Even Character Degrees and Ito–Michler Theorem

Abstract

Let $\textrm{Irr}_2(G)$ be the set of linear and even-degree irreducible characters of a finite group G. In this paper, we prove that G has a normal Sylow 2-subgroup if $\sum \limits _{\chi \in \textrm{Irr}_2(G)} \chi (1)^m/\sum \limits _{\chi \in \textrm{Irr}_2(G)} \chi (1)^{m-1} < (1+2^{m-1})/(1+2^{m-2})$ for a positive integer m, which is the generalization of several recent results concerning the well-known Ito–Michler theorem.

1 Introduction

For a finite group G, let $\textrm{Irr}(G)$ be the set of all complex irreducible characters of G. We write

$$\begin{aligned} \textrm{Irr}_2(G):= \{\chi \in \textrm{Irr}(G)\mid \chi (1) = 1 \,\ \text {or}\,\ 2\mid \chi (1) \} \end{aligned}$$

and

$$\begin{aligned} \textrm{S}^m_2(G):={\sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^m}/{\sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^{m-1}}, \end{aligned}$$

where m is a positive integer. Ito–Michler theorem [3, 5] states that if a prime p does not divide the degree of every irreducible character of a finite group G, then G has a normal abelian Sylow p-subgroup. For the prime $p=2$, the $\textrm{S}^m_2(G)$-version of the Ito–Michler theorem asserts that if $\textrm{S}^m_2(G)=1$, then G has a normal abelian Sylow 2-subgroup. In this paper, we improve this for $p=2$.

Main theorem. Let G inite group with $\textrm{S}^m_2(G) < (1+2^{m-1})/(1+2^{m-2})$. Then, G has a normal Sylow 2-subgroup.

We note that $(1+2^{m-1})/(1+2^{m-2})$ is the exactly value of $\textrm{S}^m_2(G)$ with $G = S_3$, the symmetric group of degree 3, and $S_3$ does not have a normal Sylow 2-subgroup. For G, the nonabelian group of order 8 or the alternating group $A_4$ of degree 4, the value of $\textrm{S}^m_2(G)$ respectively is $(1+2^{m-2})/(1+2^{m-3})$ and 1, which are both less than $\textrm{S}^m_2(S_3)$, indeed G has a normal Sylow 2-subgroup.

For $m=1, 2$, we obtain the following corollary.

Corollary 1.1

Let G be a finite group. If $\textrm{S}^1_2(G)< 4/3$ or $\textrm{S}^2_2(G)< 3/2,$ then G has a normal Sylow 2-subgroup.

Proof

See [1, Theorem 1.1] and [6, Theorem A].$\square$

In the following of this paper, we prove the Main theorem in Sect. 2, and discuss other variations of the Main theorem in Sect. 3.

2 Proof of Main Theorem

We denote by $n_k(G)$ the number of irreducible complex characters of degree k of G. If N is a normal subgroup of G and $\theta \in \textrm{Irr}(N)$, then $\textrm{Irr}(G|\theta )$ denotes the set of irreducible characters of G that lie over $\theta$. We write $I_G(\theta )$ to denote the inertia subgroup of $\theta$ in G. For an M-invariant subgroup N, we write $M\ltimes N$ to denote a semidirect of M and N.

The first lemma is the following observation.

Lemma 2.1

Let N be a normal subgroup of a group G contained in the derived subgroup $G'$ of G. If $\textrm{S}^m_2(G) \leqslant 2,$ then $\textrm{S}^m_2(G/N) \leqslant \textrm{S}^m_2(G)$.

