Proof. This is [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Lemma 2.1].

Proof. This is [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Corollary 2.2].

Proof. This is [Reference Giudici, Liebeck, Praeger, Saxl and Tiep3, Theorem 1].

Proof. This is [Reference Giudici, Liebeck, Praeger, Saxl and Tiep3, Theorem 3].

Proof. This is [Reference Giudici, Liebeck, Praeger, Saxl and Tiep3, Theorem 2].

Proof. We go through the cases of Theorem 2.5. Suppose first we are in case (1). If $n\leq 4$ , we get that $p=2$ , as $(a, s) \neq (1, 1)$ , and $H \neq \mathfrak A_3$ . Thus, the only possible case is $H = \mathfrak S_3$ , where we can find a rational character of degree divisible by $2$ . Hence we may assume that $n\geq 5$ . If $H=\mathfrak S_n$ , all irreducible characters of H are rational and we are done by the Itô–Michler theorem, since $\mathfrak S_n$ does not have normal Sylow p-subgroups for any prime p. Finally, if $H = \mathfrak A_n$ , by [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2.8], a counterexample could only happen if $p \neq 2$ and $n \in \{5, 6, 8\}$ . It is straightforward to check (for example, with GAP [2]) that the result holds in all of these cases.

Now suppose we are in case (2). Then, we may compute the character tables of the two groups explicitly using GAP and we see that $\mathrm{AGL}_3$ (2) possesses a rational (hence $\Omega$ -invariant) character of degree divisible by 3 and that $\mathrm{A}\Gamma\mathrm{L}_1$ (8) has two characters of degree 3; since $\Omega$ is a 3-group in this case, both must be $\Omega$ -invariant. Finally, for case (3), $p=2$ and $\Omega$ is the set of 2-elements of $\mathrm{Gal}(\mathbb{Q}_{10}/\mathbb{Q})$ . There is exactly one such element besides the identity, which is complex conjugation. Since all characters of $D_{10}$ are real, its characters of degree 2 are $\Omega$ -invariant.

Proof. Suppose first that $p = 3$ . By [Reference Landrock9, Corollary 1.6], $3 \mid |{\operatorname {Irr}}_{p'}(B_0(S))|$ . Since $\Omega $ is a $3$ -group, we have $|{\operatorname {Irr}}_{p', \Omega }(B_0(S))| \equiv |{\operatorname {Irr}}_{p'}(B_0(S))|\ \pmod {3}$ , meaning (since $1_S$ is of $3'$ -degree and $\Omega $ -invariant) there are at least two non-trivial characters in ${\operatorname {Irr}}_{p', \Omega }(B_0(S))$ . We may thus assume $p> 3$ .

For alternating groups, there is nothing to prove, since every irreducible character is $\Omega $ -invariant (as $\Omega $ is a p-group, $p> 2$ and characters of $\mathfrak A_n$ have at most $2$ Galois conjugates).

For sporadic groups (and the Tits simple group $^2F_4(2)'$ ), we may check with GAP [2] that, if $p \mid |S|$ , then there exists a rational character of $p'$ -degree in the principal p-block of S. Finally, if S is a simple group of Lie type, the result follows by [Reference Navarro, Tiep and Vallejo20, Theorems 5.4, 5.5].

Proof. This is [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Lemma 3.2].

Proof. Let $R\in \mathrm {Syl}_p(G)$ . Since $\theta $ is G-invariant and $(\theta (1)o(\theta ),|RN/N|)=1$ , $\theta $ has a canonical extension $\psi \in \mathrm { Irr}_{p',\Omega }(RN)$ and since $RN/N$ is a p-group, $\psi $ lies in $B_0(RN)$ . Now by the Alperin–Dade correspondence (see [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Lemma 3.1], for instance), we have that $\psi $ has an extension $\hat \psi \in \mathrm {Irr}_{p',\Omega }(B_0(RN\mathbf {C}_{G}(R)))$ .

Now $\Delta =\hat \psi ^G$ is $\Omega $ -invariant and has degree coprime to p. We consider $\Delta _{B_0(G)}$ , the sum of the irreducible constituents of $\Delta $ that belong to $B_0(G)$ , including multiplicities. Now write $X=RN\mathbf {C}_{G}(R)$ . By the argument in [Reference Navarro17, page 213], we have that $B_0(X)^G$ is defined. By Brauer’s Third Main Theorem, we have that $B_0(X)^G=B_0(G)$ . Now by [Reference Navarro17, Corollary 6.4], we have that

$$ \begin{align*}\Delta_{B_0(G)}(1)_p=((\hat{\psi}^G)_{B_0(G)})(1)_p=\hat{\psi}^G(1)_p=1.\end{align*} $$

Notice that $\Delta _{B_0(G)}$ is $\Omega $ -invariant. Indeed, if $\chi \in \mathrm {Irr}(\Delta )$ lies in the principal block, then, since $\Delta $ is $\Omega $ -invariant, if $\sigma \in \Omega $ , we have that $\chi ^\sigma \in \mathrm {Irr}(\Delta ^\sigma )=\mathrm {Irr}(\Delta )$ and $\chi ^\sigma $ lies in the principal block, hence $\chi ^\sigma \in \mathrm {Irr}(\Delta _{B_0(G)})$ . Now we apply [Reference Navarro and Tiep18, Lemma 2.1 (ii)] to $\Delta _{B_0(G)}$ to conclude that there exists $\chi \in \mathrm {Irr}_{p'}(\Delta _{B_0(G)})$ , $\Omega $ -invariant. Then $\chi \in \mathrm {Irr}_{p',\Omega }(B_0(G)|\theta )$ , as desired.

