Proof. Assume that $n\geqslant 2p.$ Then $p<2p\leqslant n.$ By Bertrand’s postulate, there exists a prime, say $s,$ so that $p<s<2p.$ This contradicts the hypothesis that p is the largest prime less than or equal to $n.$ Hence, $n<2p.$

Now assume that $n\geqslant rq.$ Applying Bertrand’s postulate again, we see that ${q<p<2q}$ and so $q<p<2q<3q\leqslant rq\leqslant n.$ By [Reference El Bachraoui1, Theorem 1.3], there is a prime between $2q$ and $3q.$ This is a contradiction. Thus, $n<rq.$

Proof. Assume that $n\geqslant 7.$ Then there are three consecutive primes $r,q,p$ as defined in Proposition 3.1. Thus, $n<2p$ and $n<rq.$ Consider the subset $\Delta =\{r,p,q\}\subseteq V(G)=\pi (G).$ By Lemma 2.2(2), there exists $\chi \in \mathrm {Irr}(G)$ so that $pq\mid \mathrm {cod}\,\chi $ , $rq\mid \mathrm {cod}\,\chi $ or $rp\mid \mathrm {cod}\,\chi .$ It follows from Proposition 3.1 that $n\geqslant \mathrm { cod}\,\chi \geqslant \min \{pq,rq,rp\}>n.$ This is a contradiction. Thus, $n\leqslant 6.$ Similarly, by Lemma 2.2(2), $n\not = 5.$

Proof. (1) The first part follows from Lemma 2.1(2) immediately. Notice that N is nilpotent. Then $N=P_1\times P_2\times \cdots \times P_s,$ where $P_i$ is a Sylow $p_i$ -subgroup of $N.$ Let $1_{P_i}\not =\lambda _i\in \mathrm {Irr}(P_i)$ and set $\phi =\lambda _1\times \lambda _2\times \cdots \times \lambda _s.$ Then $\phi \in \mathrm {Irr}(N)$ and $ \mathrm {cod}\,\phi =\prod _{i=1}^{s}\mathrm {cod}\,\lambda _i$ with $p_i\mid \mathrm { cod}\,\lambda _i.$ Hence, by Lemma 2.1(2), $ \prod _{i=1}^{s}p_i\mid \mathrm {cod}\,\chi $ for some $\chi \in \mathrm {Irr}(G),$ as required.

(2) It is well known that $\mathrm {Irr}(G)=\mathrm {Irr}(G/N)\bigcup \{\phi ^G\mid 1_N\not =\phi \in \mathrm {Irr}(N)\}.$ Thus, the first part is true. Notice that $\phi ^G(1)=|G:N|\phi (1).$ Then

$$ \begin{align*} \displaystyle \mathrm{cod}(\phi^G)=\frac{|G:N||N|}{|G:N|\phi(1)|\mathrm{ker}\,\phi^G|}=\frac{|N|}{\phi(1)|\mathrm{ker}\,\phi^G|} \end{align*} $$

divides $|N|,$ as required.

Proof. We work by contradiction. Assume that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ By Lemma 2.2(1) and (2), $\pi (G)=\{2,3,5\}$ and $6\in \mathrm {Cod}(G).$ First we consider the case when $|\pi (N)|=2.$ Since N is nilpotent, it follows from Proposition 3.3(1) that $\pi (N)=\{2,3\}$ and N is a direct product of an elementary abelian 2-group and an elementary abelian 3-group. Notice that the complement H is a cyclic 5-group and $\mathrm {Cod}(H)\subseteq \mathrm {Cod}(G).$ Then H must be cyclic of order 5. Since there is $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=6,$ we have $\mathrm {ker}\,\phi ^G<N$ and so $G/\mathrm {ker}\,\phi ^G=N/\mathrm { ker}\phi ^G\rtimes H\mathrm {ker}\phi ^G/\mathrm {ker}\phi ^G\cong C_6\rtimes C_5\cong C_{30}.$ Hence, $30\in \mathrm {Cod}(G),$ a contradiction.

Assume now that $|\pi (N)|=1.$ By Proposition 3.3(2), $\pi (N)=\{5\}$ and so N is elementary abelian. Since there is $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=5,$ we have $\mathrm {ker}\,\phi ^G<N.$ Let $\overline {G}=G/\mathrm {ker}\,\phi ^G.$ Then $\overline {G}=\overline {N}\rtimes \overline {H}$ is a Frobenius group with kernel $\overline {N}\cong C_5$ and $\overline {H}\cong H$ . Let $\overline {Q}$ be a Sylow 3-subgroup of $\overline {H}.$ Then $\overline {Q}$ is cyclic of order 3. It follows that $\overline {N}\overline {Q}$ is a Frobenius group of order 15. This is a contradiction since such a group does not exist.

Both cases are impossible. The proof is completed.

Proof of Theorem 1.1.

We first introduce two basic facts.

1. (A) $\mathrm {cod}\,\chi> \chi (1)$ if $1_{G}\not = \chi \in \mathrm {Irr}(G)$ .