Proof

We write $R=\textrm{S}^m_2(G)$. Then, $1 \leqslant R \leqslant 2$ and

$$\begin{aligned} R = \frac{n_1(G)+\sum \limits _{2 \mid k}k^m n_k(G)}{n_1(G)+ \sum \limits _{2 \mid k}k^{m-1}n_k(G)}. \end{aligned}$$

It follows that

$$\begin{aligned} \sum \limits _{2\mid k}(k-R)k^{m-1}n_k(G)=(R-1)n_1(G). \end{aligned}$$

Since $\textrm{Irr}(G/N) \subseteq \textrm{Irr}(G)$ and $N \subseteq G'$, we have

$$\begin{aligned} \sum \limits _{2\mid k}(k-R)k^{m-1}n_k(G/N) \leqslant \sum \limits _{2\mid k}(k-R)k^{m-1}n_k(G) \end{aligned}$$

and $n_1(G)=n_1(G/N)$. Hence $\sum \limits _{2 \mid k}(k-R)k^{m-1}n_k(G/N) \leqslant (R-1)n_1(G/N)$, which is equivalent to $\textrm{S}^m_2(G/N) \leqslant R = \textrm{S}^m_2(G)$. $\square$

Lemma 2.2

For a normal subgroup N of G,  let $\chi \in \textrm{Irr}(G)-\textrm{Irr}(G/N)$ with degree 2. If $G/\textrm{ker}(\chi )$ is nonsolvable, then $\chi$ is a primitive character of $G/\textrm{ker}(\chi )$.

Proof

We write $\overline{G}=G/\textrm{ker}(\chi )$, and suppose that $\chi$ is not a primitive character of $\overline{G}$. Then, there is a proper subgroup $\overline{H}$ of $\overline{G}$ and some character $\phi$ of $\overline{H}$ such that $\chi = \phi ^{\overline{G}}$. It follows that $2=\chi (1)=|\overline{G}:\overline{H}|\phi (1)$, $|\overline{G}:\overline{H}|=2$ and $\phi (1)=1$. Hence, $\overline{H}$ is normal in $\overline{G}$ and all the irreducible constituents of $\chi _{\overline{H}}$ are linear. Then, $[\overline{H},\overline{H}] \leqslant {\ker }(\chi )=1$, and $\overline{H}$ is abelian. So $\overline{G}$ is solvable, a contradiction. $\square$

The following lemma implies that G in our main theorem is solvable.

Lemma 2.3

Let G be a finite group with $\textrm{S}^m_2(G) < (1 + 4^m)/(1 + 4^{m-1})$. Then, G is solvable.

Proof

Suppose that G is nonsolvable and let G be a counterexample of minimal order. Let A be a minimal nonsolvable normal subgroup of G, and let N be a minimal normal subgroup of G contained in A. Then, $N \subseteq A = A' \subseteq G'$. If in addition $[A, \textrm{Rad}(A)] > 1$, then we can choose $N \subseteq [A, \textrm{Rad}(A)]$, where $\textrm{Rad}(A)$ is the solvable radical (i.e., the unique largest solvable normal subgroup) of A. We now discuss the following two cases: N is abelian or not.

(1). N is abelian.

Then, G/N is nonsolvable. Since $|G/N|<|G|$, we have $\textrm{S}^m_2(G) < (1+4^m)/(1+4^{m-1}) \leqslant \textrm{S}^m_2(G/N)$. Hence, we obtain

$$\begin{aligned} \sum \limits _{2\mid k, k\geqslant 4}(k-1+4^{m-1}k-4^m)k^{m-1}n_k(G) < 3\cdot 4^{m-1}n_1(G) + 2^{m-1}(2\cdot 4^{m-1}-1)n_2(G) \end{aligned}$$

and

$$\begin{aligned}{} & {} 3\cdot 4^{m-1}n_1(G/N) + 2^{m-1}(2\cdot 4^{m-1}-1)n_2(G/N) \\{} & {} \quad \leqslant \sum \limits _{2\mid k, k \geqslant 4}(k-1+4^{m-1}k-4^m)k^{m-1}n_k(G/N). \end{aligned}$$