Proof. We will tackle both the general case and the case where p is odd and divides $|S_1|$ at the same time, making the necessary modifications as we go. Let $H = \bigcap _{i=1}^k \mathbf N_{G}(S_i)$ and let $\overline {G} = G/H$ , which acts transitively on the set $\Lambda = \{S_1, \ldots , S_k\}$ by conjugation. Notice how $H \neq G$ , as $k> 1$ .

Suppose, first, that this action is not p-concealed, meaning there exists a subset $\Gamma $ of $\Lambda $ such that p divides $|\overline {G}:{\operatorname {Stab}}_{\overline {G}}(\Gamma )|$ . Write $S = S_1$ . In the general case, let $1_S \neq \phi \in {\operatorname {Irr}}(S)$ be an irreducible character that extends rationally to ${\operatorname {Aut}}(S)$ (such a character exists by [Reference Hung, Schaeffer Fry, Tong-Viet and Ryan Vinroot7, Lemma 4.1]). If p is odd and divides $|S|$ , let instead $1_S \neq \phi \in {\operatorname {Irr}}_{p', \Omega }(B_0(S))$ be a character as in Lemma 2.7. In any case, let $\theta $ be the character of N obtained by placing $\phi $ in the positions corresponding to $\Gamma $ and $1_S$ in the remaining positions.

Then, in the first case, $\theta $ is an $\Omega $ -invariant character of N which extends to $\psi \in {\operatorname {Irr}}(G_\theta )$ . Furthermore, by [Reference Moretó15, Theorem 7.3], for instance, we can take $\psi $ to be rational-valued. In the second case, by Lemma 2.9, there exists an irreducible character $\psi \in {\operatorname {Irr}}_{p', \Omega }(B_0(G_\theta ){\mid }\theta )$ .

Notice that, if $g \in G_\theta $ , then g stabilizes $\Gamma $ and hence $gH \in {\operatorname {Stab}}_{\overline {G}}(\Gamma )$ . Thus, $G_\theta H/H\subseteq \mathrm {Stab}_{\overline {G}}(\Gamma )$ and therefore $p \mid |G : G_\theta |$ . Then, in both cases, $\psi ^{G}$ is irreducible, $\Omega $ -invariant, and has degree divisible by p. In the second case, by [Reference Navarro17, Corollary 6.2] and Brauer’s Third Main Theorem, it also lies in the principal block of G.

Now, we may assume the action is p-concealed. Take the blocks of maximal order $\Delta _i$ that partition $\Lambda $ , in the sense of [Reference Malle, Moretó and Rizo11, Lemma 2.4]. Then, by that lemma, there exists $H \leq L \lhd G$ such that $G/L$ acts primitively on $\{\Delta _1, \ldots , \Delta _r\}$ , and this action is p-concealed. Also, as ${\mathbf O}^{p'}(G) = G$ , p divides $|G/L|$ . By Lemma 2.6, there exists an $\Omega $ -invariant character of G with degree divisible by p, which finishes off the first case.

For the second one, let $Q \in {\operatorname {Syl}}_p(S)$ and let $R = Q \times \cdots \times Q \in {\operatorname {Syl}}_p(N)$ . Let $g \in G \setminus H$ . Then, g moves at least one of the $S_i$ through its action on $\Lambda $ by conjugation, say $S_1$ . Then, if $1 \neq x \in Q$ , $(x, 1, \ldots , 1)^g \neq (x, 1, \ldots , 1)$ , and so, $g \not \in \mathbf C_G(R)$ . Consequently, $\mathbf C_G(R) \subseteq H$ . Let $U \in {\operatorname {Syl}}_p(L)$ contain R (notice how $N \subseteq H \subseteq L$ ). Then, $\mathbf C_G(U) \subseteq \mathbf C_G(R) \subseteq H \subseteq L$ . By [Reference Malle, Moretó and Rizo11, Lemma 4.2], ${\operatorname {Irr}}(G/L) \subseteq {\operatorname {Irr}}(B_0(G))$ . Finally, by Lemma 2.6, there exists $\chi \in {\operatorname {Irr}}_\Omega (G/L)$ with $p \mid \chi (1)$ , as desired.

Proof. We break the proof down into a series of steps.

Step 0. We may assume ${\mathbf O}^{p'}(G)=G$ and ${\mathbf O}_{p'}(G)=1$ . Moreover, if $1<N\lhd G$ , $G/N$ has abelian Sylow p-subgroups. Also, G has a unique minimal normal subgroup N and $p\neq 2$ .

