Proof. Given $j\geq 1$ , $m\in \mathbb {F}_q$ and $(a_1,\ldots ,a_j)\in \mathbb {F}_q^j$ with $a_1,\ldots ,a_j \neq 0$ , consider the equation $\sum _{i=1}^j a_ix_i=m$ . Call a solution $(x_1, \ldots , x_j) \in \mathbb {F}_q^j$ to this equation ‘good’ if $x_1, \ldots , x_j \neq 0$ . First, notice that, because $\mathbb {F}_q$ is a field, the number of good solutions does not depend on $a_1,\ldots ,a_j$ . Similarly, the number of good solutions depends only on whether $m=0$ or not. We may thus define $D_m(j)$ to be the number of good solutions to this and, hence, every equation of the given form. For $j\geq 2$ ,

$$ \begin{align*} D_0(j) = (q-1)D_{-a_jx_j}(j-1) = (q-1)D_1(j-1). \end{align*} $$

It follows that, for $j\geq 2$ ,

$$ \begin{align*} D_1(j) = (q-2)D_1(j-1)+ D_0(j-1). \end{align*} $$

Combining these two equations yields, for $j \geq 3$ ,

$$ \begin{align*} D_1(j) = (q-2)D_1(j-1)+(q-1)D_1(j-2). \end{align*} $$

This is a second-order linear difference equation for $D_1(j)$ ; the initial conditions $D_1(1)=1$ and $D_1(2)=q-2$ yield

$$ \begin{align*} D_1(j) = \frac{(q-1)^{j}-(-1)^j}{q} \quad \text{and hence} \quad D_0(j) = (q-1)\frac{(q-1)^{j-1}-(-1)^{j-1}}{q} \end{align*} $$

for all $j\geq 1$ . The first assertion of the lemma follows upon observing that the number of $(u_1,\ldots ,u_d)\in U$ such that $u_1,\ldots ,u_d \neq 0$ is equal to $(q-1)^{d-k}D_0(k)$ . Since $a\neq 0$ , we have $k\geq 1$ and the second assertion is then easily verified.

Proof. The assertion that $d \geq 2$ is immediate, because condition (1) is never satisfied if $d\leq 1$ . Let $W_1\in \mathcal {W}$ . If $W_1\neq \{0\}$ , then, by condition (2), there exists $W_2\in \mathcal {W}$ such that $W_1\nleq W_2$ . It follows that $W_1+W_2=V$ and

$$ \begin{align*}\dim(W_1\cap W_2)=\dim(W_1)+\dim(W_2)-\dim(V)=2(d-1)-d=d-2,\end{align*} $$

so $W_1\cap W_2$ has codimension $2$ in V. If $W_1\cap W_2\neq \{0\}$ , then, by condition (2), there exists $W_3\in \mathcal {W}$ such that $W_1\cap W_2\nleq W_3$ . Since $W_3$ has codimension $1$ in V, this implies that $W_3+(W_1\cap W_2)=V$ and, by a similar calculation as earlier, ${W_1\cap W_2\cap W_3}$ has codimension $3$ in V. Repeating this procedure, we find that $\mathcal {W}$ contains d subspaces $W_1,\ldots ,W_d$ with $\bigcap _{i\in \{1,\ldots ,d\}} W_i = \{0\}$ . By choosing an appropriate basis for V, we can assume that $W_i$ is defined by the linear equation $x_i=0$ . Note that $\bigcup _{i\in \{1,\ldots ,d\} } W_i$ consists of the vectors in V with at least one coordinate equal to  $0$ , so $|\!\bigcup _{i\in \{1,\ldots ,d\} } W_i|=q^d-(q-1)^d$ . This leaves $(q-1)^d$ elements of V to ‘cover’ by adjoining further subspaces from $\mathcal {W}$ , to satisfy condition (1). Lemma 2.2 implies that adjoining one further subspace from $\mathcal {W}$ covers at most $(q-1)^{d-1}$ further elements. We must therefore adjoin at least another $q-1$ subspaces to satisfy condition (1), so $|\mathcal {W}| \geq d + (q-1)$ .