Since $n_1(G/N)=n_1(G)$ and $n_k(G/N) \leqslant n_k(G)$ for $k \geqslant 2$, we obtain $n_2(G/N) < n_2(G)$. Now there is $\chi \in \textrm{Irr}(G)-\textrm{Irr}(G/N)$ such that $\chi (1)=2$. Since $N \nsubseteq \textrm{ker}(\chi )$, we know that $A\textrm{ker}(\chi )/\textrm{ker}(\chi )$ is a nontrivial subgroup of $G/\textrm{ker}(\chi )$, and thus $G/\textrm{ker}(\chi )$ is nonsolvable. Now by Lemma 2.2, we know that $G/\textrm{ker}(\chi )$ is a nonsolvable primitive linear group of degree 2. Write $C/\textrm{ker}(\chi )=\textrm{Z}(G/\textrm{ker}(\chi ))$. According to the proof in [6, Theorem 3.1], we know that $G=AC$ is a central product, $G/C\cong A_5$, $N=A\cap C \cong \mathbb {Z}_2$, $A \cong SL(2,5)$, $n_1(G)=n_1(C/N)$, $n_2(G)=n_2(C/N)+2n_1(G)$, $n_4(G) \geqslant 2n_1(G)$, and $n_6(G) \geqslant n_1(G) + 2n_2(C/N)$. Now

$$\begin{aligned}&\sum \limits _{2\mid k, k\geqslant 4}(k-1+4^{m-1}k-4^m)k^{m-1}n_k(G)\\&\quad \geqslant 3\cdot 4^{m-1}n_4(G) + (5+2\cdot 4^{m-1})\cdot 6^{m-1}n_6(G)\\&\quad \geqslant 6\cdot 4^{m-1}n_1(G) + (5+2\cdot 4^{m-1})6^{m-1}\cdot (n_1(G)+2n_2(C/N))\\&\quad =(6\cdot 4^{m-1}+ 5\cdot 6^{m-1}+2\cdot 24^{m-1})n_1(G) + (10+4^m)6^{m-1}n_2(C/N) \\&\quad =(6\cdot 4^{m-1}+ 5\cdot 6^{m-1}+2\cdot 24^{m-1})n_1(G) + (10+4^m)6^{m-1}(n_2(G)-2n_1(G)) \\&\quad =(6\cdot 4^{m-1}- 15\cdot 6^{m-1} -6 \cdot 24^{m-1})n_1(G) + (10+4^m)6^{m-1}n_2(G) \\&\quad =(6\cdot 4^{m-1}- 15\cdot 6^{m-1} -6 \cdot 24^{m-1})n_1(G) + [(10+4^m)6^{m-1}\\&\qquad -2^{m-1}(2\cdot 4^{m-1}-1)]n_2(G)+2^{m-1}(2\cdot 4^{m-1}-1)n_2(G)\\&\quad \geqslant (6\cdot 4^{m-1}- 15\cdot 6^{m-1} -6 \cdot 24^{m-1})n_1(G)+ 2[(10+4^m)6^{m-1} \\&\qquad -2^{m-1}(2\cdot 4^{m-1}-1)]n_1(G)+2^{m-1}(2\cdot 4^{m-1}-1)n_2(G)\\&\quad =(5\cdot 6^{m-1} + 2\cdot 24^{m-1} + 6 \cdot 4^{m-1} -4\cdot 8^{m-1}+ 2^m)n_1(G) \\&\qquad + 2^{m-1}(2\cdot 4^{m-1}-1)n_2(G) \\&\quad \geqslant 3\cdot 4^{m-1}n_1(G)+2^{m-1}(2\cdot 4^{m-1}-1)n_2(G). \end{aligned}$$

Hence, $\textrm{S}^m_2(G) \geqslant (1+4^m)/(1+4^{m-1})$, a contradiction.

(2). N is not abelian.