First, suppose ${\mathbf O}^{p'}(G) < G$ . If there exists $\theta \in {\operatorname {Irr}}_{\Omega }(B_0({\mathbf O}^{p'}(G)))$ of degree divisible by p, then, by Lemma 2.8, we could find $\chi \in {\operatorname {Irr}}_{\Omega }(B_0(G)\mid \theta )$ . Since $\theta (1)$ divides $\chi (1)$ , $\chi (1)_p> 1$ , which goes against our hypothesis. Thus, ${\mathbf O}^{p'}(G)$ also fulfills our hypothesis, meaning it has abelian Sylow p-subgroups by induction. As $|G:{\mathbf O}^{p'}(G)|$ is not divisible by p, this implies G has abelian Sylow p-subgroups and we are done.

Now, let $1 \neq N \lhd G$ . Then, ${\operatorname {Irr}}_{\Omega }(B_0(G/N)) \subseteq {\operatorname {Irr}}_{\Omega }(B_0(G))$ , meaning $G/N$ also satisfies the hypothesis. By induction, it has abelian Sylow p-subgroups. In particular, assume ${\mathbf O}_{p'}(G)> 1$ . Then, $G/{\mathbf O}_{p'}(G)$ has abelian Sylow p-subgroups and if $P \in {\operatorname {Syl}}_p(G)$ , then $P \cong P{\mathbf O}_{p'}(G)/{\mathbf O}_{p'}(G) \in {\operatorname {Syl}}_p(G/{\mathbf O}_{p'}(G))$ , meaning P is also abelian, as we wanted to prove.

Finally, let $M, N$ be distinct minimal normal subgroups of G. Then, the diagonal map composed with the canonical projections gives an embedding of G into $G/N \times G/M$ . By our previous paragraph, both of these have abelian Sylow p-subgroups. Since the Sylow p-subgroups of G embed into those of $G/N \times G/M$ , they are abelian as well. The fact that we can assume $p \neq 2$ follows from [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2].

Step 1. Let N be the unique minimal normal subgroup of G. We may assume that N is not semisimple; in particular, ${\mathbf O}_p(G)> 1$ .

Suppose $N = S_1 \times \cdots \times S_t$ is a product of isomorphic non-abelian simple groups of order divisible by p, which are transitively permuted by G-conjugation. If $t> 1$ , then the result follows by Step 0 and Lemma 2.10. Now, if $t = 1$ , by the uniqueness of $N = S$ , $\mathbf C_G(S) = 1$ and G is an almost simple group with socle S, without proper normal subgroups of $p'$ -index. Then, the result holds by [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2.17].

Step 2. If $N\subseteq \mathbf Z(G)$ , then we are done.

Suppose that $N\subseteq \mathbf Z(G)$ (so $|N|=p$ ). Then, ${\mathbf O}_{p'}(G/N) = N/N$ . Otherwise, if $N \subsetneqq K$ is such that $K/N = {\mathbf O}_{p'}(G/N)$ , then $K = XN$ , by the Schur–Zassenhaus Theorem. But the centrality of N implies $X \unlhd K$ , and thus X is characteristic in K, hence normal in G, contradicting N being the unique minimal normal subgroup of G.

Since $G/N$ has abelian Sylow p-subgroups, by [Reference Malle, Moretó and Rizo11, Theorem 4.1], we have that G is the central product of some normal subgroups X, $S_1,\ldots , S_t$ , where $X/N$ is an abelian p-group and, for each i, $S_i/N$ is a non-abelian simple group with abelian Sylow p-subgroups. Since N is the unique minimal normal subgroup of G, $N\subseteq S_i'$ . Then, as $S_i/N$ is non-abelian simple, we have that $S_i$ is perfect, hence $S_i$ is a quasi-simple group with center N, for every i.

Let $1_N\neq \lambda \in {\operatorname {Irr}}(N)$ . By [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2.9] there exists $\psi _i \in {\operatorname {Irr}}_\Omega (B_0(S_i))$ of degree divisible by p lying over $\lambda $ (if necessary, replacing $\psi _i$ by a Galois conjugate). Now, letting $\xi \in \mathrm {Irr}(X|\lambda )$ , then $\xi \in \mathrm {Irr}_\Omega (B_0(X)|\lambda )$ . By [Reference Malle and Navarro13, Lemma 4.1] we have that the central product of characters

$$ \begin{align*}\chi=\xi\star \psi_1\star\psi_2\star \cdots\star \psi_t\end{align*} $$

lies in the principal block of G. Hence $\chi \in \mathrm { Irr}_\Omega (B_0(G))$ has degree divisible by p, a contradiction.

Step 3. We may assume that $\mathbf F(G)=\mathbf F^*(G)$ .

Since ${\mathbf O}_{p'}(G)=1$ we have that $F=\mathbf F(G)={\mathbf O}_p(G)>1$ . Suppose that $E=\mathbf E(G)>1$ and let $Z=\mathbf Z(E)$ . Since N is the unique minimal normal subgroup of G, $N\subseteq Z$ (notice that $Z>1$ since otherwise $\mathbf F^*(G)=\mathbf F(G)\times E$ in contradiction with the fact that $N\subseteq E$ ). We claim that $E/Z=S_1/Z\times \cdots \times S_n/Z$ , where $S_i\trianglelefteq G$ for every i. Let $W/Z$ be a (non-abelian) minimal normal subgroup of $G/Z$ contained in $E/Z$ . By the Schur–Zassenhaus Theorem and Step 0, we know that $|W/Z|$ is divisible by p. Now, since $G/Z$ has abelian Sylow subgroups, we can once again use [Reference Malle, Moretó and Rizo11, Theorem 4.1]. Then, if $K/Z = {\mathbf O}_{p'}(G/Z)$ , $K \cap W = Z$ and $WK/K$ is a minimal normal subgroup of $G/K$ , which is the direct product of an abelian p-group and non-abelian simple groups of order divisible by p. Since $WK/K \cong W/Z$ , we have that $W/Z$ is simple and the claim follows.

