The order-$p^3$ unitriangular group is an M-group

By [F1] the set $U$ with the displayed multiplication is a group of order $p^3$, its identity is $(0,0,0)$, its inverses are $(x,y,z{)}^{-1}=(-x,-y,-z+xy)$, and $e_1,e_2,e_3$ generate $U$ with $e_1^p=e_2^p=e_3^p=1$.

$H=\{(0,y,z)\}$ is a subgroup: for $(0,y,z),(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }})\in H$ the product is $(0,y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }}+0\cdot y^{\mathrm{\prime }})=(0,y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }})\in H$, and for $h=(0,y,z)$ the inverse computed from the formula of step 1.1 is $(0,-y,-z)$ (since $(-0,-y,-z+0\cdot y)=(0,-y,-z)$), which lies in $H$; so $H$ is closed under products and inverses, and $\mathrm{\mid }H\mathrm{\mid }=p^2$ because $y,z$ range over $\mathbb{Z}\mathrm{/}p$. The subset $Z=\{(0,0,z)\}$ is contained in $H$ and has $\mathrm{\mid }Z\mathrm{\mid }=p$.

For $(a,b)\in (\mathbb{Z}\mathrm{/}p{)}^2$ define ${\chi }_{a,b}(x,y,z):={\omega }^{ax+by}$. This is well defined on the triple $(x,y,z)\in U$, and it is a homomorphism: by step 1.1 the first two coordinates of a product add, so ${\chi }_{a,b}((x,y,z)(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }}))={\omega }^{a(x+x^{\mathrm{\prime }})+b(y+y^{\mathrm{\prime }})}={\chi }_{a,b}(x,y,z){\chi }_{a,b}(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})$. Different pairs give different characters, because ${\chi }_{a,b}(e_1)={\omega }^a$ and ${\chi }_{a,b}(e_2)={\omega }^b$ determine $a,b$ by the distinctness of the powers of $\omega$ in [F3]. Hence $U$ has at least $p^2$ characters of degree $1$; each of them is an irreducible character of $U$ and, being a linear character of the subgroup $U$ itself, is monomial in the sense of [F7]: in the covariant-function model, evaluation at $1$ identifies ${\mathrm{Ind}}_U^U{\chi }_{a,b}$ with its one-dimensional space, with inverse $v\mapsto (g\mapsto {\chi }_{a,b}(g{)}^{-1}v)$.

Characters of $H$: for $(c,d)\in (\mathbb{Z}\mathrm{/}p{)}^2$ define ${\mu }_{c,d}(0,y,z):={\omega }^{cy+dz}$. Each ${\mu }_{c,d}$ is a homomorphism, because the coordinates of $H$ multiply by adding by step 2.1: ${\mu }_{c,d}((0,y,z)(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }}))={\omega }^{c(y+y^{\mathrm{\prime }})+d(z+z^{\mathrm{\prime }})}={\mu }_{c,d}(0,y,z){\mu }_{c,d}(0,y^{\mathrm{\prime }},z^{\mathrm{\prime }})$. The $p^2$ characters ${\mu }_{c,d}$ are pairwise distinct, since ${\mu }_{c,d}(0,1,0)={\omega }^c$ and ${\mu }_{c,d}(0,0,1)={\omega }^d$ are determined by $(c,d)$ by [F3]. Since the abelian group $H$ has only one-dimensional irreducible complex characters by [F2], and every such character is a homomorphism determined by its two values on $(0,1,0)$ and $(0,0,1)$, each of which is a $p$-th root of unity by [F3], there are exactly $p^2$ of them, so $\mathrm{Irr}(H)=\{{\mu }_{c,d}:(c,d)\in (\mathbb{Z}\mathrm{/}p{)}^2\}\mathrm{.}$ Moreover ${\mu }_{c,d}(Z)=1$ exactly when $d=0$.

Conjugation formula: for $u=(x,y,z)\in U$ and $h=(0,s,t)\in H$ one has $uh{u}^{-1}=(0,s,t+xs)$. Indeed $uh=(x,y+s,z+t+xs)$ by the multiplication law, and multiplying by the inverse $u^{-1}=(-x,-y,-z+xy)$ from step 1.1 gives first coordinate $x-x=0$, second coordinate $(y+s)-y=s$, and third coordinate $(z+t+xs)+(-z+xy)+x(-y)=t+xs$. This formula gives $uH{u}^{-1}=H$, so $H$ is normal. An element $(x,y,z)$ commuting with $e_2$ must have $x=0$ by this formula; comparison of its products with $e_1$ then forces $y=0$. Conversely every $(0,0,z)$ commutes with all triples by the multiplication law. Thus $Z$ is exactly the centre, and $(x,y,z)Z\mapsto (x,y)$ identifies $U\mathrm{/}Z$ with the additive group $(\mathbb{Z}\mathrm{/}p{)}^2$.