Assume that $N \ncong A_5$. By [1, Theorem 2.2], there exists $\phi \in \textrm{Irr}(N)$ of even degree such that $\phi (1) \geqslant 8$ and $\phi$ is extendible to $I:= I_G(\phi )$. By [1, Proposition 2.3], we have $n_1(G) \leqslant n_d(G)|G: I|$ and $n_2(G) \leqslant n_{2d}(G)|G: I| + \frac{1}{2}n_d(G)|G: I|$, where $d=\phi (1)|G: I| \geqslant 8|G: I|$. Now

$$\begin{aligned}&3 \cdot 4^{m-1}n_1(G) + 2^{m-1}(2\cdot 4^{m-1}-1)n_2(G) \\&\quad \leqslant 3\cdot 4^{m-1} n_d(G)|G: I| + 2^{m-1}(2 \cdot 4^{m-1}-1)n_{2d}(G)|G: I| \\&\qquad + 2^{m-2}(2\cdot 4^{m-1}-1)n_d(G)|G: I|\\&\quad = (3\cdot 4^{m-1} + 8^{m-1}-2^{m-2})n_d(G)|G:I| + 2^{m-1}(2 \cdot 4^{m-1}-1)n_{2d}(G)|G:I|\\&\quad \leqslant (3\cdot 4^{m-1} + 8^{m-1}-2^{m-2})n_d(G)\frac{d}{8} + 2^{m-1}(2 \cdot 4^{m-1}-1)n_{2d}(G)\frac{d}{8}\\&\quad = \left( \frac{3}{2}\cdot 4^{m-2} + 8^{m-2}-2^{m-5}\right) dn_d(G) + 2^{m-4}(2 \cdot 4^{m-1}-1)dn_{2d}(G)\\&\quad \leqslant (d-1 + 4^{m-1}\cdot d-4^m)d^{m-1}n_d(G) + (2d-1+4^{m-1}\cdot 2d -4^m)(2d)^{m-1}n_{2d}(G) \\&\quad \leqslant \sum \limits _{2\mid k, k\geqslant 4}(k-1+4^{m-1}k-4^m)k^{m-1}n_k(G), \end{aligned}$$

so we have $\textrm{S}^m_2(G) \geqslant (1+4^m)/(1+4^{m-1})$, a contradiction.

Assume that $N \cong A_5$. Since the irreducible character of N of degree 4 is extendible to G, by [1, Proposition 2.3] again, we have $n_1(G) \leqslant n_4(G)$ and $n_2(G) \leqslant n_8(G)$. Now

$$\begin{aligned}&3 \cdot 4^{m-1}n_1(G) + 2^{m-1}(2\cdot 4^{m-1}-1)n_2(G) \\&\quad \leqslant 3 \cdot 4^{m-1}n_4(G) + 2^{m-1}(2\cdot 4^{m-1}-1)n_8(G) \\ {}&\quad \leqslant 3 \cdot 4^{m-1}n_4(G) + (7+4^m)8^{m-1}n_8(G)\\&\leqslant \sum \limits _{2\mid k, k\geqslant 4}(k-1+4^{m-1}k-4^m)k^{m-1}n_k(G), \end{aligned}$$

and thus $\textrm{S}^m_2(G) \geqslant (1+4^m)/(1+4^{m-1})$. This contradiction completes the proof.$\square$

Lemma 2.4

Let $G=M\ltimes N$. If $\lambda$ is a G-invariant linear character of N,  then $\lambda$ is extendible to G.

Proof

Clearly, we may assume that $\textrm{ker}(\lambda )=1$. Then, N is cyclic. Let $\chi \in \textrm{Irr}(\lambda ^G)$, then $\chi _N=\chi (1)\lambda$. Hence, $N \subseteq \textrm{Z}(\chi )=\{g \in G\big | |\chi (g)|=\chi (1)\}$ and $N\textrm{ker}(\chi )/\textrm{ker}(\chi ) \subseteq \textrm{Z}(\chi )/\textrm{ker}(\chi ) = \textrm{Z}(G/\textrm{ker}(\chi ))$. Then, $[G, N] \subseteq \textrm{ker}(\chi )$. It follows that $[G, N] \subseteq N \cap {\ker }(\chi ) = \textrm{ker}(\lambda ) = 1$. Hence, $N \subseteq \textrm{Z}(G)$. Now $G = M \times N$. Let $\psi =1_m \times \lambda$. Then, $\psi \in \textrm{Irr}(G)$ and $\psi _N=\lambda$. So $\lambda$ is extendible to G. $\square$

The following lemma is important in the proof of our Main theorem.