Write $S=S_1$ , so that $S'$ is a quasi-simple normal subgroup of G. Using again that N is the unique minimal normal subgroup of G, we have that $N\subseteq \mathbf Z(S)\cap S'\subseteq \mathbf Z(S')$ . Looking at the Schur multipliers of the simple groups [Reference Gorenstein, Lyons and Solomon4], if $p\geq 5$ , we deduce that $\mathbf Z(S')$ has cyclic Sylow p-subgroups. We claim that $\mathbf Z(S')$ has cyclic Sylow p-subgroups for $p=3$ as well.

Indeed, it can be checked in [2] that the unique simple group S whose Schur multiplier has a non-cyclic Sylow $3$ -subgroup is $\operatorname {PSU}_4(3)$ , but this group does not have abelian Sylow $3$ -subgroups.

In all cases, we conclude that N is cyclic and hence $|N|=p$ . Now, the order of $G/\mathbf C_G(N)$ divides $p-1$ . Using Step 0 we conclude that N is central in G and we are done by Step 2. Therefore, we may assume that $E=1$ , so that $\mathbf F(G)=\mathbf F^*(G)$ , as desired.

Step 4. We may assume that G has a unique p-block.

Since ${\mathbf O}_{p'}(G)=1$ and we have that ${\mathbf O}_p(G)=\mathbf F(G)=\mathbf F^*(G)$ by Step 3, then $\mathbf {C}_{G}(\mathbf{O}_p(G))\subseteq {\mathbf O}_p(G)$ , by Hall–Higman’s Lemma 1.2.3 ([Reference Isaacs8, Theorem 3.21], for instance). In this situation, G has a unique p-block, since, by [Reference Navarro17, Theorems 4.8, 4.14], each p-block of G is induced from a p-block of ${\mathbf O}_p(G)$ , and there is only one such block.

Step 5. We may assume that if $\lambda \in \mathrm {Irr}(N)$ , then p does not divide $|G:G_\lambda |$ .

We first prove that there exists an $\Omega $ -invariant $\chi \in \mathrm {Irr}(G_\lambda )$ . Let $R\in \mathrm {Syl}_p(G_\lambda )$ and let $\psi \in \mathrm { Irr}(R|\lambda )$ of minimum degree among the characters in $\mathrm {Irr}(R|\lambda )$ . Let $\Delta =\psi ^{G_\lambda }$ , so $\Delta $ is an $\Omega $ -invariant character (notice that $\psi $ is $\Omega $ -invariant because R is a p-group). Now notice that if $\chi \in \mathrm {Irr}(\Delta )$ , then $\chi $ lies over $\psi $ and hence over $\lambda $ . Since $\lambda $ is $G_\lambda $ -invariant, if $\xi \in \mathrm {Irr}(R)$ is a constituent of $\chi _R$ , then $\xi $ lies necessarily over $\lambda $ and then $\xi (1)\geq \psi (1)$ , and since they are p-powers, $\psi (1)\mid \xi (1)$ . Hence, if $\xi $ is an irreducible constituent of $\Delta _R$ , we have that $\psi (1)\mid \xi (1)$ . Write $\Delta _R=a_1\xi _1+\ldots + a_k\xi _k$ , with $\xi _i\in \mathrm {Irr}(R)$ .

Then

$$ \begin{align*}|G_\lambda:R|\psi(1)=\Delta(1)=a_1\xi_1(1)+\ldots +a_k\xi_k(1)=\psi(1)\left(a_1\frac{\xi_1(1)}{\psi(1)}+\ldots + a_k\frac{\xi_k(1)}{\psi(1)}\right),\end{align*} $$

and hence

$$ \begin{align*}|G_\lambda:R|=a_1\frac{\xi_1(1)}{\psi(1)}+\ldots + a_k\frac{\xi_k(1)}{\psi(1)},\end{align*} $$

and, since $|G_\lambda :R|$ is a $p'$ -number, we conclude that there exists i such that $a_i$ is not divisible by p. Now by [Reference Navarro and Tiep18, Lemma 2.1 (i)] applied to $G_\lambda $ , R, $A=\Omega $ and $\Delta $ , we conclude that there exists an $\Omega $ -invariant $\tau \in \mathrm {Irr}(G_\lambda )$ with $[\Delta ,\tau ]\neq 0$ . In particular, $\tau $ lies over $\lambda $ and is $\Omega $ -invariant. Then $\tau ^G\in \mathrm {Irr}(G)$ is $\Omega $ -invariant and by our hypothesis, combined with Step $4$ , $p\nmid \tau ^G(1)$ . Then $|G:G_\lambda |$ is a $p'$ -number, as wanted.

