Proof. Fix an index i. The character χ_i of X ⁿ may be extended to its inertia group $I_G(\chi_i)=X \wr I_{H}(\chi_{i})$ in G by [Reference James and Kerber13, p. 154]. Thus the number of irreducible characters of $I_G(\chi_i)$ lying above χ_i is $k(I_{H}(\chi_{1}))$, by Gallagher’s theorem [Reference Isaacs12, Corollary 6.17]. The identity now follows from Clifford’s correspondence [Reference Isaacs12, Theorem 6.11].

Proof. Let h be an arbitrary element of H. The number of characters in ${{\operatorname{Irr}}}(X^n)$ fixed by h is $k^{\sigma(h)}$. The statement follows from the orbit-counting lemma.

Proof. In the first case, we have $(1 - \alpha(H))n \geq \log_{k}(2kn|H|^{2})$ and so

\begin{equation*}2kne \leq 2kn|H| \leq \frac{k^{(1-\alpha(H))n}}{|H|}.\end{equation*}

This and (2.1) imply the desired inequality. If the second condition holds then $\alpha(H)n \leq n-2 \leq n - \log_{k}(2kn|H|^{2})$, and the lemma follows as well. We note that if a primitive permutation group of degree n contains a transposition then it is $\mathsf{S}_{n}$.

Proof. Let H be transitive of order at most $2^{\sqrt{n}/4}$. Observe that if

(2.4)\begin{equation} n - \frac{n}{2\log_{2}|H|} \leq n - \log_{k}\left(2kn|H|^{2}\right), \end{equation}

then we are done by using Lemma 2.3. Inequality (2.4) is equivalent to the inequality $n \geq 2 (\log_{2}|H|) (\log_{k}(2kn|H|^{2}))$. Since the right-hand side is at most $10 (\log_{2}|H|)^{2}$ and $|H| \leq 2^{\sqrt{n}/4}$, inequality (2.4) is satisfied, finishing the proof of the theorem.

Proof. The lower bound follows from Lemma 2.1. Without the classification of finite simple groups Sun and Wilmes [Reference Sun and Wilmes20, Corollary 1.6] proved that

(2.5)\begin{equation} |H| \leq {{\operatorname{exp}}} \left(O(n^{1/3} \log^{7/3}n )\right), \end{equation}

unless H is a family of primitive groups appearing in (i), (ii) or (iii) of the statement of the theorem. Since the right-hand side of (2.5) is less than $2^{\sqrt{n}/4}$ for every sufficiently large n, the result follows from Theorem 2.4 for every sufficiently large n. When n is bounded and $k \to \infty$, the statement follows from the paragraph after (2.3).

Proof. Let n be bounded by an absolute constant. For every sufficiently large k and for any n at least 3, we have

\begin{equation*}k(G) \leq 2 \cdot k(X \wr \mathsf{A}_n ) \leq 2 \cdot \Big(1 + \frac{1}{kn}\Big) \frac{k^{n}}{|\mathsf{A}_n|} \lt k^{n}\end{equation*}

by Lemma 2.3. When n = 2 (and $H = \mathsf{S}_2$) the action is regular and this case was treated earlier (see [Reference Schmid19, Proposition 8.5d], [Reference Garzoni and Gill8, Lemma 4.3] or Section 2). Let $n \geq 5$. Observe that $n(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n)$ is exactly the number of k-tuples $(x_1,...,x_k)\in{\mathbb{Z}}_{\geq 0}^k$ such that $n=x_1+\cdots +x_k$. (These tuples are often referred to as weak k-compositions of n, and their number is given by ${n+k-1 \choose k-1}$.) We have

\begin{equation*}n(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n) = {n+k-1 \choose k-1} \leq \min \Big\{ (n+1)^{k-1}, k \cdot {\Big( \frac{k+1}{2} \Big)}^{n-1} \Big\}.\end{equation*}

If k is bounded by an absolute constant, then

\begin{equation*}k(G) \leq 5^{n/3} \cdot 2 \cdot n(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n) \leq 5^{n/3} \cdot 2 \cdot (n+1)^{k-1} \lt k^{n},\end{equation*}

for every sufficiently large n, by Lemma 2.1 and [Reference Garonzi and Maróti7]. Let $k \geq 100$. We have

\begin{equation*}k(G) \leq 5^{n/3} \cdot 2 \cdot k \cdot {\Big( \frac{k+1}{2} \Big)}^{n-1} \leq k^n,\end{equation*}

and the proof is complete.