Lemma 2.5

Let $G=M\ltimes N,$ where $N \leqslant G'$ is an abelian group. Assume that no nontrivial irreducible character of N is invariant under M. If $\textrm{S}^m_2(G) < (1+2^{m-1})/(1+2^{m-2}),$ then there is no orbit of even size in the action of M on the set of irreducible characters of N.

Proof

Let $\{\theta _0=1_N, \theta _1, \cdots , \theta _t\}$ be a set of representatives of M-orbits on $\textrm{Irr}(N)$. Let $I_i= I_G(\theta _i)$ for $i \in [1, t]$. By hypothesis, we have $I_i < G$ for $i \geqslant 1$. Suppose that there is some orbit of even size in the action of M on $\textrm{Irr}(N)$. Then, we can find an integer k such that $2\mid |G:I_k|$. For $0 \leqslant i \leqslant t$, we set $n_{i,1}=n_1(I_i/N)$ and $\textrm{T}_{i, m} = \sum \limits _{\lambda \in \textrm{Irr}(I_i/N), 2\mid \lambda (1)}\lambda (1)^m$.

By Lemma 2.4, every $\theta _i$ has an extension $\psi$ to $I_i$. By Gallagher theorem [2, Corollary 6.17], we have bijections $\lambda \mapsto \lambda \psi _i$ from $\textrm{Irr}(I_i/N)$ to $\textrm{Irr}(I_i|\theta _i)$. By Clifford correspondence, we have a bijections $\lambda \psi _i \mapsto (\lambda \psi _i)^G$ from $\textrm{Irr}(I_i|\theta _i)$ to $\textrm{Irr}(G|\theta _i)$. Observe that $(\lambda \theta _i)^G(1)=|G:I_i|\lambda (1)$ is even if and only if $|G:I_i|$ is even or $|G: I_i|$ is odd and $\lambda (1)$ is even. Then,

$$\begin{aligned} \sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^{m-1}&= n_1(G/N) + \sum \limits _{2\mid |G:I_i|}|G:I_i|^{m-1}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m-1}\\&\quad + \sum \limits _{2 \not \mid |G:I_i|}|G:I_i|^{m-1}\textrm{T}_{i, m-1} \end{aligned}$$

and

$$\begin{aligned}&\sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^{m} = n_1(G/N) + \sum \limits _{2\mid |G:I_i|}|G:I_i|^{m}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m} + \sum \limits _{2 \not \mid |G:I_i|}|G:I_i|^{m}\textrm{T}_{i,m}\\&\quad \geqslant n_1(G/N) + \sum \limits _{2\mid |G:I_i|}|G:I_i|^{m}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m-1} + \sum \limits _{2 \not \mid |G:I_i|}|G:I_i|^{m}\textrm{T}_{i, m-1}. \end{aligned}$$

From $\sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^{m}=\textrm{S}^m_2(G)\sum \limits _{\chi \in \textrm{Irr}_2(G)}\chi (1)^{m-1}$, we have

$$\begin{aligned}&\quad \textrm{S}^m_2(G)(n_1(G/N) + \sum \limits _{2\mid |G:I_i|}|G:I_i|^{m-1}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m-1} + \sum \limits _{2 \not \mid |G:I_i|}|G:I_i|^{m-1}\textrm{T}_{i, m-1})\\&\geqslant n_1(G/N) + \sum \limits _{2\mid |G:I_i|}|G:I_i|^{m}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m-1} + \sum \limits _{2 \not \mid |G:I_i|}|G:I_i|^{m}\textrm{T}_{i, m-1}. \end{aligned}$$