Proof. Let $\varphi _i$ be the natural projection from K to $T_i$ and let

$$ \begin{align*}K_i := \mathrm{ker}\,\varphi_i=T_1\times\cdots \times T_{i-1}\times 1 \times T_{i+1}\times\cdots\times T_n.\end{align*} $$

Note that $\operatorname {\mathrm {Inn}}(T_1)$ is the unique minimal normal subgroup of $G_{T_1}^{T_1}$ , because $\operatorname {\mathrm {Inn}}(T_1)\leq G_{T_1}^{T_1}\leq \operatorname {\mathrm {Aut}}(T_1)$ . Since $G\cap K$ is a normal subgroup of $G_{T_1}$ , it follows that $(G\cap K)^{T_1}$ is normal in $G_{T_1}^{T_1}$ and so either $(G\cap K)^{T_1}=1$ or $\operatorname {\mathrm {Inn}}(T_1)\leq (G\cap K)^{T_1}$ . If $(G\cap K)^{T_1}=1$ , then $G\cap K\leq K_1$ . Since G acts transitively on $\{T_1,\ldots ,T_n\}$ , this implies that ${G\cap K \leq K_i}$ for every $i\in \{1,\ldots ,n\}$ , and thus $G\cap K=1$ , as required. We may thus assume that $\operatorname {\mathrm {Inn}}(T_1)\leq (G\cap K)^{T_1}$ , that is, the restriction $\varphi _1:G\cap K\to T_1$ is surjective and thus $(G\cap K)/(G\cap K_1)\cong T$ .

For $i\in \{1,\ldots ,n\}$ , let $G_i=G\cap K_i$ . Define an equivalence relation $\sim $ on $\{T_1,\ldots ,T_n\}$ by $T_i \sim T_j$ if and only if $G_i = G_j$ . This equivalence relation is G-invariant so, by condition (2), the induced partition is either the universal one or the partition into singletons. We now consider these two cases separately.

Case 1: $ {G_i = G_j}$  for all  ${i,j\in \{1,\ldots ,n\}}$ . Since $K_1\cap \cdots \cap K_n=1$ , we have $G_1=1$ , so the restriction $\varphi _1:G\cap K\to T_1$ is injective. We saw earlier that it is surjective, and hence it is an isomorphism and $G\cap K\cong T$ , as required.

Case 2: $ {G_i \neq G_j}$  for all  $ {i,j}$  with  ${i\neq j}$ . We saw earlier that $(G\cap K)/G_1\cong T$ . Together with condition (2), this implies that $(G\cap K)/G_i\cong T$ for all $i\in \{1,\ldots ,n\}$ . We now proceed by induction. Suppose that $1\leq m<n$ is such that ${(G\cap K)/(G_1 \cap \cdots \cap G_m) \cong T^m}$ . Let $N=(G_1 \cap \cdots \cap G_m) G_{m+1}$ . Note that $G_1 \cap \cdots \cap G_m$ and $G_{m+1}$ are both normal subgroups of $G\cap K$ , and hence so is N. Since $(G\cap K)/G_{m+1}\cong T$ is simple, we must have $N=G\cap K$ or $N = G_{m+1}$ . If $N = G_{m+1}$ , then $G_1 \cap \cdots \cap G_m \leq G_{m+1}$ . Since $(G\cap K) /(G_1 \cap \cdots \cap G_m) \cong T^m$ has precisely m normal subgroups of index $|T|$ , it follows that $G \cap K$ has precisely m normal subgroups of index $|T|$ containing $G_1 \cap \cdots \cap G_m$ . The latter are precisely $G_1,\ldots ,G_m$ , so we must have $G_{m+1}=G_i$ for some $i \in \{1,\ldots ,m\}$ , which is a contradiction. Therefore, $N = G \cap K$ . It follows that

$$ \begin{align*} &(G\cap K)/(G_1 \cap \cdots \cap G_{m+1}) \\ & \quad \cong (G\cap K)/(G_1 \cap \cdots \cap G_m)\times (G\cap K)/G_{m+1} \cong T^m \times T \cong T^{m+1}. \end{align*} $$

This completes the induction and yields $(G\cap K)/(G_1 \cap \cdots \cap G_n) \cong T^n$ . Since $G \cap K \leq K \cong T^n$ , it follows that $G \cap K=K$ and hence $K \leq G$ , as required.