Step 6. We may assume that ${\mathbf O}_{p'}(G/N)>1$ .

Recall that by Step 0, $G/N$ has abelian Sylow p-subgroups. Suppose ${\mathbf O}_{p'}(G/N)=1$ . Then, since $G/N$ has abelian Sylow p-subgroups, using [Reference Malle, Moretó and Rizo11, Theorem 4.1], there exist $X,Y\trianglelefteq G$ containing N such that $X/N$ is an abelian p-group, $Y/N$ is a direct product of non-abelian simple groups $V_i/N$ of order divisible by p and $G/N=X/N\times Y/N$ . Furthermore $V_i/N$ is normal in $G/N$ , since it is normal in $Y/N$ and commutes with all elements of $X/N$ . Notice that X is a p-group, so $1<\mathbf Z(X)\lhd G$ . Therefore, $N\subseteq \mathbf Z(X)$ . Then $X\subseteq \mathbf C_G(N)$ . Moreover, since $V_i/N$ is simple, $\mathbf C_{V_i}(N)=V_i$ or $\mathbf C_{V_i}(N)=N$ . If the former happens for some i, then $N\subseteq \mathbf Z(V_i)$ and hence, as in Step 3, it is cyclic. Then $|N|=p$ , and $|G:\mathbf {C}_{G}(N)|$ divides $p-1$ . By Step 0 we conclude that $N\subseteq \mathrm {\textbf {Z}}(G)$ and hence we are done by Step 2. Therefore $\mathbf C_{G}(N)=X$ .

Using Step 5, we now apply Theorem 2.3 and Theorem 2.4 to the action of $G/X\cong Y/N$ on ${\operatorname {Irr}}(N)$ . Suppose first that this action is primitive, so that Theorem 2.3 applies. Since $G/X$ is not solvable, we are not in case (2). Suppose now that we are in case (3). In subcases (i) and (v), $p=2$ in contradiction with Step 0. Since $G/X\cong Y/N$ is a direct product of simple groups, subcase (ii) does not occur. In subcases (iii) and (iv), we may assume that $G/X=\operatorname {PSL}_2(11)$ or $G/X=M_{11}$ and that $p=3$ , respectively. In subcase (iii), there exists $\chi \in {\operatorname {Irr}}_\Omega (G/X)$ such that $\chi (1)=12$ , a contradiction. Suppose that $G/X\cong Y/N=M_{11}$ and $p=3$ . Then there is a rational character of degree 45 in $\mathrm {Irr}(G/X)$ , another contradiction.

Finally, assuming we are in case (1), we may argue as in Step 8 of the proof of [Reference Malle, Moretó and Rizo11, Theorem 4.6]. Doing so, we see that all of the cases we need to consider – namely, the simple groups with abelian Sylow p-subgroup that appear in [Reference Liebeck10, Appendix 1] – require $p = 2$ , and this goes against Step 0.

Now, we may assume that the action of $L=Y/N$ on N is imprimitive. We apply Theorem 2.4 (recall that $L={\mathbf O}^{p'}(L)$ by Step 0). Let $N=N_1\oplus \cdots \oplus N_n$ be an imprimitivity decomposition for N. Therefore, $\mathbf N_L(N_i)$ is transitive on $N_i\setminus \{0\}$ and $M:=L/\bigcap \mathbf N_L(N_i)$ induces a primitive p-concealed subgroup of $\mathfrak S_n$ . Note that M is a factor group of G. By Lemma 2.6, this group has an $\Omega $ -invariant irreducible character $\chi $ such that $\chi (1)_p>1$ and the result follows in this case.

Step 7. We may assume that $N=\mathbf F(G)$ ; in particular, $\mathbf C_G(N)=N$ .

Let $K/N={\mathbf O}_{p'}(G/N)$ . By Step 6, we may assume that $K>N$ . By the Schur–Zassenhaus Theorem, there exists a p-complement H of K, so $K=HN$ and $H\cap N=1$ . Then by the Frattini argument we have $G=N\mathbf N_G(H)$ . Write $L=\mathbf N_G(H)$ . Now, since N is abelian and normal in G, $\mathbf N_N(H)$ is also normal in $G=N\mathbf N_G(H)=NL$ , and so either $\mathbf N_N(H) =1$ or $\mathbf N_N(H) = N$ . If $\mathbf N_N(H)=N$ , then $H\lhd G$ , and we get a contradiction since ${\mathbf O}_{p'}(G)=1$ by Step 0. Thus, $L\cap N=\mathbf N_N(H) =1$ and L is a complement of N in G.

Let $F=\mathbf F(G)={\mathbf O}_p(G)$ . We claim that $F=N$ . Notice that $N\subseteq \mathbf Z(F)$ since $\mathbf Z(F)>1$ . Let $F_1=F\cap L$ . Then $F_1\lhd L$ and since $G=NL$ , we have that $F_1\lhd G$ . Since N is the unique minimal normal subgroup of G, this forces $F_1=1$ , so $F=N$ as claimed. Also, since $N=\mathbf F(G)=\mathbf F^*(G)$ by Step 3, we have $\mathbf C_G(N)\subseteq N$ , as wanted.

Step 8. Completion of the proof.