Then,

$$\begin{aligned}&\sum \limits _{2 \mid |G:I_i|}(|G:I_i|-\textrm{S}^m_2(G))|G:I_i|^{m-1}\sum \limits _{\lambda \in \textrm{Irr}(I_i/N)}\lambda (1)^{m-1}\\&\qquad + \sum \limits _{2 \not \mid |G:I_i|}(|G:I_i|-\textrm{S}^m_2(G))|G:I_i|^{m-1}\textrm{T}_{i,m-1}\\&\quad \leqslant (\textrm{S}^m_2(G)-1)n_1(G/N). \end{aligned}$$

In particular, for k, we obtain

$$\begin{aligned} (|G:I_k|-\textrm{S}^m_2(G))|G:I_k|^{m-1}\sum \limits _{\lambda \in \textrm{Irr}(I_k/N)}\lambda (1)^{m-1} \leqslant (\textrm{S}^m_2(G)-1)n_1(G/N). \end{aligned}$$

Since $n_1(G/N)=|G/N:G'/N|$ and $n_{k,1}=n_1(I_k/N)=|I_k/N, (I_k/N)'|$, so $n_1(G/N) \leqslant |G:I_k|n_{k,1}$. In addition, $\sum \limits _{\lambda \in \textrm{Irr}(I_k/N)}\lambda (1)^{m-1} \geqslant n_1(I_k/N)$. Hence, we get that

$$\begin{aligned} (|G:I_k|-\textrm{S}^m_2(G))|G:I_k|^{m-1}n_1(I_k/N) \leqslant (\textrm{S}^m_2(G)-1)|G:I_k|n_1(I_k/N). \end{aligned}$$

It then follows that

$$\begin{aligned} \textrm{S}^m_2(G) \geqslant \frac{|G:I_k|^m + |G:I_k|}{|G:I_k|^{m-1} + |G:I_k|} \geqslant \frac{2^{m-1}+1}{2^{m-2}+1}, \end{aligned}$$

which is a contradiction.$\square$

We now prove our main theorem, which is restated.

Theorem 2.6

Let G be a finite group. Suppose that $\textrm{S}^m_2(G) < (1+2^{m-1})/(1+2^{m-2})$. Then, G has a normal Sylow 2-subgroup.

Proof

Suppose the theorem is false and let G be a counterexample of minimal order. By Lemma 2.3, G is solvable. Let N be a minimal normal subgroup of G contained in the derived subgroup $G'$ of G. Then, N is elementary abelian. By Lemma 2.1, $\textrm{S}^m_2(G/N) \leqslant \textrm{S}^m_2(G) < (1+2^{m-1})/(1+2^{m-2})$, and by the choice of G, G/N has a normal Sylow 2-subgroup R/N. If N is a 2-group, then R is a normal Sylow 2-subgroup of G, a contradiction. Hence, N is a $2'$-group. By the Schur–Zassenhaus theorem, $R = P \ltimes N$, where P is a Sylow 2-subgroup of R. By the Frattini argument, $G=RN_G(P) = NN_G(P)$. Clearly, $N \nsubseteq N_G(P)$. Since $N_G(P)\cap N < N$ and $N_G(P)\cap N \unlhd G= NN_G(P)$, we obtain $N_G(P)\cap N=1$, and thus $G=N_G(P)\ltimes N$. If $N\subseteq Z(G)$, then $R= P \times N$ and $P \unlhd G$, a contradiction. Hence, $N \nsubseteq Z(G)$. Then, $[N,G]=N$. Let $\lambda \in \textrm{Irr}(N)-1_N$. Assume that $\lambda$ is G-invariant. Then by Lemma 2.4, $\lambda$ has an extension $\chi$ to G. Now we have $\chi _N = \lambda$, $N \subseteq \textrm{ker}(\chi )$, and $\lambda = 1_N$, this contradiction implies that no nontrivial irreducible character of N is G-invariant. By Lemma 2.5, there is no orbit of even size in the action of P on N, that is, P acts trivially on N, hence $R=P \times N$ and $P \unlhd G$, this final contradiction completes the proof. $\square$