Proof. For each $i \in \{1,\ldots ,k\}$ , let $K_i$ be the kernel of the action homomorphism from G to $\operatorname {\mathrm {Sym}}(\Omega _i)$ . We have

$$ \begin{align*} \operatorname{\mathrm{\nu}}(G)=\frac{|\!\bigcup_{\omega \in \Omega} G_\omega|}{|G|} & \leq \sum_{i\in\{1,\ldots,k\}} \frac{|\!\bigcup_{\omega \in \Omega_i} G_\omega|}{|G|} = \sum_{i\in\{1,\ldots,k\}} \frac{|\!\bigcup_{\omega \in \Omega_i} G_\omega^{\Omega_i}|\cdot |K_i|}{|G^{\Omega_i}|\cdot |K_i|}\\ &=\sum_{i\in\{1,\ldots,k\}} \frac{|\!\bigcup_{\omega \in \Omega_i} G_\omega^{\Omega_i}|}{|G^{\Omega_i}|}=\sum_{i\in\{1,\ldots,k\}} \operatorname{\mathrm{\nu}}(G^{\Omega_i}), \end{align*} $$

as required.

Proof. Note that the transitive groups of degree at most $30$ are known [Reference Hulpke6] and readily accessible in Magma [Reference Bosma, Cannon and Playoust2]. The proof is supported by the Magma code available on the first author’s GitHub [Reference Lee8]; the process that we now describe is carried out for each degree $n \in \{2,\ldots ,30\}$ using the function HalfTransitiveDerangements. If G has no derangements, then [Reference Ellis and Harper5, Corollary 4] implies that there exist transitive but imprimitive permutation groups $G_1$ and $G_2$ of degree n such that $G^{\Omega _i} \cong G_1$ for ${i \in \{1,2\}}$ . We begin by constructing each transitive but imprimitive permutation group H of degree n and directly counting the number of nonderangements in H (by computing the support of a representative of each conjugacy class of H). By Lemma 2.6, the proportion of nonderangements in G is bounded above by the sum of the proportions of the nonderangements in $G_1$ and $G_2$ , which we compute for each pair $(G_1,G_2)$ using the aforementioned data. If this proportion is less than $1$ , then G certainly has a derangement, so we discard the pair $(G_1,G_2)$ . For each remaining pair $(G_1,G_2)$ , we know that G must be a subdirect product of $G_1$ and $G_2$ . The property of being a subdirect product is preserved when taking supergroups, so we simply traverse the upper layers of the subgroup lattice of $G_1 \times G_2$ , stopping whenever we find a group that is not a subdirect product, until we have constructed all subdirect products of $G_1$ and $G_2$ (up to conjugacy in $G_1\times G_2$ ). Finally, we exhibit a derangement in each subdirect product either by random search or by checking each conjugacy class.

Proof. By Lemma 2.1, the minimum length of a $P^{\Omega _i}$ -orbit is $p^k$ . Since $b<p$ , we have $n<p^{k+1}$ , and hence every $P^{\Omega _i}$ -orbit has length exactly $p^k$ and the result follows.

Proof. Note that $\bigcap _{\omega \in \Omega } P_\omega =1$ because G is a permutation group on $\Omega $ . By Lemma 3.2, P has $2b$ orbits of length p. It follows that for every $\omega \in \Omega $ , we have $|P:P_\omega |=p$ , so $P_\omega $ is a normal (and maximal) subgroup of P. This implies that P is elementary abelian (because P has trivial Frattini subgroup) and that $|\{P_\omega \mid \omega \in \Omega \}|\leq 2b$ . If P does not contain a derangement, then $\bigcup _{\omega \in \Omega } P_\omega =P$ and Lemma 2.3 implies that $2\leq d\leq |\{P_\omega \mid \omega \in \Omega \}|-p+1$ .

Proof. Assume for a contradiction that G does not have a derangement. Let P be a Sylow p-subgroup of G and write $|P|=p^d$ . By Lemma 3.3, P has $2q$ orbits of length p, P is elementary abelian, $|\{P_\omega \mid \omega \in \Omega \}|\leq 2q$ and

$$ \begin{align*}2\leq d\leq |\{P_\omega \mid \omega \in \Omega \}|-p+1\leq 2q-p+1.\end{align*} $$

Suppose first that $|P^{\Omega _i}|\leq p$ for some $i\in \{1,2\}$ . This implies that $P_\omega =P_{\omega '}$ for every $\omega ,\omega '\in \Omega _i$ , namely $|\{P_\omega \mid \omega \in \Omega _i\}|=1$ , and hence $|\{P_\omega \mid \omega \in \Omega \}|\leq q+1$ , which contradicts the facts that $2\leq d \leq |\{P_\omega \mid \omega \in \Omega \}|-p+1$ and $q<p$ . Therefore, ${|P^{\Omega _i}|\geq p^2}$ for both $i\in \{1,2\}$ .