Proof. According to [Reference Dona, Maróti and Pyber3, Lemma 2.1], whenever $x \in \mathsf{A}_n$ satisfies $(1-(1-\epsilon)\gamma)n \leq \sigma(x)$, then $|x^{\mathsf{S}_n}| \leq 2 \cdot |x^{\mathsf{A}_n}| \lt 2 \cdot |\mathsf{A}_n|^{\gamma} \lt 2 \cdot |\mathsf{S}_n|^{\gamma}$.

Let $x \in \mathsf{S}_n \setminus \mathsf{A}_{n}$ satisfy the inequality $(1-(1-\epsilon)\gamma)n \leq \sigma(x)$. In the disjoint cycle decomposition of x there is a cycle π of even length, say 2r. Let c_r and $c_{2r}$ be the number of cycles of lengths r and 2r respectively in the disjoint cycle decomposition of x. We have $|{\mathbf C}_{\mathsf{S}_{n}}(x)| = a \cdot r^{c_{r}} \cdot c_{r}! \cdot (2r)^{c_{2r}} \cdot c_{2r}!$ for some positive integer a. Let $y \in \mathsf{A}_n$ be the permutation obtained from x by replacing π by π ². We have $|{\mathbf C}_{\mathsf{S}_{n}}(y)| = a \cdot r^{c_{r}+2} \cdot (c_{r}+2)! \cdot (2r)^{c_{2r}-1} \cdot (c_{2r}-1)!$. It is easy to see that $|{\mathbf C}_{\mathsf{S}_n}(y)| \leq n^{4} \cdot |{\mathbf C}_{\mathsf{S}_{n}}(x)|$ from which it follows that $|x^{S_{n}}| \leq n^{4} \cdot |y^{\mathsf{S}_{n}}|$. Since $\sigma(y) = \sigma(x) + 1$, we have $(1-(1-\epsilon)\gamma)n \leq \sigma(y)$ by hypothesis and so $|y^{\mathsf{S}_n}| \lt 2 \cdot |\mathsf{S}_n|^{\gamma}$ by the first paragraph. This gives $|x^{\mathsf{S}_n}| \lt 2 \cdot n^{4} \cdot |\mathsf{S}_n|^{\gamma}$.

Second proof of Theorem 3.1

Let ϵ and γ be such that $0 \lt \epsilon \lt 1$ and $0 \lt \gamma \lt 1$ such that $\delta := 1 - (1-\epsilon)\gamma \lt 1-\log_2(5)/3$. Let β be such that $\gamma \lt \beta \lt 1$. There exists by Lemma 3.2 an integer N such that whenever $n \geq N$ the inequality $\delta n \leq \sigma(x)$ (for $x \in \mathsf{S}_n$) implies $|x^{\mathsf{S}_n}| \lt |\mathsf{S}_n|^{\beta}$. It follows that the number of elements $x \in \mathsf{S}_n$ such that $\sigma(x) \geq \delta n$ is less than $|\mathsf{S}_{n}|^{\beta} p(n)$ where p(n) denotes the number of partitions of n.

By Lemma 2.2 we have

\begin{equation*}n(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n) =\frac{1}{|\mathsf{S}_n|}\sum_{h\in \mathsf{S}_n}k^{\sigma(h)} = \frac{1}{|\mathsf{S}_n|}\underset{\sigma(h) \lt \delta n}{\sum_{h\in \mathsf{S}_n}}k^{\sigma(h)} + \frac{1}{|\mathsf{S}_n|} \underset{\sigma(h) \geq \delta n}{\sum_{h\in \mathsf{S}_n}}k^{\sigma(h)}\end{equation*}

\begin{equation*} \lt k^{\delta n} + \frac{1}{|\mathsf{S}_n|} k^{n} + |\mathsf{S}_n|^{\beta -1} p(n) k^{n-1}.\end{equation*}