3 Variations of Main Theorem

Recall that a character $\chi \in \textrm{Irr}(G)$ is real if $\chi (g) \in \mathbb {R}$ for every element $g \in G$, and $\chi \in \textrm{Irr}(G)$ is strongly real if $\chi$ is afforded by a real representation, or equivalently, its Frobenius–Schur indicator $v_2(\chi )$ is 1. We denote by $n_{k,+}$ the number of irreducible strongly real characters of degree k of G. We write

$$\begin{aligned}&\textrm{Irr}_{2,\mathbb {R}}(G):= \{\chi \in \textrm{Irr}_2(G)\mid \chi \,\ \text {is real} \},\\&\textrm{S}^m_{2, \mathbb {R}}(G):={\sum \limits _{\chi \in \textrm{Irr}_{2,\mathbb {R}}(G)}\chi (1)^m}/{\sum \limits _{\chi \in \textrm{Irr}_{2,\mathbb {R}}(G)}\chi (1)^{m-1}},\\&\textrm{Irr}_{2,+}(G):= \{\chi \in \textrm{Irr}_2(G)\mid v_2(\chi )=1 \},\\&\textrm{S}^m_{2, +}(G):={\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^m}/{\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^{m-1}}. \end{aligned}$$

Similar to Lemma 2.1, we have the following observation.

Lemma 3.1

Let N be a normal subgroup of a group G such that $N \subseteq G'$. If $\textrm{S}^m_{2, +}(G) \leqslant 2$, then $\textrm{S}^m_{2,+}(G/N) \leqslant \textrm{S}^m_{2, +}(G)$.

Proof

Similar to Lemma 2.1. $\square$

For a normal subgroup with odd index, we have the next lemma.

Lemma 3.2

Let N be a normal subgroup of a group G with G/N odd order. If $\textrm{S}^m_{2, +}(G) \leqslant 2,$ then $\textrm{S}^m_{2, +}(N) \leqslant \textrm{S}^m_{2, +}(G)$.

Proof

Firstly, every strongly real linear character of G restricts to a strongly real linear character of N. Secondly, by [4, Lemma 2.1], every strongly real linear character of N lies under a unique strongly real linear character of G. Thus, we obtain $n_{1, +}(N) = n_{1, +}(G)$. In addition, from [4, Lemma 2.1], we also have

$$\begin{aligned} \sum \limits _{2\mid k, k\geqslant 2}n_{k, +}(G) \geqslant \sum \limits _{2\mid k, k\geqslant 2}n_{k, +}(N). \end{aligned}$$

Denote by $V = \textrm{S}^m_{2, +}(G)$. Then

$$\begin{aligned} V = \frac{n_{1,+}(G)+\sum \limits _{2\mid k}k^m n_{k, +}(G)}{n_{1,+}(G)+\sum \limits _{2\mid k}k^{m-1} n_{k, +}(G)}, \end{aligned}$$

and we obtain

$$\begin{aligned} (V-1)n_{1,+}(G) = \sum \limits _{2\mid k}(k^m-Vk^{m-1}) n_{k, +}(G). \end{aligned}$$

It then follows that

$$\begin{aligned} (V-1)n_{1,+}(N) = \sum \limits _{2\mid k}(k^m-Vk^{m-1}) n_{k, +}(G) \geqslant \sum \limits _{2\mid k}(k^m-Vk^{m-1}) n_{k, +}(N), \end{aligned}$$

which is equivalent to $\textrm{S}^m_{2, +}(N) \leqslant V = \textrm{S}^m_{2, +}(G)$, the proof is complete. $\square$

Now we give a variation of Main theorem for strongly real character.

Theorem 3.3

Let G be a finite group. Suppose that $\textrm{S}^m_{2,+}(G) < (1+2^{m-1})/(1+2^{m-2})$. Then, G has a normal Sylow $2-$subgroup.