2. (B) If G is abelian, then $\mathrm {cod}\,\chi $ is equal to the order of $\chi $ in the group $\mathrm {Irr}(G)\cong G.$

There is nothing to prove when $n=1.$ Assume that $n\geqslant 2.$ Applying fact (A), together with $2\in \mathrm {Cod}(G),$ we see that there exists a linear character $\chi \in \mathrm {Irr}(G)$ such that $\mathrm {cod}\,\chi =2$ and hence $\chi \in \mathrm {Irr}(G/G').$ Then it follows from fact (B) that $2\mid |G:G'|.$

If $n=2,$ then by facts (A) and (B), G is an elementary abelian $2$ -group and (2) follows.

Assume that $n=3.$ Then by Lemma 2.2(1) and (3), $\pi (G)=\{2,3\}$ and G is a Frobenius group or a 2-Frobenius group. First suppose that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ Since $2\mid |G:G'|,$ it follows that H is a 2-group and N is a 3-group. By Proposition 3.3, $\mathrm {Cod}(N)=\{1,3\}$ and $\mathrm {Cod}(G/N)=\mathrm {Cod}(H)=\{1,2\}.$ Therefore, N is an elementary abelian 3-group and H is an elementary abelian 2-group. Notice that the complement H must be cyclic or a generalised quaternion group (see [Reference Grove2, Theorem 9.2.10]). Hence, H is cyclic of order 2. Since $6\not \in \mathrm { Cod}(G),$ we have $G'=N.$ To complete the proof of (3), we only need to show that G is not a 2-Frobenius group. Assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. It follows from Proposition 3.3 that $\mathrm {Cod}(G/N)=\mathrm {Cod}(M)=\{1,2,3\}.$ Similarly, $G/N\cong C_3^s\rtimes C_2$ and $M\cong C_3^t\rtimes C_2$ for some positive integers s and $t.$ This is a contradiction. Hence, (3) follows.

Now we show that $n\not =4.$ If $n=4,$ then $\pi (G)=\{2,3\}$ and G is a Frobenius group or a 2-Frobenius group. First assume that $G=N\rtimes H$ is a Frobenius group with kernel $N.$ Then by a proof similar to that above, N is an elementary abelian 3-group and H is a 2-group with $\mathrm {Cod}(H)=\{1,2,4\}.$ Together with the fact that the complement H must be cyclic or a generalised quaternion group, we have $H\cong C_4$ or $Q_8.$ Notice that there exists $\phi \in \mathrm {Irr}(N)$ so that $\mathrm {cod}(\phi ^G)=|N|/|\mathrm {ker}\,\phi ^G|=3.$ It is obvious that $\mathrm {ker}\,\phi ^G<N.$ Then $G/\mathrm {ker}\,\phi ^G=N/\mathrm {ker}\,\phi ^G\rtimes H\mathrm {ker}\,\phi ^G/\mathrm {ker}\,\phi ^G\cong C_3\rtimes C_4$ or $C_3\rtimes Q_8$ is a Frobenius group, which is a contradiction. Thus, G cannot be a Frobenius group. Assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. Since $G/N$ is Frobenius and $\mathrm {Cod}(G/N) \subseteq \mathrm {Cod}(G),$ we have $\mathrm {Cod}(G/N)=\{1,2,3\}.$ Similarly, $\mathrm {Cod}(M)=\{1,2,3\}.$ This cannot happen by statement (2) of this theorem. Hence, $n\not =4.$

By Proposition 3.2, it remains to show that $n\not =6.$ If $n=6,$ then $\pi (G)=\{2,3,5\}$ and G is a Frobenius group or a 2-Frobenius group. It follows by Proposition 3.4 that G is not a Frobenius group. We may assume that $G=\mathrm {Frob}_2(G,M,N)$ is a 2-Frobenius group. By Proposition 3.3(1) and (2), $\mathrm {Cod}(G/N) \subseteq \mathrm {Cod}(G)$ and $\mathrm {Cod}(M/N) \subseteq \mathrm {Cod}(M) \subseteq \mathrm { Cod}(G).$ Since both $G/N$ and M are Frobenius groups, it follows by Proposition 3.4 that $|\pi (G/N)|=|\pi (M)|=2.$ Write $\overline {G}=G/N$ and then $\overline {G}=\overline {M}\rtimes \overline {K},$ where $\overline {K}$ is the Frobenius complement. First consider the case when $\pi (\overline {K})\not =\{2\}.$ As $\overline {K}$ is cyclic and $\mathrm {Cod}(\overline {K}) \subseteq \mathrm {Cod}(G),$ we have $G'\leqslant M$ and $\overline {K}$ is cyclic of order 3 or 5. Notice that $2\mid |G:G'|.$ Then $\overline {K}\cong C_3$ and $\overline {M}$ is a 2-group. Since M is Frobenius and $\overline {M}$ is isomorphic to its complement, it follows that $\overline {M}$ is cyclic or a generalised quaternion group and hence $\overline {M}\cong C_2, C_4$ or $Q_8.$ But $\overline {G}\cong \overline {M}\rtimes C_3$ is Frobenius, which is impossible. Assume now that $\pi (\overline {K})=\{2\}.$ It is obvious that $\pi (M)=\{3,5\}.$ Let $M=N\rtimes H$ be a Frobenius group with kernel $N.$ By a similar proof, there exists $\phi \in \mathrm {Irr}(N)$ so that $G/\mathrm { ker}\phi ^G\cong N/\mathrm {ker}\phi ^G\rtimes H\cong C_{15}.$ It follows that $15\in \mathrm {Cod}(G),$ a contradiction. Hence, $n\not =6.$ The proof is completed.