Since $p(n) \lt 13.01^{\sqrt{n}}$ by [Reference Erdös5], it follows that

\begin{equation*}n(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n) \lt \frac{k^{n}}{2\cdot 5^{n/3}}\end{equation*}

for every sufficiently large n or k. By Lemma 2.1 and [Reference Garonzi and Maróti7], it follows that

\begin{equation*}k(G) = k(X \wr H) \leq 2 \cdot 5^{n/3}n\left(\mathsf{S}_n,{{\operatorname{Irr}}}(X)^n\right) \lt 2 \cdot 5^{n/3}\frac{k^{n}}{2\cdot 5^{n/3}} = k^{n}\end{equation*}

for every sufficiently large n or k, as wanted.

We will fix the following notation when working with large base groups:

1. (i) Ω is the set of $\ell$-subsets of $\{1,\ldots,m\}$,

2. (ii) $B_t:={{\operatorname{Irr}}}(X)^{{m\choose\ell}^t}$,

3. (iii) $B:=B_1={{\operatorname{Irr}}}(X)^{{m\choose\ell}}$.

Proof. In Lemma 3.2, let us take $\gamma=2/5$ and $1-(1-\epsilon)\gamma=3/4$ (that is, $\epsilon=3/8$). Assume that $m\geq N(\epsilon,\gamma)$ and $j \gt 3m/4$. The set $\{\pi \in \mathsf{S}_m|\sigma(\pi)=j\}$ is a union of conjugacy classes of $\mathsf{S}_m$. Since all elements in $\{\pi \in \mathsf{S}_m|\sigma(\pi)=j\}$ satisfy that $ \sigma(\pi)=j \gt 3m/4=(1-(1-\epsilon)\gamma)m$, we deduce that

\begin{equation*}\mathcal{S}(j,m)\leq 2m^4p(m)|\mathsf{S}_m|^{\gamma} \lt 2m^413.01^{\sqrt{m}}|\mathsf{S}_m|^{2/5},\end{equation*}

and the lemma follows.

Proof of Lemma 4.4

For each i with $1 \leq i \leq m$, let α_i be the number of cycles in π of length i. We have $\sigma(\pi)=\sum_i \alpha_i$ and $m=\sum_i i\alpha_i$. Let $\mathcal{P}(\ell)$ denote the set of partitions of the integer $\ell$. For each $\lambda\in \mathcal{P}(\ell)$, let λ_i denote the number of parts of λ equal to i. Then

\begin{equation*} |\operatorname{fix}(\pi)|=\sum_{\lambda\in\mathcal{P}(\ell)} {\alpha_1\choose \lambda_1}\cdot {\alpha_2\choose \lambda_2}\cdots \end{equation*}

Using the well-known estimate ${a\choose b}\cdot{c\choose d}\leq {a+c\choose b+d}$, we obtain

(4.1)\begin{equation} |\operatorname{fix}(\pi)|\leq \sum_{\lambda\in\mathcal{P}(\ell)} {\sum\alpha_i\choose \sum\lambda_i} = \sum_{\lambda\in\mathcal{P}(\ell)} {\sigma(\pi)\choose l(\lambda)}, \end{equation}

where $l(\lambda)$ is the number of parts of λ.

I. Assume first that $\ell\leq \sigma(\pi)/2$. Then ${\sigma(\pi)\choose l(\lambda)}\leq {\sigma(\pi)\choose \ell}$ for every $\lambda\in\mathcal{P}(\ell)$. It follows from (4.1) that

\begin{equation*} |\operatorname{fix}(\pi)|\leq p(\ell) \cdot {\sigma(\pi)\choose \ell}. \end{equation*}

Assume furthermore that $m \gt \ell+\sigma(\pi)$. We then have

\begin{align*} {m\choose \ell}&=\frac{m(m-1)\cdots (\sigma(\pi)+1)}{(m-\ell)(m-1-\ell)\cdots(\sigma(\pi)+1-\ell)}\cdot{\sigma(\pi)\choose\ell}\\ &=\frac{m(m-1)\cdots(m-\ell+1)}{\sigma(\pi)(\sigma(\pi)-1)\cdots (\sigma(\pi)-\ell+1)}\cdot{\sigma(\pi)\choose\ell}\\ &\geq\left(\frac{4}{3}\right)^\ell\cdot{\sigma(\pi)\choose\ell}, \end{align*}

where the last inequality follows from the hypothesis on $\sigma(\pi)$. The lemma then follows if $(4/3)^{\ell-1} \gt p(\ell)$. This is true when $\ell\geq 59$, by using the bound for the partition function in [Reference Erdös5].