Proof

Suppose that the theorem is not true, and let G be a counterexample of minimal order. Let N be a minimal normal subgroup of G contained in $G'$. Observe that

$$\begin{aligned} \frac{\sum \limits _{\chi \in \textrm{Irr}_{2,+}(G)}\chi (1)}{|\textrm{Irr}_{2,+}(G)|}&\leqslant \frac{n_{1,+}(G) + \sum \limits _{2 \mid k}k^2n_{k,+}(G)}{n_{1,+}(G) + \sum \limits _{2 \mid k}kn_{k,+}(G)}\\&\leqslant \cdots \\&\leqslant \frac{n_{1,+}(G) + \sum \limits _{2 \mid k}k^mn_{k,+}(G)}{n_{1,+}(G) + \sum \limits _{2 \mid k}k^{m-1}n_{k,+}(G)}\\&= \textrm{S}^m_{2,+}(G)\\&< \frac{1+2^{m-1}}{1+2^{m-2}} < 2, \end{aligned}$$

by [1, Theorem 5.1] and Lemma 3.1, we know that G is solvable, N is an elementary abelian group with odd order, G/N has a normal Sylow 2-subgroup R/N, and $G = R = P \ltimes N$, where $P \in \textrm{Syl}_2(G)$.

From the proof of [6, Theorem 5.1], we have that $n_{1, +}(G) \leqslant |N| - 1 \leqslant \chi _1(1) + \cdots + \chi _s(1)$, where $\chi _1, \cdots , \chi _s$ are the strongly real irreducible characters of G. Now

$$\begin{aligned} \frac{n_{1,+}(G)+\sum \limits _{i=1}^s\chi _i(1)^mn_{\chi _i(1), +}}{n_{1,+}(G)+\sum \limits _{i=1}^s\chi _i(1)^{m-1}n_{\chi _i(1), +}}&\geqslant \frac{n_{1,+}(G)+\sum \limits _{i=1}^s\chi _i(1)^m}{n_{1,+}(G)+\sum \limits _{i=1}^s\chi _i(1)^{m-1}}\\&\geqslant \frac{n_{1,+}(G)+ 2^{m-1}\sum \limits _{i=1}^s\chi _i(1)}{n_{1,+}(G)+2^{m-2}\sum \limits _{i=1}^s\chi _i(1)}\\&\geqslant \frac{n_{1,+}(G)+ 2^{m-1}n_{1,+}(G)}{n_{1,+}(G)+2^{m-2}n_{1,+}(G)}\\&= \frac{1+2^{m-1}}{1+2^{m-2}}. \end{aligned}$$

Then for any $\chi \in \textrm{Irr}_{2,+}(G)$ with $\chi (1) \geqslant 2 \geqslant (1+2^{m-1})/(1+2^{m-2})$, we have

$$\begin{aligned} \textrm{S}^m_{2,+}(G)=\frac{{\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^m}}{{\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^{m-1}}} \geqslant \frac{1+2^{m-1}}{1+2^{m-2}}, \end{aligned}$$

and this contradiction completes the proof. $\square$

Note that for any $\chi \in \textrm{Irr}_{2,\mathbb {R}}(G) - \textrm{Irr}_{2, +}(G)$, we have $\chi (1) \geqslant 2$. If $\textrm{S}^m_{2,+}(G) \leqslant 2$, then we have

$$\begin{aligned} \frac{\sum \limits _{\chi \in \textrm{Irr}_{2,+}(G)}\chi (1)^m}{\sum \limits _{\chi \in \textrm{Irr}_{2,+}(G)}\chi (1)^{m-1}} \leqslant \frac{\sum \limits _{\chi \in \textrm{Irr}_{2,\mathbb {R}}(G)}\chi (1)^m-\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^m}{\sum \limits _{\chi \in \textrm{Irr}_{2,\mathbb {R}}(G)}\chi (1)^{m-1}-\sum \limits _{\chi \in \textrm{Irr}_{2, +}(G)}\chi (1)^{m-1}}. \end{aligned}$$

Therefore, we obtain the following corollary, which is a variation of Main theorem for real characters.

Corollary 3.4

Let G be a finite group with $\textrm{S}^m_{2,\mathbb {R}}(G) < (1+2^{m-1})/(1+2^{m-2})$. Then, G has a normal Sylow $2-$subgroup.