Inducing representation for groups of order $p^3$ For groups $G$ such that $|G|=p^3$ one can show that
$(1)$ $Z=Z(G)\cong C_p$
$(2)$ $G'=Z$
$(3)$ $G/Z \cong C_p \times C_p$
Take any $x \in G/Z$. Then $N=\langle x,Z \rangle$ is an abelian normal subgroup of order $p^2$.
It can be shown that $G$ has $p-1$ $p$-dimensional representations induced from $1$-dimensional representations of $N$.
Pick $\chi=$ $1$-dimensional character of $N$ such that $\chi(z)=\xi \neq 1$ for $\xi$ some pth root of (there will be $p-1$ choices).
$\psi-\mathrm{Ind}_H^G \chi$ and $\psi(g)=\frac{1}{|N|}\sum_{h \in G}\chi^{\circ}(hgh^{-1})$
$\underline{g=e}$ (representative of conjugacy class) $\psi(g)=\frac{|G|}{|N|}=p$
$\underline{g=z^i}$    $\displaystyle \psi(g)=\frac{1}{|N|}\sum_{h \in G}\chi^{\circ}(hz^ih^{-1})=\frac{1}{|N|}|G|\chi(z^i)$

$\underline{g=x^j}$    $\displaystyle \psi(g)=\frac{1}{|N|}\left[p^2\chi(x^j)+\dots+p^2\chi(x^jz^{p-1})\right]$

How have the conjugacy classes been evaluated for the highlighted area? I don't see any rules as it is just a general group of order $p^3$. Moreover where does the $p^2$ come from?
 A: Let $\tau\in G\setminus N$, $t_i=\tau^i$ for $i\in\{0,\ldots,p-1\}=:I$, then $t_0,\ldots,t_{p-1}$ is a transversal of $N$ in $G$. If $[x,\tau]=1$, then $\tau$ commutes with $x,\tau$ and all elements of $Z$, hence $\tau\in C_G(\langle \tau,x,Z\rangle)$. But $\langle \tau,x,Z\rangle=G$, hence $\tau\in Z$ - a contradiction. Therefore $\zeta:=[x,\tau]\neq 1$. Since $[x,\tau]\in G'=Z$, then $\zeta\in Z\setminus\{1\}$, thus $Z=\langle \zeta\rangle$, because $|Z|$ is a prime number. Note that $\zeta=[x,\tau]=x^{-1}x^{\tau}$, hence $x^\tau=x\zeta$ and by induction $x^{t_i}=x\zeta^i$ for $i\in I$. It follows that for $i\in I$
$$
(x^j)^{t_i}=(x^{t_i})^j=(x\zeta^i)^j=x^j \zeta^{ij}.
$$
Denote $z=\zeta^j$, then $(x^j)^{t_i}=x^j z^{i}$ for all $i\in I$. By the well known formula about induced characters through the transversals we have 
$$
\psi(g)=\sum_{i\in I}\chi^\circ(g^{t_i})=\sum_{i\in I}\chi(x^j z^{i})=
\frac{1}{p^2}\sum_{i\in I}p^2\chi(x^j z^{i})=
\frac{1}{ |N|}\sum_{i\in I}p^2\chi(x^j z^{i}).
$$
