Proof. Suppose that $\chi_N\not\in{{\operatorname{Irr}}}(N)$. Then there exists a central extension $G^*$ of G and $\alpha,\beta\in{{\operatorname{Irr}}}(G^*)$ such that $\chi=\alpha\beta$, where $\alpha,\beta$ are primitive nonlinear irreducible characters of $G^*$. Without loss of generality, we may assume that $\alpha(1)\leq\chi(1)^{1/2}$. Since $1_{G^*}$ is an irreducible constituent of $\alpha\overline{\alpha}$, the minimality of $\chi(1)$ implies that $\alpha\overline{\alpha}$ is a sum of linear characters. Hence $(G^*)'\leq\operatorname{Ker}(\alpha\overline{\alpha})$. In particular, $(G^*)'\leq{\mathbf{Z}}(\alpha)$. By Lemma 2.27 of [Reference Isaacs1], ${\mathbf{Z}}(\alpha)/\operatorname{Ker}\alpha\leq{\mathbf{Z}}(G^*/\operatorname{Ker}\alpha)$. If follows that $G^*/\operatorname{Ker}\alpha$ is nilpotent (of class at most 2). But α is nonlinear and primitive. This contradicts Theorem 6.22 of [Reference Isaacs1].

Proof. We may assume that $\chi(1) \gt 1$. By Theorem 2.17 of [Reference Isaacs3], χ factors as a product of p-special characters, where p runs over the set of prime divisors of $\chi(1)$. Since $\chi(1)=m(G)$, it follows that $\chi(1)=p^n$ is a power of a prime p. This proves the first part of the lemma.

Suppose now that p > 2. Let q ≠ p be a prime. Then $\chi_{\mathbf{O}_q(G)}$ is not irreducible. It follows from Lemma 2.1 that $\mathbf{O}_q(G)$ is central in G. Hence ${\mathbf{F}}(G)=E{\mathbf{Z}}(G)$, where $E=\mathbf{O}_p(G)$. Furthermore, using again Lemma 2.1, every normal abelian subgroup of G is central. Since G has a faithful irreducible character, Theorem 2.32 of [Reference Isaacs1] implies that ${\mathbf{Z}}(G)$ is cyclic. Now, Corollary 1.10 of [Reference Manz and Wolf5] implies that E is extraspecial of exponent p. Since $\chi_E\in{{\operatorname{Irr}}}(E)$, we necessarily have that $|E|=p^{2n+1}$.

Note that ${\mathbf{C}}_G(E)={\mathbf{C}}_G({\mathbf{F}}(G))={\mathbf{Z}}(G)$, so $G/{\mathbf{Z}}(G)$ is isomorphic to a subgroup of ${{\operatorname{Aut}}}_{{\mathbf{Z}}(E)}(E)$. By [Reference Winter8], using again that p > 2, we deduce that $G/{\mathbf{F}}(G)$ is isomorphic to a subgroup of $\operatorname{Sp}(2n,p)$. By [Reference Landazuri and Seitz4], $\operatorname{Sp}(2n,p)$ has a faithful irreducible representation of dimension $(p^n-1)/2$. Hence $G/{\mathbf{F}}(G)$ has a faithful character of degree $\leq (p^n-1)/2$. Since $m(G)=p^n$, we conclude that $G/{\mathbf{F}}(G)$ has a faithful character that is a sum of linear characters. We conclude that $G/{\mathbf{F}}(G)$ is abelian, as we wanted to prove.

Proof. Let $H\leq G$ and $\beta\in{{\operatorname{Irr}}}(H)$ primitive such that $\beta^G=\chi$. Suppose first that $\beta(1)=1$, so that $\chi(1)=|G:H|$. Since 1_G is an irreducible constituent of $(1_H)^G$ and $m(G)=|G:H|=(1_H)^G(1)$, we deduce that $(1_H)^G$ is a sum of linear characters. Hence $G'\leq\operatorname{Ker}(1_H)^G\leq H$. Thus $H\trianglelefteq G$ and by Clifford’s theorem (Theorem 6.2 of [Reference Isaacs1]), χ_H is a sum of conjugates of β. In particular, χ_H is a sum of linear characters, so $H'\leq\operatorname{Ker}\chi=1$. Hence G is metabelian and the result follows.

Now, we may assume that $\beta(1) \gt 1$ is odd. First, we will see that $H\trianglelefteq G$ and $G'=H'$. Note that $\chi(1)=|G:H|\beta(1) \gt |G:H|$. Hence $(1_H)^G$ is a sum of linear characters and $G'\leq H$, as before. In particular, $H\trianglelefteq G$. Thus H ^ʹ is also normal in G and all the irreducible characters of $G/H'$ have degree divisible by $|G:H| \lt \chi(1)=m(G)$. Hence, ${{\operatorname{Irr}}}(G/H')$ is a set of linear characters, and we conclude that $G'=H'$, as desired.

Now, we claim that $\beta(1)=m(H)$. Let $\mu\in{{\operatorname{Irr}}}(H)$ be nonlinear. Hence, there exists $\nu\in{{\operatorname{Irr}}}(G)$ nonlinear such that $[\mu^G,\nu]\neq 0$. Thus

\begin{equation*} |G:H|\beta(1)=\chi(1)=m(G)\leq\nu(1)\leq\mu^G(1)=|G:H|\mu(1). \end{equation*}

We conclude that $\beta(1)\leq\mu(1)$. The claim follows.

Thus β is a primitive faithful minimal character of $H/\operatorname{Ker}\beta$. By Lemma 2.2, we have that $\beta(1)$ is a power of a prime p. Let $\beta=\beta_1,\dots,\beta_t$ be the G-conjugates of β. Let $K_i=\operatorname{Ker}\beta_i$. Since $\chi=\beta^G$ is faithful, Lemma 5.11 of [Reference Isaacs1] implies that $\bigcap_{i=1}^tK_i=1$. Since p > 2, by the second part of Lemma 2.2, we have that $H/K_i$ is nilpotent-by-abelian. Write $F_i/K_i={\mathbf{F}}(H/K_i)$, so that $G'=H'\leq F_i$ for every i. By Proposition 9.5 of [Reference Manz and Wolf5], $\bigcap_{i=1}^tF_i={\mathbf{F}}(H)$. Therefore, $G'\leq {\mathbf{F}}(H)$ and we conclude that G is nilpotent-by-abelian, as wanted.