Showing that an induced representation on the group of unitriangular $3\times 3$ matrices is irreducible Let $G$ be the group of unitriangular $3\times3$ matrices over $\mathbb{F}_p$ (i.e. upper triangular matrices with $1$ on the main diagonal). For any nontrivial $\phi : \mathbb{F}_p \rightarrow \mathbb{C}^\times$, the mapping
$$ \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 &1\end{pmatrix} \mapsto \phi(a)$$
is a representation of the subgroup $H$ of $G$ of all matrices of that specific form (i.e. the entry at $(1,2)$ is zero). Let $V'$ be the corresponding $\mathbb{C}H$-module.
I want to show that the induced representation on $G$ is irreducible.  Here's how far I have gotten:
The induced representation has dimension less than or equal to $p$, as $|G:H| = p^3/p^2 = p$. If the corresponding $\mathbb{C}G$-module $V = \operatorname{Ind}_H^G V'$ can be decomposed into $V = U \oplus W$ such that $U$ is irreducible, then $\dim U < p$ and $\dim U \mid p^3$, so $\dim U = 1$. That is, if $V$ can be decomposed into irreducible $\mathbb{C}G$-modules, then each of these modules has dimension 1. I want to somehow get a contradiction from this, but I don't see how.
I also thought about using reciprocity: We have
$$ (\chi_V, \chi_V) = ( \operatorname{Ind}_H^G \chi_{V'}, \operatorname{Ind}_H^G \chi_{V'}) = (\chi_{V'}, \operatorname{Res}_H(\operatorname{Ind}_H^G V')), $$
but the problem here is that I don't really know what $\operatorname{Res}_H(\operatorname{Ind}_H^G V')$ looks like.
 A: Let $W=\mathrm{Ind}_H^G(V),$ so you would like to show that $W$ is irreducible. You know $\dim W=p$ (i.e. there is no ``less than'') since we always have $\dim\mathrm{Ind}^G_H(V)=|G:H|\dim(V).$
There are many ways to prove $W$ irreducible. For example, since $|G|=p^3,$ irreducible complex characters of $G$ have degree $1$ or $p,$ so if $W$ is not irreducible, it is a sum of linear characters. By Frobenius reciprocity, each such character must restrict to $\phi$ on $H,$ and this means in turn that $\phi|_{H\cap G^\prime}=1,$ where $G^\prime$ is the commutator subgroup. But $G^\prime$ consists of the matrices of the form $$\begin{pmatrix}1&0&a\\0&1&0\\0&0&1\end{pmatrix}$$ where $a\in\mathbb F_p,$ so in fact $G^\prime\subseteq H$ and $\phi(a)$ must be $1$ for all $a,$ contrary to your hypothesis that $\phi$ is nontrivial.
Another way to check $W$ is irreducible is to compute its character. Let this be $\chi.$ A set of coset representatives for $H$ in $G$ is furnished by the matrices $$\begin{pmatrix}1&t&0\\0&1&0\\0&0&1\end{pmatrix}$$ for $t\in\mathbb F_p.$ Then $$
\begin{pmatrix}1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & -t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=
\begin{pmatrix}1 & 0 & a+tb \\ 0 & 1 & b \\ 0 & 0 & 1\end{pmatrix}
$$ (the first and third matrices on the left are inverse), so by the formula for induced characters \begin{equation}\tag{1}\chi(g)=\sum_{t=0}^{p-1}\phi{(a+tb)}.\end{equation} where
$g=\left(\begin{smallmatrix} 0 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{smallmatrix}\right)$ To evaluate the sum on the right, recall that $\phi$ is given by $\phi(x)=\zeta_m^x$ for $x\in\mathbb F_p,$ where $\zeta_m=e^{2\pi im/p}$ for some $0<m\le p-1.$ Using this it's easy to check from (1) that $\chi(g)$ is zero unless $b=0,$ i.e. unless $g\in G^\prime,$ the commutator subgroup described above, and $\chi(g)=p\zeta_m^a,$ so in particular $|\chi(g)|^2=p^2$ for $g\in G^\prime.$ Finally, $\chi$ vanishes off $H,$ because $H$ is normal in $G$ and $\chi$ is induced from $H.$ Now we calculate $$[\chi,\chi]=\frac 1{p^3}\sum_{g\in G}|\chi(g)|^2=\frac 1{p^3}|G^\prime|p^2=1,$$ so $\chi$ is indeed irreducible.
Note here $\chi$ vanishes on all of $G$ except the center $Z$ (which is equal to $G^\prime$ for this group). This is an important situation in character theory; $\chi$ is said to be fully ramified with respect to $G/Z.$