We therefore may assume that $\ell\leq 58$. By (4.1) and the hypothesis,

(4.2)\begin{equation} |\operatorname{fix}(\pi)|\leq \sum_{\lambda\in\mathcal{P}(\ell)} {[3m/4]\choose l(\lambda)}. \end{equation}

Thus, we are done if

\begin{equation*} \sum_{\lambda\in\mathcal{P}(\ell)} {[3m/4]\choose l(\lambda)} \lt \frac{3}{4}{m\choose \ell}. \end{equation*}

For each fixed $\ell$ with $\ell \leq 58$, we observe that the right-hand side is a polynomial (in m) of degree $\ell$, while the left-hand side is a polynomial of degree at most $\ell$ and if equal to $\ell$ then with smaller leading coefficient. Therefore the inequality holds for every sufficiently large m, and we are done.

Now assume that $m\leq \ell+\sigma(\pi)$. Then

\begin{align*} {m\choose \ell}\geq\left(\frac{m}{m-\ell}\right)^{m-\sigma(\pi)}\cdot{\sigma(\pi)\choose\ell}\geq \left(\frac{m}{m-\ell}\right)^{m/4}\cdot{\sigma(\pi)\choose\ell}\geq \left(\frac{4}{3}\right)^{m/4}\cdot{\sigma(\pi)\choose\ell}. \end{align*}

As above, the desired inequality follows from these bounds for every sufficiently large m.

II. Next we consider the case $\ell \gt \sigma(\pi)/2$. Then ${\sigma(\pi)\choose l(\lambda)}\leq {\sigma(\pi)\choose [\sigma(\pi)/2]}$ for every $\lambda\in\mathcal{P}(\ell)$. As in the previous case, it follows from (4.1) that

\begin{equation*} |\operatorname{fix}(\pi)|\leq p(\ell) \cdot {\sigma(\pi)\choose [\sigma(\pi)/2]}. \end{equation*}

On the other hand, by the hypothesis on $\sigma(\pi)$ and $\ell$, we have

\begin{align*} {m\choose \ell}&=\frac{(m-[\sigma(\pi)/2])\cdots(m-\ell+1)}{\ell(\ell-1)\cdots([\sigma(\pi)/2]+1)}\cdot{m\choose[\sigma(\pi)/2]}\\ &\geq \left(\frac{5}{4}\right)^{\ell-[\sigma(\pi)/2]}\cdot{m\choose[\sigma(\pi)/2]}. \end{align*}

The lemma now follows by similar estimates as in the previous case, but for ${m\choose[\sigma(\pi)/2]}$ instead of ${m\choose \ell}$.

Proof. Recall from Notation 4.2 that Ω is the set of $\ell$-subsets of $\{1, \ldots, m\}$. Let $\pi\in \mathsf{S}_m$. Let $\operatorname{fix}(\pi)$ denote the set of $\ell$-subsets fixed by π, as in Lemma 4.4. Note that

(4.3)\begin{equation} \sigma'(\pi)\leq |\operatorname{fix}(\pi)|+\frac{1}{2}\left({m\choose\ell}-|\operatorname{fix}(\pi)|\right)=\frac{1}{2}\left({m\choose\ell}+|\operatorname{fix}(\pi)|\right). \end{equation}

Furthermore, by Lemma 2.2,

\begin{equation*} n(\mathsf{S}_m,B)=\frac{1}{|\mathsf{S}_m|} \sum_{\pi\in \mathsf{S}_m} {k^{\sigma'(\pi)}}. \end{equation*}