The action of $U$ on $\mathrm{Irr}(H)$: by the definition [F4] and step 3.2, $({}^u{\mu }_{c,d})(0,s,t)={\mu }_{c,d}(u^{-1}(0,s,t)u)={\mu }_{c,d}(0,s,t-xs)={\omega }^{cs+d(t-xs)}={\mu }_{c-dx,d}(0,s,t)$ for every $(0,s,t)\in H$, where $u=(x,y,z)$. Since characters are determined by their values, ${}^u{\mu }_{c,d}={\mu }_{c-dx,d}\mathrm{.}$

Orbits and inertia groups. Fix $(c,d)$ and let $u$ run over $U$, so that $x$ runs over $\mathbb{Z}\mathrm{/}p$ while the remaining coordinates are arbitrary. If $d\mathrm{\ne }0$, then $x\mapsto c-dx$ is injective by [F8] on the $p$-element set $\mathbb{Z}\mathrm{/}p$, hence bijective, and the orbit of ${\mu }_{c,d}$ is $\{{\mu }_{c^{\mathrm{\prime }},d}:c^{\mathrm{\prime }}\in \mathbb{Z}\mathrm{/}p\}$, of size $p$; the stabilizer is $\{u:x=0\}=H$ by step 3.2, so $I_U({\mu }_{c,d})=H$. If $d=0$, then ${}^u{\mu }_{c,0}={\mu }_{c,0}$ for all $u\in U$ by step 4.1, so $I_U({\mu }_{c,0})=U$. Hence the $p(p-1)$ characters ${\mu }_{c,d}$ with $d\mathrm{\ne }0$ split into $p-1$ orbits of size $p$ (the sets with a fixed $d\mathrm{\ne }0$), and the $p$ characters with $d=0$ are fixed points.

Characters of degree $p$. Let $d\mathrm{\ne }0$ and let $\theta ={\mu }_{c,d}$; by step 5.1 its inertia group is $I_U(\theta )=H$. Since $H$ is abelian with irreducible characters exactly the ${\mu }_{c,d}$ by step 3.1, the only irreducible character of $H$ lying over $\theta$ is $\theta$ itself, so by the Clifford correspondence [F5] applied to $N=H$ and $\theta$ the set $\mathrm{Irr}(U\mid \theta )$ consists of the single character $\mathrm{\Theta }={\mathrm{Ind}}_H^U\theta$; in particular $\mathrm{\Theta }$ is irreducible, of degree $[U:H]\cdot 1=p$ (a covariant function is specified by one scalar at each of the $p$ left-coset representatives), and it is monomial, being the induction of the linear character $\theta$ of the subgroup $H$ by [F7]. This construction is well defined on orbits: the two members of an orbit have the same inertia group and induce isomorphic characters, while distinct orbits have disjoint sets $\mathrm{Irr}(U\mid \cdot )$ by [F5], so the $p-1$ orbits of step 5.1 produce $p-1$ pairwise distinct irreducible characters of $U$, all of degree $p$ and all induced from the abelian subgroup $H$ of index $p$.

Completeness. The $p^2$ linear characters ${\chi }_{a,b}$ of step 2.2 and the $p-1$ characters of degree $p$ of step 6.1 are pairwise distinct irreducible characters of $U$, of degrees $1$ and $p$. Since ${\sum }_{\chi \in \mathrm{Irr}(U)}\chi (1{)}^2=\mathrm{\mid }U\mathrm{\mid }=p^3$ by [F6] and by step 1.1, and the sum of squares over the characters listed so far is $p^2\cdot 1^2+(p-1)\cdot p^2=p^2+p^3-p^2=p^3,$ the list already attains the total: there is no further irreducible character of $U$, and the listed ones are exactly $\mathrm{Irr}(U)$. In particular the $p^2$ linear and the $p-1$ degree-$p$ characters of steps 2.2 and 6.1 are precisely the irreducible characters of $U$.

Therefore every irreducible complex character of $U$ is monomial: the $p^2$ linear characters are induced from $U$ itself by step 2.2, and the $p-1$ characters of degree $p$ are induced from the linear character ${\mu }_{c,d}$ of the abelian subgroup $H$ of index $p$ by step 6.1. By the definition [F7] the group $U={\mathrm{UT}}_3(\mathbb{Z}\mathrm{/}p)=H_p$ of order $p^3$ is an $M$-group, with exactly $p^2$ linear characters and exactly $p-1$ irreducibles of degree $p$.

The example is verified: for every prime $p$, including $p=2$, the Heisenberg group $U=H_p={\mathrm{UT}}_3(\mathbb{Z}\mathrm{/}p)$ has exactly $p^2$ linear characters, namely the ${\chi }_{a,b}$, and exactly $p-1$ irreducible characters of degree $p$, namely the ${\mathrm{Ind}}_H^U{\mu }_{c,d}$ with $d\mathrm{\ne }0$; the linear characters are induced from $U$ itself, of index $1$, and the degree-$p$ characters are induced from the abelian subgroup $H$, of index $p$, so $U$ is an $M$-group.