Proof. This is [Reference Qian and Yang8, Lemma 2.2].

Proof. $|G:G^{c}| \leq |G:G^{c}||G^{c}\cap N|/|H^{c} \cap N|$ $= |G:G^{c} N||N:H^{c} \cap N| \leq |H:H^{c} N|| N:H^{c} \cap N|=|H:H^{c}|$.

Proof. This follows from the first half of the proof of [Reference Keller and Yang6, Theorem 2.3], where it reduces $G$ to be nilpotent. In fact, the proof shows that for any finite solvable group $G$, we have $|G:G^{c}| \leq |F:F^{c}|$ where $F$ is the Fitting subgroup of $G$.

Proof. We may assume $G$ is non-solvable, and we work by induction on $|G|$. Let $E$ be a minimal normal subgroup of $G$ and write $D/E=O_p(G/E)$. Assume that $G/D$ is non-solvable. By induction, $G/D$ has a solvable subgroup $B/D$ with $O_p(B/D)=1$ such that $|(G/D):(G/D)^{c}| \leq |(B/D):(B/D)^{c}|$. By Lemma 2.3, we have

\[ |G:G^{c}| \leq |B:B^{c}|. \]

Since $O_p(D)\leq O_p(G)=1$ and $O_p(B/D)=1$, we have $O_p(B)=1$. Suppose that $B$ is solvable. Then $B$ meets the requirements. Suppose that $B$ is non-solvable. By induction, $B$ has a solvable subgroup $H$ with $O_p(H)=1$ such that $|B:B^{c}| \leq |H:H^{c}|$. Then

\[ |G:G^{c}| \leq |B:B^{c}| \leq |H:H^{c}|, \]

so $H$ meets the requirements. Hence, we may assume that $G/D$ and $G/E$ are solvable for all minimal normal subgroups $E$ of $G$.

Since $G$ is non-solvable, it forces that $E$ is non-solvable and is the unique minimal normal subgroup of $G$. In particular, $C_G(E)=1$. By Lemma 2.4, $G/E$ has a nilpotent subgroup $U/E$ such that

\[ |G/G^{c}|=|G/G^{c} E| \leq |U/U^{c} E|=|U/U^{c}|. \]

Observe that $O_p(U)\leq C_G(E)=1$. Suppose that $U< G$, applying the inductive hypothesis to $U$, we get the required result. Hence, we may assume that $G/E$ is nilpotent.

Write $E=S_1\times \cdots \times S_t$, where $S_1,\, \ldots ,\, S_t$ are isomorphic non-abelian simple groups. Let us investigate $D={\bigcap} _{i=1}^{t} N_G(S_i)$. Since all $S_i$ are normal subgroups of $D$, by Lemma 2.2 there exists an abelian $p'$-subgroup $H$ of $D$ such that $2^{t}|D/E| \leq |H|$. Observe that $G$ acts on $\{S_1,\, \ldots ,\, S_t\}$ with the kernel $D$, it follows that $G/D$ is a nilpotent permutation group of degree $t$. As is well known (see, for example, [Reference Dixon2]), we have that $|G/D|\leq 2^{t-1}$. Now

\[ |G:G^{c}| \leq 2|G/E| \leq |H|=|H:H^{c}|, \]

so $H$ meets the requirements.

Proof. (1) Set $K=G\ltimes V$. It is clear that $F(K)$ is nilpotent, so we can write $F(K) = P \times Q$, where $P$ is a $p$-group and $Q$ is a $p'$-group. Note that both $P$ and $Q$ are normal in $K$, which implies that $PV/V$ is a normal $p$-subgroup of $K/V \cong G$ and $[Q,\,V] \leq Q \cap V = 1$. Now the assumptions that $O_p(G) = 1$ and ${{\mathbf {C}}}_G(V) = 1$ guarantee that $P \leq V$ and $Q = 1$, so $F(K) = P \leq V$. But the converse containment is clear, so $\varPhi (K) \leq F(K) = V$. If $\varPhi (K) = V$, then $K = GV = G \varPhi (K)$, which forces $K = G$ and hence $V = 1$, a trivial case. Thus, we have $F(K)=V$ and $\varPhi (K)< V$.

Set

\[ \overline{G}=G\varPhi(K)/\varPhi(K), \overline{V}=V/\varPhi(K), \overline{K}=K/\varPhi(K) =\overline{G} \ltimes \overline{V}. \]

Observe that $\overline {G}$ also acts faithfully on $\overline {V}$. We note this assertion is equivalent to saying that $G$ acts faithfully on $\overline {V}$. Let $C = {{\mathbf {C}}}_G(\overline {V})$, and take $A$ to be a $p'$-subgroup of $C$. Then $[V,\,A] \leq \varPhi (K)$. Since $A$ acts coprimely on $V$, we have $V = [V,\, A] {{\mathbf {C}}}_V(A) = \varPhi (K) {{\mathbf {C}}}_V(A)$ and thus $K = GV = \varPhi (K) G {{\mathbf {C}}}_V(A)$. This forces $GV = G{{\mathbf {C}}}_V(A)$ and hence $V = {{\mathbf {C}}}_V(A)$. Note that we are assuming that ${{\mathbf {C}}}_G(V) = 1$, so $A = 1$, which means that $C$ is a $p$-group. But $O_p(G) = 1$ by hypotheses, so $C = 1$, as required.

Assume that $\varPhi (K)>1$. By induction,

\[ |G:G^{c}|=|\overline{G}:\overline{G}^{c}|<|\overline{V}|<|V|, \]

and we are done. Therefore, we may assume that $\varPhi (K)=1$. By Proposition 2.5, we may assume that the group $G$ is solvable. Since $F(K) = V$, $V$ is a faithful and completely reducible $G$-module over a field of characteristic $p$ by Gaschütz's theorem. Now the result follows by [Reference Keller and Yang6, Theorem 1.1].

(2) Suppose that $V$ is a finite, faithful and completely reducible $G$-module. We work by induction on $|G|+|V|$. Assume that $V=V_1\oplus V_2$, where $V_1,\, V_2$ are non-trivial $G$-submodules of $V$. Let $C=C_G(V_1)$. Observe that $V_1$ is a faithfully and completely reducible $G/C$-module, while $V_2$ is a faithfully and completely reducible $C$-module (note that $C$ is normal in $G$). By induction,

\[ |G/C : (G/C)^{c}| \leq |V_1| \]

and

\[ |C: C^{c}| \leq |V_2|. \]

Thus, $|G:G^{c}| \leq |G/C : (G/C)^{c}| \cdot |C: C^{c}| < |V_1| |V_2|=|V|$, and we are done.

Hence, we may assume that $V$ is an irreducible $G$-module over a finite field of characteristic $p$. Clearly, $O_p(G)=1$ because $G$ acts faithfully on $V$. Now the required result follows by (1).