Character table of modular Heisenberg groups Let $p$ be a prime number and let $G$ be the modular Heisenberg group of order $p^3$
$$ G = \left\{\, \begin{bmatrix} 1 & b & c \\ 0 & 1 & a \\ 0 & 0 & 1 \end{bmatrix} : a, b, c \in \mathbb{Z}/p\mathbb{Z} \,\right\} $$
so that $G$ is the extra-special group of order $p^3$ and exponent $p$ if $p$ is an odd prime. What is the character table of $G$?
When $p = 2$ direct computation shows that the character table of $G$ is the following.
$$
\begin{array}{cccccc}
\hline
|g^G| & 1 &  1 &  2 &  2 &  2 \\
g
&
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} 
&
\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
&
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} 
&
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} 
&
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} 
\\
\hline
\chi_1 & 1 &  1 &  1 &  1 &  1 \\
\chi_2 & 1 &  1 & -1 &  1 & -1 \\ 
\chi_3 & 1 &  1 &  1 & -1 & -1 \\ 
\chi_4 & 1 &  1 & -1 & -1 &  1 \\ 
\chi_5 & 2 & -2 &  0 &  0 &  0 \\ 
\hline
\end{array}
$$
The first four linear characters comes from the abelianization $G/D(G) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. The last non-linear character is obtained from the orthogonality relations.
 A: Elements
Let's introduce three matrices
$$
x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}, 
\quad
y = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},
\quad
z = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.
$$
It's straightforward to check that $x^p = y^p = z^p = 1$, $z = y^{-1}x^{-1}yx$, and
$$
x^ay^bz^c = \begin{bmatrix} 1 & b & c \\ 0 & 1 & a \\ 0 & 0 & 1 \end{bmatrix}.
$$
In particular, we get $G = \langle x, y \rangle = \{\, x^ay^bz^c : a, b, c \in \mathbb{Z}/p\mathbb{Z} \,\}$.
Conjugacy classes
The center $Z(G)$ of $G$ is simple and is generated by $z$. So $G$ has $p$ central conjugacy classes and they are parametrized by $c \in \mathbb{Z}/p\mathbb{Z}$ as follows.
$$
(z^c)^G = \left\{ \begin{bmatrix} 1 & 0 & c \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right\}.
$$
Take a non-central element $g$ of $G$. Since
$$
p = |Z(G)| = |\langle z \rangle| < |\langle g, z \rangle| \le |C(g)| < |G| = p^3,
$$
the centralizer $C(g)$ of $g$ has order $p^2$ and the conjugacy class $g^G$ of $g$ has $p = p^3/p^2$ elements. Counting argument proves that $G$ has $(p^3 - p)/p = p^2 - 1$ non-central conjugacy classes and they are parametrized by $(a, b) \in \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} - \{ (0, 0) \}$ as follows.
$$
(x^ay^b)^G = \left\{ \begin{bmatrix} 1 & b & c \\ 0 & 1 & a \\ 0 & 0 & 1 \end{bmatrix} : c \in \mathbb{Z}/p\mathbb{Z} \right\}.
$$
Irreducible characters
There are $p^2$ linear characters comes from the abelianization $\overline G = G/D(G) \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$. Since $D(G) = Z(G)$, the quotient is generated by two elements $\overline x$ and $\overline y$. Take a primitive $p$-th root of unity $\zeta = e^{2\pi i/p}$. Then we have $p^2$ linear characters
$
\rho_{s, t} \colon \overline G \to \mathbb{C}^\times
$
defined by
$
\overline x \mapsto \zeta^s
$
and
$
\overline y  \mapsto \zeta^t
$
for $(s, t) \in \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$. These yield $p^2$ linear characters
$$\sigma_{s, t} \colon G \to \mathbb{C}^\times, \qquad (s, t) \in \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$$
of $G$.
To get non-linear characters, we consider characters of a subgroup $H = \langle y, z \rangle$ which is isomorphic to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ and induce them to $G$. Let
$$\theta_u \colon H \to \mathbb{C}^\times, \qquad u \in \mathbb{Z}/p\mathbb{Z} - \{ 0 \}$$
be a  linear character of $H$ that is defined by $y \mapsto 1$ and $z \mapsto \zeta^u$.
If $a \neq 0$ then no elements of $(x^a y^b)^G$ lies in $H$ and we have
$
\theta_u^G(x^a y^b) = 0
$.
If $a = 0$ then
$$
\begin{align}
\theta_u^G(x^a y^b)
&= \theta_u^G(y^b)
= \frac{1}{|H|} \sum_{g \in G} \dot\theta_u(g^{-1}y^bg) 
= \sum_{g \in [G/H]} \dot\theta_u(g^{-1}y^bg)
\\
&= \sum_{k = 0}^{p - 1} \dot\theta_u(x^{-k}y^bx^k)
= \sum_{k = 0}^{p - 1} \theta_u(y^bz^{bk})
= \sum_{k = 0}^{p - 1} \zeta^{ubk}
= 0.
\end{align}
$$
Finally,
$$
\theta_u^G(z^c)
= \frac{1}{|H|} \sum_{g \in G} \dot\theta_u(g^{-1}z^cg) = p \theta_u(z^c) = p \zeta^{uc}.
$$
Character table
What we get so far is summarized in the next table.
$$
\begin{array}{ccc}
\hline
|g^G| & 1 &  p \\
g & z^c & x^ay^b
\\
\hline
\sigma_{s, t} & 1 & \zeta^{sa + tb} \\
\theta_{u}^G & p\zeta^{uc} & 0 \\
\hline
\end{array}
$$
From the orthogonality relations, one can check that these characters form all irreducible characters of $G$. Hence this is the character table of $G$ in general.