By [Reference Ellis and Harper5, Corollary 4], each $G^{\Omega _i}$ is imprimitive and so preserves a block system consisting of either p blocks of size q or q blocks of size p. In the former case, ${G^{\Omega _i} \leq S_q \wr S_p}$ , so $|P^{\Omega _i}|\leq p$ , which contradicts $|P^{\Omega _i}|\geq p^2$ . (Here, we abuse notation and write $\leq $ to mean ‘is isomorphic to a subgroup of’.) Therefore, each $G^{\Omega _i}$ preserves a system $\mathcal {B}_i$ of q blocks of size p, so $G^{\Omega _i} \leq S_p \wr S_q$ . Choose $B\in \mathcal {B}_i$ . Then we have ${G^{\Omega _i}\leq (G^{\Omega _i})_B^B\wr (G^{\Omega _i})^{\mathcal {B}_i}}$ , with $(G^{\Omega _i})_B^B$ a transitive subgroup of $\text {Sym}(B) \cong S_p$ and $(G^{\Omega _i})^{\mathcal {B}_i}$ a transitive subgroup of $\text {Sym}(\mathcal {B}_i) \cong S_q$ . Since p is prime, it follows from classical results of Burnside [Reference Burnside4, Ch. IX, Theorem IX] that either $(G^{\Omega _i})_B^B \leq \operatorname {\mathrm {AGL}}(1,p)$ or $(G^{\Omega _i})_B^B$ is almost simple.

Suppose first that $(G^{\Omega _i})_B^B$ is almost simple, and note that the socle T of $(G^{\Omega _i})_B^B$ is transitive and, in particular, p divides $|T|$ . Let $K=\operatorname {\mathrm {soc}} ((G^{\Omega _i})_B^B\wr S_q)\cong T^q$ . By Proposition 2.5, either $G^{\Omega _i}\cap K=1$ , $G^{\Omega _i}\cap K\cong T$ or $K\leq G^{\Omega _i}$ . If $G^{\Omega _i}\cap K=1$ or $G^{\Omega _i}\cap K\cong T$ , then $|P^{\Omega _i}|\leq p$ , which contradicts $|P^{\Omega _i}|\geq p^2$ . If $K\leq G^{\Omega _i}$ , then $p^q$ divides $|G^{\Omega _i}|$ and $q\leq d$ , contradicting $d \leq 2q-p+1$ , $q<p$ and $q\neq p-1$ .

We must therefore have $(G^{\Omega _i})_B^B \leq \operatorname {\mathrm {AGL}}(1,p)$ , hence $G^{\Omega _i}\leq \operatorname {\mathrm {AGL}}(1,p)\wr S_q$ for both $i\in \{1,2\}$ . Let Q be a Sylow q-subgroup of G. Since q does not divide $p-1$ , it does not divide $|\operatorname {\mathrm {AGL}}(1,p)|=p(p-1)$ , hence $|Q^{\Omega _i}|=q$ . It follows from Lemma 2.1 that all orbits of $Q^{\Omega _i}$ have length q and thus $|Q:Q_\omega |=q$ for every $\omega \in \Omega $ . Let $\omega _i\in \Omega _i$ and $g\in Q_{\omega _i}$ . Note that g stabilises the block of $\mathcal {B}_i$ containing $\omega _i$ , and hence, since g is a q-element, g must also stabilise the remaining $q-1$ blocks of $\mathcal {B}_i$ . It follows that $g^{\Omega _i}\in \operatorname {\mathrm {AGL}}(1,p)^q$ , but q does not divide $|\operatorname {\mathrm {AGL}}(1,p)|$ , hence $g^{\Omega _i}=1$ . This implies that $g\in Q_\omega $ for every $\omega \in \Omega _i$ , so $Q_{\omega _i} \leq Q_\omega $ for every $\omega \in \Omega _i$ . Since $|Q_\omega |=|Q_{\omega _i}|$ , we have $Q_\omega =Q_{\omega _i}$ for every $\omega \in \Omega _i$ . This implies that Q has at most two distinct point stabilisers, so it must contain a derangement (because it cannot be equal to the union of two proper subgroups).

Proof. If $p=q$ , then the result follows from [Reference Ellis and Harper5, Corollary 5], so we assume that $p>q$ . If q does not divide $p-1$ , then the result follows from Proposition 3.5, so we assume that q divides $p-1$ . If $q=p-1$ , then $p=3$ and $q=2$ , and the result follows from Proposition 3.1. Otherwise, $q\leq (p-1)/2$ and the result follows from Corollary 3.4.