Let $K/N={\mathbf O}_{p'}(G/N)$ . By Step 6, we may assume that $K>N$ . Again, by Step 0 we have that there exist $X,Y\trianglelefteq G$ containing N such that $X/K$ is an abelian p-group, $Y/K$ is a direct product of non-abelian simple groups $V_i/K\trianglelefteq G/K$ of order divisible by p and $G/K=X/K\times Y/K$ .

As in Step 7, let H be a p-complement of K and let $L=\mathbf N_G(H)$ . Then $G=NL$ and $N\cap L=1$ . By Step 7, we may assume that L acts faithfully and irreducibly on ${\operatorname {Irr}}(N)$ . By Step 5, this action can be assumed to be p-exceptional.

Suppose first that the action of L on ${\operatorname {Irr}}(N)$ is primitive. We apply Theorem 2.3. In case (2), G is solvable and we are done by [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 3]. Assume that we are in case (3). In subcases (i) and (v) have that $p=2$ , in contradiction with Step 0. In subcases (ii), (iii) and (iv) $G/N$ has a normal subgroup $V/N=\operatorname {SL}_2(5), \operatorname {PSL}_2(11)$ or $M_{11}$ respectively and in all cases $p=3$ . Note that one of the simple direct factors of $Y/K$ is then $\mathfrak A_5,\operatorname {PSL}_2(11)$ or $M_{11}$ , respectively. Since they all have $\Omega $ -invariant irreducible characters of degree divisible by $3$ , the result follows in these cases.

Now, we may assume that we are in case (1). So L is transitive on ${\operatorname {Irr}}(N)\setminus \{1_N\}$ and L is one of the groups from Hering’s theorem. Again, we use the description in [Reference Liebeck10, Appendix 1]. We start with the infinite classes described in (A). Since $G/N$ is not solvable and has abelian Sylow p-subgroups, it follows that $G/N$ has a normal subgroup $V/N\cong \operatorname {SL}_2(p^n)$ , where $|N|=p^{2n}>3$ . Now, write $G/K=X/K\times Y/K=X/K\times Y_1/K\times \cdots \times Y_t/K$ . Since $\operatorname {PSL}_2(p^n)$ is a non-abelian composition factor of G of order divisible by p, it is a composition factor of $G/K$ . Hence, there must be some i such that $Y_i/K\cong \operatorname {PSL}_2(p^n)$ . Write $M/K=Y_i/K$ and let $\varphi \in \mathrm {Irr}_\Omega (M/K)$ of degree divisible by p (using [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2.8], for instance). Then $\chi =1\times \cdots \times 1\times \varphi \times 1\times \cdots \times 1\in \mathrm { Irr}(G/K)\subseteq \mathrm {Irr}(G)$ is $\Omega $ -invariant and of degree divisible by p.

Next, we consider the extra-special classes described in (B). The first four groups in [Reference Liebeck10, Table 10] are solvable, so they do not occur. The last case of this table (where $|N|=3^4)$ can be handled with [2]. In this case we have that L ( $G_0$ in the notation of Liebeck) has a normal $2$ -subgroup R such that $L/R$ is isomorphic to a subgroup of $\mathfrak S_5$ . Since G is not solvable, $L/R$ cannot be solvable and hence $L/R\cong \mathfrak A_5$ or $L/R\cong \mathfrak S_5$ . As before, this means that one of the simple direct factors of $Y/K$ is isomorphic to $\mathfrak A_5$ and we know it possesses an $\Omega $ -invariant character of degree divisible by $3$ .

Finally, we consider the exceptional classes described in (C). In the cases where $p^d\in \{11^2,19^2,29^2,59^2\}$ in [Reference Liebeck10, Table 11], we have that $G/N$ (and hence, G) is p-solvable, so we are done by [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 3]. Since $G/N$ has abelian Sylow p-subgroups, we are left with the first and the last cases of Table 11. In the last case, $p=3$ and $G/N\cong \operatorname {SL}_2(13)$ . Here, there is $\chi \in {\operatorname {Irr}}(G/N)$ of degree 6 which is $\Omega $ -invariant. The first case can be handled as above, since we again have $\mathfrak A_5$ as a simple direct factor of $Y/K$ .

Now, assume that the action of L on N is imprimitive. Arguing as in the last paragraph of Step 6, we find $\chi \in {\operatorname {Irr}}_\Omega (L)$ such that $\chi (1)_p>1$ . This final contradiction completes the proof.

Proof. By Theorem A, G has abelian Sylow p-subgroups. The result then follows by [Reference Malle, Moretó and Rizo11, Theorem 4.1]. The fact that the simple direct factors of $Y/N$ do not possess any $\Omega $ -invariant characters of degree divisible by p in the principal block follows from Lemma 2.8 and the fact that $Y/N \cong {\mathbf O}^{p'}(G)/X$ .

