Proof (1)By Zsigmondy’s prime theorem (see [Reference Huppert and Blackburn2, Theorem IX.8.3]) and the condition that $r>3$ , there exists a prime q dividing $p^{2r}-1$ but not $p^{i}-1$ for all ${i=1,\ldots , 2r-1}$ . It follows that $2r$ is the order of p modulo q, establishing (1).

(2) Write $S=\operatorname {\mathrm {Sp}}(2r,p)$ for $p>2$ and $S=\text {O}_{-}(2r,2)$ for $p=2$ . Let $\Lambda $ denote the subgroup of $\operatorname {\mathrm {Aut}}(P)$ consisting of those automorphisms of P acting trivially on $Z(P)$ . By the construction of P, it is well known that $\Lambda /\kern-1.5pt\operatorname{\mathrm {Inn}}(P)$ is isomorphic to S (see, for example, [Reference Winter10, Theorem 1]). Since the orders of $C_{q}\rtimes C_{r}$ and $\operatorname {\mathrm {Inn}}(P)$ are relatively prime (as p does not divide $qr$ ), it suffices to show that S contains a subgroup isomorphic to $C_{q}\rtimes C_{r}$ by the Schur–Zassenhaus theorem.

To do this, we consider the finite field ${\mathbb {F}}_{p^{2r}}$ of $p^{2r}$ elements and let $V={\mathbb {F}}_{p^{2r}}$ be the vector space over ${\mathbb {F}}_{p}$ of dimension $2r$ . We need to construct a nonsingular symplectic form $\langle \;, \rangle $ on V for $p>2$ and a nonsingular quadratic form Q on V for $p=2$ .

Fix an element $a\in {\mathbb {F}}_{p^{2}}-{\mathbb {F}}_{p}$ . Then $a\not \in {\mathbb {F}}_{p^{r}}$ since r is odd. Let $\operatorname {\mathrm {tr}}:{\mathbb {F}}_{p^{r}}\to {\mathbb {F}}_{p}$ be the trace map. It is easy to see that

$$ \begin{align*}\langle v,w\rangle=\operatorname{\mathrm{tr}}((a-a^{p^{r}})(vw^{p^{r}}-v^{p^{r}}w)),\quad v,w\in V,\end{align*} $$

defines a nonsingular symplectic form on V and the corresponding symplectic group is S for $p>2$ (see [Reference Manz and Wolf5, Example 8.5]). When $p=2$ ,

$$ \begin{align*}Q(v)=\operatorname{\mathrm{tr}}(v^{1+p^{r}}),\quad v\in V,\end{align*} $$

defines a nonsingular quadratic form on V and the corresponding orthogonal group is $S=\text {O}_{-}(2r,2)$ .

Now, let $\Gamma _{0}(V)=\{v\mapsto av\ |\ 0\neq a\in V\}$ , consisting of multiplications, and let $\sigma :v\mapsto v^{p}$ be the Frobenius field automorphism of ${\mathbb {F}}_{p^{2r}}$ . Then $\sigma $ acts naturally on $\Gamma _{0}(V)$ , and we consider the corresponding semi-direct product $\Gamma _{0}(V)\rtimes \langle \sigma \rangle $ . Observe that $\Gamma _{0}(V)$ is cyclic of order $p^{2r}-1$ and the order of $\sigma $ is $2r$ . Recall that the prime q divides $p^{r}+1$ and let C be the unique subgroup of order q in $\Gamma _{0}(V)$ . It is clear that C is invariant under $\sigma $ , and furthermore, by elementary Galois theory, the fixed point of $\sigma ^{2}$ in C is trivial since q does not divide $p^{2}-1$ . So we can form the semi-direct product $C\rtimes \langle \sigma ^{2}\rangle $ , which is clearly isomorphic to $C_{q}\rtimes C_{r}$ . What remains is to show that $C\rtimes \langle \sigma ^{2}\rangle \le S$ .

A simple calculation shows that both the symplectic form $\langle \; ,\;\rangle $ and the quadratic form Q defined above are preserved by the map on V induced by multiplication by an element of order $p^{r}+1$ , and thus the unique cyclic subgroup of $\Gamma _{0}(V)$ of order $p^{r}+1$ must be contained in S. In particular, we have $C\le S$ . To prove $\sigma ^{2}\in S$ , we distinguish two cases. For $p=2$ , since the Galois group of ${\mathbb {F}}_{p^{r}}/{\mathbb {F}}_{p}$ can be identified with $\langle \sigma ^{2}\rangle $ (as r is odd), we conclude that $\sigma ^{2}$ must preserve the trace map from ${\mathbb {F}}_{p^{r}}$ to ${\mathbb {F}}_{p}$ and hence lies in S. For $p>2$ , we need to establish that $\langle v^{\sigma ^{2}},w^{\sigma ^{2}}\rangle =\langle v,w\rangle $ for all $v,w\in V$ . Let $b=a-a^{p^{r}}$ and $x=vw^{p^{r}}-v^{p^{r}}w$ . It suffices to prove $\operatorname {\mathrm {tr}}(bx^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ . Since $(bx)^{p^{r}}=(-b)(-x)=bx$ , we have $bx\in {\mathbb {F}}_{p^{r}}$ . It follows that $\operatorname {\mathrm {tr}}((bx)^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ . By the choice of a, we know that $b\in {\mathbb {F}}_{p^{2}}$ and hence b is fixed by $\sigma ^{2}$ . Thus, $\operatorname {\mathrm {tr}}(bx^{\sigma ^{2}})=\operatorname {\mathrm {tr}}(bx)$ and $\sigma ^{2}\in S$ , as required.

Proof. Note that $Z(P)$ is cyclic of order p, and thus we can choose a faithful linear character $\lambda $ of $Z(P)$ . Then it is well known that $\lambda ^{P}=p^{r}\theta $ for some $\theta \in \operatorname {\mathrm {Irr}}(P)$ . Since A acts trivially on $Z(P)$ , it fixes $\lambda $ and hence $\theta $ is A-invariant. It follows that $\theta $ extends to some $\chi \in \operatorname {\mathrm {Irr}}(G)$ by Gallagher’s theorem (see [Reference Isaacs3, Corollary 6.28]), so that $\chi (1)=\theta (1)=p^{r}$ . Also, let B be the unique subgroup of A of order q. Then $P/Z(P)$ is irreducible as an ${\mathbb {F}}_{p}[B]$ -module because $|P/Z(P)|=p^{2r}$ and the order of p modulo q is exactly $2r$ . From this, we conclude that $Z(P)$ is the unique minimal normal subgroup of G, and since $\theta $ is faithful, we have $\operatorname {\mathrm {Ker}}\chi \cap P=\operatorname {\mathrm {Ker}}\theta =1$ . Obviously, P is the Fitting subgroup of the solvable group G, and thus $\operatorname {\mathrm {Ker}}\chi $ contains no minimal normal subgroup of G, which forces $\operatorname {\mathrm {Ker}}\chi =1$ . Therefore, we have $\chi ^{c}(1)=|G|/\chi (1)=p^{r+1}qr$ .

Finally, if G has an element of order divisible by all the primes $p, q, r$ , then A contains an element of order $qr$ and thus A is cyclic, which is not the case by the choice of primes $q, r$ . The proof is now complete.