We decompose this into two smaller sums, depending on whether $\sigma(\pi)$ is smaller or larger than $3m/4$:

\begin{equation*}n(\mathsf{S}_m,B)=n_1(\mathsf{S}_m,B_1)+n_2(\mathsf{S}_m,B_1),\end{equation*}

where

\begin{equation*} n_1(\mathsf{S}_m,B)=\frac{1}{|\mathsf{S}_m|} \sum_{\sigma(\pi)\leq 3m/4} {k^{\sigma'(\pi)}} \text{and } n_2(\mathsf{S}_m,B)=\frac{1}{|\mathsf{S}_m|} \sum_{\sigma(\pi) \gt 3m/4} {k^{\sigma'(\pi)}}. \end{equation*}

By using (4.3), we get

\begin{equation*} n_1(\mathsf{S}_m,B)\leq \frac{1}{m!}\sum_{\sigma(\pi)\leq 3m/4}{k^{\frac{1}{2}\left({m\choose\ell}+|\operatorname{fix}(\pi)|\right)}}, \end{equation*}

and it follows from Lemma 4.4 that

\begin{equation*} n_1(\mathsf{S}_m,B)\leq k^{\frac{7}{8}{m\choose\ell}} \end{equation*}

for every $m\geq N_1$ for some positive integer N ₁.

We now work on $n_2(\mathsf{S}_m,B)$. Recall that $\mathcal{S}(j,m)$ denotes the number of elements of $\mathsf{S}_m$ with precisely j cycles. So

\begin{equation*} n_2(\mathsf{S}_m,B) \leq \frac{1}{m!}\sum_{j \gt 3m/4}\mathcal{S}(j,m)k^{{m \choose \ell}}. \end{equation*}

Using the bound for $\mathcal{S}(j,m)$ in Lemma 4.3, we deduce that

\begin{equation*} n_2(\mathsf{S}_m,B)\leq\frac{1}{4}m(m!)^{-0.59}k^{m\choose\ell} \end{equation*}

for every $m\geq N_2$ for some positive integer N ₂. Now taking $N_3:=\max\{N_1,N_2\}$, we arrive at

\begin{equation*} n(\mathsf{S}_m,B)\leq k^{\frac{7}{8}{m\choose\ell}}+\frac{1}{4}m(m!)^{-0.59}k^{m\choose\ell}\end{equation*}

for every $m\geq N_3$, and the result readily follows.

Proof. Let $D:= (\mathsf{S}_m)^t\cap H$ be the ‘diagonal’ subgroup of H. By Lemma 2.2, the number of orbits of H acting on $B_{t} ={{\operatorname{Irr}}}(X^n)$ is

(5.1)\begin{equation} n(H,B_{t})=\frac{1}{|H|}\sum_{x\in H}k^{\gamma(x)}=\frac{1}{|H|}\sum_{x\in D}k^{\gamma(x)}+\frac{1}{|H|}\sum_{x\in H\setminus D}k^{\gamma(x)}. \end{equation}

For $x \in H$, we write $\operatorname{fix}(x)$ to denote the set of elements in $\Omega^t$ fixed by x. By (2.3) we have

\begin{equation*}\gamma(x)\leq \frac{n}{2}(1+ \operatorname{fpr}(x)),\end{equation*}

where $\operatorname{fpr}(x)=|\operatorname{fix}(x)|/n$ is the fixed point ratio of x. Let $x \in H\setminus D$ and let us write $x=(x_1,\ldots,x_t)\pi\in H$ for $x_1,\ldots, x_t\in \mathsf{S}_m$ and $1\neq \pi\in \mathsf{S}_t$. We know that π must contain a cycle of length r for some $r \geq 2$. A straightforward computation shows that $\operatorname{fix}(x)\leq |\Omega|^{t-(r-1)}$ (see also [Reference Burness and Guralnick1, Proposition 6.1] for a similar argument) and thus

\begin{equation*}\operatorname{fpr}(x)\leq |\Omega|^{1-r}\leq (r+1)^{-1} \leq 1/3,\end{equation*}

where the second inequality holds since $|\Omega|\geq 5$. Thus, the second term in the far-right-hand-side sum in (5.1) is bounded by $k^{2n/3}$.