Proof. Let G be a minimal counterexample to the assertion in the theorem. Write, as in Corollary 3.1, ${\mathbf O}^{p'}(G)/N = X/N \times Y/N$ , where $N = {\mathbf O}_{p'}({\mathbf O}^{p'}(G))$ . Notice that $X = HN$ , where H is an abelian p-group; from this, it follows that ${\mathbf O}^{p'}(G) = HY$ . Notice also that ${\mathbf O}_p(G)\subseteq {\mathbf O}^{p'}(G)$ and hence ${\mathbf O}_p(G)N/N\subseteq X/N={\mathbf O}_p({\mathbf O}^{p'}(G)/N)$ . It follows that ${\mathbf O}_p(G) \subseteq X$ ; in particular, it is contained in H.

Step 1. ${\mathbf O}^{p'}(G) = G$ .

Assume otherwise. By Lemma 2.8, ${\mathbf O}^{p'}(G)$ also satisfies our hypothesis, since each of its $\Omega $ -invariant irreducible characters lies under some $\Omega $ -invariant irreducible character of G. Then, by the minimality of G, we have ${\mathbf O}^{p'}(G) = {\mathbf O}_p({\mathbf O}^{p'}(G)) \times K$ , for some $K \unlhd {\mathbf O}^{p'}(G)$ in the conditions of the statement of the lemma, as ${\mathbf O}^{p'}({\mathbf O}^{p'}(G)) = {\mathbf O}^{p'}(G)$ . But also ${\mathbf O}_p(G) \subseteq {\mathbf O}^{p'}(G)$ . Thus ${\mathbf O}^{p'}(G) = {\mathbf O}_p(G) \times K$ . Finally, since $K/{\mathbf O}_{p'}(K)$ is a (possibly empty) direct product of non-abelian finite simple groups of order divisible by p, ${\mathbf O}^p(K) = K$ and thus $K = {\mathbf O}^p({\mathbf O}^{p'}(G))$ , meaning K is normal in G. This violates G being a counterexample.

Step 2. ${\mathbf O}_p(G) = 1$ .

If $M={\mathbf O}_p(G)>1$ , write $\bar {G}=G/M$ . Since $\bar {G}$ satisfies our hypothesis, we have, by the minimality of G, that ${\mathbf O}^{p'}(\bar {G})=\bar {K}$ , where $\bar {K}/{\mathbf O}_{p'}(\bar {K})$ is a (possibly empty) direct product of non-abelian simple groups of order divisible by p. Notice that ${\mathbf O}^{p'}(\bar {G})={\mathbf O}^{p'}(G)/M = G/M$ . Since $X/N = {\mathbf O}_p(G/N)$ and $MN/N$ is a normal p-subgroup of G, $MN \subseteq X$ . But also $|X:MN|$ is a power of p and $G/X$ is a direct product of non-abelian simple groups of order divisible by p, meaning $G/NM$ has no proper normal subgroups of $p'$ -order. Consequently, $MN/M = {\mathbf O}_{p'}(G/M)$ . We conclude that $G/MN$ is the direct product of non-abelian simple groups and therefore, $X=MN$ which implies that $H=M$ , from which $G = M \times Y$ , which violates it being a counterexample.

Step 3. If $M \subseteq N$ is a minimal normal subgroup of G, then M is not abelian.

First, notice that if $N=1$ we have the desired result, so we can exclude this case. Now, suppose that M is abelian. Let P be a Sylow p-subgroup of G containing H. Then, by Theorem 4.3, we have that

$$ \begin{align*} 1<\mathbf N_G(P)\cap M=\mathbf C_M(P)\leq \mathbf C_M(HM)=M\cap \mathbf Z(HM). \end{align*} $$

We claim that $\mathbf Z(HM)$ is normal in G. Indeed, by induction $G/M={\mathbf O}^{p'}(G/M)={\mathbf O}_p(G/M)\times K_1/M$ . Notice that ${\mathbf O}_p(G/M)\subseteq X/M$ and therefore ${\mathbf O}_p(G/M)\subseteq HM/M$ .

Notice that $N \subseteq K_1$ , and hence ${\mathbf O}_{p'}(K_1) = N$ . Then $K_1/N$ is a (possibly empty) direct product of non-abelian simple groups of order divisible by p. Also notice that $K_1 = {\mathbf O}^p(G)$ , as $K_1 = {\mathbf O}^p(K_1)$ and $|G:K_1|$ is a power of p. Since the same is true of Y, it follows that $Y = K_1$ .

Therefore, we conclude that $HM/M={\mathbf O}_p(G/M)$ , by comparing orders. Hence $HM$ is normal in G, and so is $\mathbf Z(HM)$ , as wanted. Since $M\cap \mathbf Z(HM)>1$ , we conclude that $M\subseteq \mathbf Z(HM)$ . Hence H is normal in $HM$ and therefore characteristic, so it is normal in G. As ${\mathbf O}_p(G) = 1$ , $H = 1$ and $G = Y$ , a contradiction to G being a counterexample.

Step 4. Final contradiction.

By Steps 2 and 3, M is the product of non-abelian simple groups of order not divisible by p, transitively permuted by G-conjugation. Write $M = S_1 \times \cdots \times S_k$ . By Lemma 2.10, $k = 1$ , meaning $M = S$ , a non-abelian simple group of $p'$ -order. In particular, $G/\mathbf C_G(S)$ is an almost simple group with socle $S\mathbf C_G(S)/\mathbf C_G(S)\cong S$ .

