}}$ is a morphism, if $g_1, g_2 \in G$ are conjugate, so must be $\pi(g_1)$ and $\pi(g_2)$. Because $G^{\mathrm{ab}}$ is abelian, it entails that $\pi(g_1) = \pi(g_2)$. So each conjugacy class of $G$ is included in one of the $p^2$ fibers $\pi^{-1}[\{x\}]$, $x \in G^{\mathrm{ab}}$ of the morphism $\pi$, whose cardinal is $\lvert G \rvert/\lvert G^{\mathrm{ab}}\rvert = p$.
Futhermore, the size of a conjugacy class of $G$ must divide $\lvert G \rvert = p^3$, so it is either $1$ or $p$. The first case occurs exactly when we are in the center $ZG$. This accounts for $p$ conjugacy classes of size $1$, whose union is $ZG = \ker \pi$. All other fibers of $\pi$ are then size-$p$ conjugacy classes.
So, in a nutshell, $G$ has exactly $p^2 + p - 1$ conjugacy classes.


*Degree of the missing irreps.
Because $G$ has $p^2+p-1$ conjugacy classes, it must have $p^2 + p - 1$ irreducible representations. You already know the $p^2$ which are of degree one, so we are looking for $p-1$ "missing" representations. Let's denote by $n_1, \ldots, n_{p-1}$ their degrees. We have $1^2 + \cdots + 1^2 + n_1^2 + \cdots + n_{p-1}^2 = p^3$, that is, $\sum_i n_i^2 = p^3 - p^2 = p^2(p-1)$.
A classical argument (easily proven "by hand", but already pertaining to induction theory) says that if $A$ is an abelian subgroup of a group $G$, then the dimension of the irreducible representations of $G$ are $\leq [G:A]$. Here, we have the (non-normal) abelian subgroup $A = \left\{ \begin{pmatrix} 1 & 0 & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{pmatrix} \right\}$ of index $p$, so $\forall i, n_i \leq p$. Therefore, $\sum_i n_i^2 \leq (p-1) p^2$.
Because this inequality is an equality, all constituents of it must be equalities too. So we have $n_1 = \cdots = n_{p-1} = p$.
So you really are looking for $p-1$ missing representations of degree $p$. 


*Characters of the missing irreps.
I won't construct the missing irreps (see below for a sketch of construction based on induction theory), but I'll explain how we can get their characters $\chi_1, \ldots, \chi_{p-1}$.
First, the orthogonality relations (w.r.t. the $p^2$ degree-$1$ characters) easily give that $\chi_i$ vanishes outside of $ZG$. In particular, $\chi_i$ is entirely determined by its restriction $\chi_i' = {\chi_i}_{| ZG}$, which is the character of the restriction of the representation.
As $\chi_i$ is an irreducible character of $G$, we have $\langle \chi_i, \chi_i \rangle_G = 1$. We can then compute
$$\begin{align*} \langle \chi_i',  \chi_i' \rangle_{ZG} &= \frac 1{p} \sum_{g\in H} \lvert \chi_i'(g) \rvert^2 \\ &= \frac 1p \sum_{g\in G} \lvert \chi_i(g) \rvert^2 \\ &= p^2 \langle \chi_i, \chi_i \rangle_G = p^2\end{align*}.$$
If we denote by $\lambda_1, \ldots, \lambda_p$ the number of times the various irreps (all of degree $1$, since $ZG$ is abelian) of $ZG$ appear in $\chi_i$, we should have $\lambda_1 + \cdots + \lambda_p = p$ (degree identification) and $\lambda_1^2 + \cdots + \lambda_p^2 = p^2$, because of the previous computation. This is only possible if one of the $\lambda_j$ is $p$, and the others are $0$.
In other words, $\chi'_i$ is simply $p$ times one of the irreducible characters of $ZG$. Because the $\chi'_i$ are distinct, the corresponding irreducible characters of $ZG$ must be, too. 
The only thing left to do is to determine which of the $p$ irreducible characters of $ZG$ appear this way or, better said, which is the one not appearing. Not surprisingly, the trivial character is the one not appearing here, because 
$$\langle 1, \chi'_i \rangle_H = \frac 1p \sum_{g\in H} \chi'_i(g) = \frac 1p \sum_{g\in G} \chi_i(g) = p^2 \langle 1 , \chi_i \rangle_G = 0.$$
To put it in a nutshell, the $p-1$ "missing" characters vanish outside of $ZG$, and on it they are simply $p$ times the nontrivial characters of $ZG$. Our work here is done.


*Spoiler alert.
Induction theory in fact proves that all irreps of $G$ appear in the representations induced from the irreps of any subgroup. If we look at $ZG$, the picture is quite pretty: it's easy to see that if $\tilde \chi$ is the character of a representation $\tilde\rho$ of $ZG$, then the character of the induced representation $\rho = \mathrm{Ind}_{ZG}^G (\tilde \rho)$ is 
$$\chi : g \mapsto \begin{cases} p^2\, \tilde\chi(g) & \text{if } g \in ZG \\ 0 & \text{if } g \not\in ZG. \end{cases}$$
From that you can check that 


*

*If we start from the trivial representation of $ZG$, the induced representation (which is the permutation representation associated to the action of $G$ on $G/ZG$) splits off as the direct sum of all degree-$1$ irreps of $G$.

*If we start from another irrep of $ZG$ (there are $p-1$ of them, all of degree $1$), the induced representation splits off as the sum of $p$ copies of a degree-$p$ irrep of $G$: here are the $p-1$ representations which were missing!


*Finale.
Just for fun, here is the complete character table, in the case $p = 3$ (so $p^2 + p - 1 = 11$) with some meaningless bars to insist on the structure of the table. The first three columns are the central elements, the trivial element leading the way. Here, $j = e^{i 2\pi/3}$.
$$\begin{array}{|c|cc|cccccccc|}\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & j & j & j & \bar\jmath & \bar\jmath & \bar\jmath \\  1 & 1 & 1 & 1 & 1 & \bar\jmath & \bar\jmath & \bar\jmath & j & j & j \\  1 & 1 & 1 & j & \bar\jmath & 1 & j & \bar\jmath & 1 & j & \bar\jmath \\  1 & 1 & 1 & j & \bar\jmath & j & \bar\jmath & 1 & \bar\jmath & 1 & j \\  1 & 1 & 1 & j & \bar\jmath & \bar\jmath & 1 & j & j & \bar\jmath & 1 \\  1 & 1 & 1 & \bar\jmath & j & 1 & \bar\jmath & j & 1 & \bar\jmath & j \\  1 & 1 & 1 & \bar\jmath & j & j & 1 & \bar\jmath & \bar\jmath & j & 1 \\  1 & 1 & 1 & \bar\jmath & j & \bar\jmath & j & 1 & j & 1 & \bar\jmath \\  \hline 3 & 3j & 3\bar \jmath & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \hline 3 & 3\bar \jmath & 3j & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \hline \end{array}$$