On the other hand, for the first term, we have

\begin{equation*} \frac{1}{|H|}\sum_{x\in D}k^{\gamma(x)}\leq \frac{1}{|\mathsf{A}_m|^t}\sum_{x\in D}k^{\gamma(x)}\leq \frac{1}{|\mathsf{A}_m|^t}\sum_{x\in (\mathsf{S}_m)^t}k^{\gamma(x)}=2^tn((\mathsf{S}_m)^t,B_t). \end{equation*}

We have shown that

\begin{equation*} n(H,B_{t})\leq 2^tn((\mathsf{S}_m)^t,B_t)+k^{2n/3}. \end{equation*}

Note that $H\leq \mathsf{S}_m\wr \mathsf{S}_t$ and $\mathsf{S}_m\wr \mathsf{S}_t$ may be viewed as a subgroup of $\mathsf{S}_{mt}$. It follows that every subgroup of H has at most 5^mt classes, by [Reference Kovács and Robinson16, Theorem 1.2]. The desired bound now follows by using Lemma 2.1.

Proof. Observe that $B_t=(B_1)^t$ and an element $x=(x_1,\ldots,x_t)\in(\mathsf{S}_m)^t$ fixes $(\chi_1,\ldots,\chi_t)\in B_t$ if and only if each $x_i\in\mathsf{S}_m$ fixes $\chi_i\in B_1$ for every i. Now,

\begin{align*} n((\mathsf{S}_m)^t,B_t)&=\frac{1}{|\mathsf{S}_m|^t} \sum_{x\in(\mathsf{S}_m)^t} |\operatorname{fix}(x,B_t)|\\ &=\frac{1}{|\mathsf{S}_m|^t} \sum_{x_1\in\mathsf{S}_m}\cdots \sum_{x_t\in\mathsf{S}_m}|\operatorname{fix}(x_1,B_1)|\cdots |\operatorname{fix}(x_t,B_1)|\\ &=\frac{1}{|\mathsf{S}_m|^t} \left(\sum_{x_1\in\mathsf{S}_m}|\operatorname{fix}(x_1,B_1)|\right)^t\\ &=n(\mathsf{S}_m,B_1)^t, \end{align*}

and the lemma follows.

Proof of Theorem 5.1

By Proposition 5.2 and Lemma 5.3, we have

(5.2)\begin{equation} k(G) \lt 5^{mt}\left(2^tn(\mathsf{S}_m,B_1)^t+k^{2n/3}\right). \end{equation}

Obviously, $n(\mathsf{S}_m,B_1)\leq |B_1|=k^{{m\choose\ell}}$. Hence

\begin{equation*} k(G) \lt 5^{mt}2^tk^{{m\choose\ell}t} + 5^{mt}k^{2n/3}. \end{equation*}

It is straightforward to see that, as $(t,\ell)\neq (1,1)$, both terms on the right-hand side are less than $\frac{1}{2}k^{{m\choose\ell}^t}$ for every sufficiently large t. We assume from now on that t is bounded.

Next, using Proposition 4.6 together with (5.2), we have that there exists a positive integer N such that, for every $m\geq N$,

\begin{align*} k(G)& \lt 5^{mt}4^t(\max\{k^{\frac{7}{8}{m\choose\ell}},(m!)^{-0.58}k^{{m\choose\ell}}\})^t+5^{mt}k^{2n/3}\\ &\leq 5^{mt}4^tk^{\frac{7t}{8}{m\choose\ell} t} + 5^{mt}4^t(m!)^{-0.58t}k^{{m\choose\ell} t}+5^{mt}k^{2n/3}. \end{align*}

With t being bounded, each of these three terms is less than $\frac{1}{3}k^{{m\choose\ell}^t}$, and therefore $k(G)\leq k^n$, for every sufficiently large m.

We now assume that both t and m are bounded, or equivalently, that n is bounded. Given the hypothesis that H is a primitive group that is different from $\mathsf{S}_n$, it follows that H does not contain a transposition. In this case, the remark following (2.3) shows that $k(G) \lt k^n$ for all sufficiently large k. This completes the proof.

Proof of Theorem 1.1

Let H be a semiprimitive permutation group. This is a transitive permutation group all of whose normal subgroups are transitive or semiregular. We may assume at this point that H is not primitive. The group H acts on the set Ω of factors of Xⁿ. Let

\begin{equation*}Y := X^{n/r}\end{equation*}

for some divisor $r\leq n/2$ of n such that H acts primitively on the set $\overline{\Omega}$ of factors of Y^r. Let the kernel of this action be K. Since this is an intransitive normal subgroup of H, it must be semiregular on Ω.