Now, $G/S\mathbf C_G(S)\cong \frac {G/\mathbf C_G(S)}{S\mathbf C_G(S)/\mathbf C_G(S)}$ is solvable by Schreier’s conjecture. Write $W/S={\mathbf O}_p(G/S)$ , and notice that, by induction $G/S=W/S\times K_1/S$ for some $K_1$ such that $\frac {K_1/S}{{\mathbf O}_{p'}(K_1/S)}$ is a product of non-abelian simple groups. Let $L=W\mathbf C_G(S)$ . Then, $G/L$ is solvable and hence $K_1/(K_1\cap L)$ is solvable. Thus, $K_1=(K_1\cap L)N$ and we have that $K_1/(K_1\cap L)$ is a $p'$ -group. But since $G=WK_1$ , we have that $G=LK_1$ and hence $G/L\cong K_1/(K_1\cap L)$ is a $p'$ -group. As ${\mathbf O}^{p'}(G)=G$ , we obtain that $G=L$ and hence $G/S\mathbf C_G(S)\cong W/(W\cap S\mathbf C_G(S))$ is a p-group.

By the proof of [Reference Malle, Moretó, Rizo and Schaeffer Fry12, Theorem 2.8], if $G \neq S\mathbf C_G(S)$ , since p does not divide $|S|$ , there exists $\chi \in {\operatorname {Irr}}_{\Omega }(G/\mathbf C_G(S))$ of degree divisible by p, which goes against our hypothesis. Thus, $G = S\mathbf C_G(S)$ . But this is a contradiction, as $S \cap \mathbf C_G(S) = 1$ and $|S|$ is not divisible by p.

Proof. By Lemma 4.4, we may write ${\mathbf O}^{p'}(G) = {\mathbf O}_p(G) \times K$ where $K/{\mathbf O}_{p'}(K)$ is a direct product of non-abelian finite simple groups of order divisible by p. If these groups have an $\Omega $ -invariant character of degree divisible by p, then so does $K/{\mathbf O}_{p'}(K)$ , and hence, so does ${\mathbf O}^{p'}(G)/{\mathbf O}_p(G)$ . This means that ${\mathbf O}^{p'}(G)$ has an $\Omega $ -invariant character of degree divisible by p and, by Lemma 2.8, so does G, a contradiction.

Proof. By what we did before in Lemma 4.4 and Lemma 4.5, ${\mathbf O}^{p'}(G) = {\mathbf O}_p(G) \times Y$ , $Y/N$ is a direct product of non-abelian finite simple groups of order divisible by p without $\Omega $ -invariant characters of degree divisible by p, where $N = {\mathbf O}_{p'}({\mathbf O}^{p'}(G))$ , and ${\mathbf O}^{p'}(G) = XY$ , where $X = {\mathbf O}_p(G)N$ . Also, $Y/N = {\mathbf O}^p({\mathbf O}^{p'}(G)/N)$ , meaning $Y/N$ is normal in $G/N$ ; hence, Y is normal in G.

All that is left is to show that ${\mathbf O}_{p'}(Y)$ is solvable. In view of that, we may assume $p \neq 2$ by the Feit–Thompson theorem. If ${\mathbf O}^{p'}(G) \neq G$ , then, by Lemma 2.8, we get the desired result by induction. Hence, we may assume $G = {\mathbf O}^{p'}(G)$ . Along the same lines, as $G/{\mathbf O}_p(G) \cong Y$ , if Y had an $\Omega $ -invariant character of degree divisible by p, so would G. Thus, Y also satisfies our hypothesis.

Notice that ${\mathbf O}^{p'}(Y) = Y$ , since $Y \cong G/{\mathbf O}_p(G)$ and ${\mathbf O}^{p'}(G) = G$ . Consequently, if $Y \neq G$ , by induction, we get that ${\mathbf O}_{p'}(Y)$ is solvable, which is our desired result. So we may assume $Y = G$ . In this situation, ${\mathbf O}_p(G) = 1$ , $N = {\mathbf O}_{p'}(G)$ and $G/N$ is a direct product of non-abelian simple groups of order divisible by p whose $\Omega $ -invariant characters are of $p'$ -degree. Let $R = \mathbf {R}(N)$ be the solvable radical of N and suppose $R \neq 1$ . Then, by induction on $G/R$ , we get that ${\mathbf O}_{p'}(G/R) = N/R$ is solvable, meaning so, too, is N. Thus, we may assume $R = 1$ .

Let M be a minimal normal subgroup of G contained in N. By the previous paragraph, $M = S_1 \times \cdots \times S_k$ , a product of non-abelian simple groups transitively permuted by G-conjugation and of order not divisible by p. If $k> 1$ , then the result follows by Lemma 2.10. Thus, we may assume $k = 1$ . Then, $M = S$ is a non-abelian simple group and $G/\mathbf C_G(S)$ is an almost simple group with socle isomorphic to S. By the Schreier Conjecture, $G/S\mathbf C_G(S)$ is solvable. In particular, due to the structure of G, $|G:S\mathbf C_G(S)|_p = 1$ . Consequently, $G = S\mathbf C_G(S)$ as ${\mathbf O}^{p'}(G) = G$ . But this is impossible, as $p \nmid |S|$ and S is a quotient of G.