Let h be an element of H. Let the number of cycles of h acting on Ω and $\overline{\Omega}$ be denoted by $\sigma_{\Omega}(h)$ and $\sigma_{\overline{\Omega}}(h)$, respectively. Observe that $\sigma_{\Omega}(h) \leq (n/r) \cdot \sigma_{\overline{\Omega}}(h)$.

We have

\begin{align*} \alpha(H) := \max_{1 \not= h \in H} \frac{\sigma_{\Omega}(h)}{n} &= \max \left\{\max_{1 \not= h \in K} \frac{\sigma_{\Omega}(h)}{n}, \max_{h \in H \setminus K} \frac{\sigma_{\Omega}(h)}{n}\right\}\\ & \leq \max \left\{\frac{1}{2}, \max_{h \in H \setminus K} \frac{\sigma_{\overline{\Omega}}(h)}{r}\right\}. \end{align*}

It follows that if r is bounded by an absolute constant, then $\alpha(H)$ is a fixed number less than 1 and so $k(G) \leq k^n$ for every sufficiently large n or k by (2.1). We may therefore assume that $r \to \infty$, in particular, $n \to \infty$.

If $H/K$ is not a large base group (see Definition 4.1), then

\begin{equation*}|H| = |H/K| |K| \leq |H/K| \cdot n = {{\operatorname{exp}}}(O(n^{1/3}\log^{7/3}n))\end{equation*}

by (2.5). In this case the result follows from Theorem 2.4.

For general $H/K$, we have

\begin{equation*}n(H, {{\operatorname{Irr}}}(X^{n})) = \frac{1}{|H|} \sum_{h \in H} k^{\sigma_{\Omega}(h)} = \frac{1}{|H|} \Big( \sum_{h \in H \setminus K} k^{\sigma_{\Omega}(h)} + \sum_{h \in K} k^{\sigma_{\Omega}(h)} \Big) \leq\end{equation*}

\begin{equation*}\leq \frac{1}{|H|} \Big( |K| \sum_{1 \not= \bar{h} \in H/K} k^{\sigma_{\overline{\Omega}}(\bar{h}) \cdot (n/r)} + k^{n} + (|K|-1) k^{n/2} \Big) \lt \end{equation*}

\begin{equation*} \lt n\left(H/K, {{\operatorname{Irr}}}(Y^{r})\right) + \frac{k^{n}}{|H|} + \frac{n k^{n/2}}{|H|},\end{equation*}

where $n(H/K, {{\operatorname{Irr}}}(Y^{r}))$ is the number of orbits of $R = H/K$ on ${{\operatorname{Irr}}}(Y^{r}) = {{{\operatorname{Irr}}}(Y)}^{r}$ where the action of R is defined from the action of the primitive permutation group R acting on the set of factors of Y ^r (that is, R acts primitively on the set of factors of ${{{\operatorname{Irr}}}(Y)}^{r}$).

The number k(G) is equal to the sum of the numbers of conjugacy classes of $n(H, {{\operatorname{Irr}}}(X^{n}))$ inertia subgroups, by Lemma 2.1. As before, let e be the maximum of these numbers. Since K is semiregular, at most $\sum_{1 \not= h \in K} k^{\sigma_{\Omega}(h)} \lt (n/r) k^{n/2}$ of the inertia subgroups intersect K nontrivially. These numbers contribute less than $e (n/r) k^{n/2} \leq (n/r)^{2} \cdot 5^{r/3} k^{n/2}$ to k(G). (For $H/K \in \{ \mathsf{S}_r, \mathsf{A}_r \}$ this follows from [Reference Nagao17] and from the statement in the second paragraph of Section 3, otherwise $|H/K| \leq 5^{r/3}$ for every sufficiently large r by [Reference Sun and Wilmes20, Corollary 1.6].) Since $r \leq n/2$, we have $(n/r)^{2} \cdot 5^{r/3} k^{n/2} \leq n^{2} \cdot 5^{n/6} k^{n/2}$ and this is less than $k^{n}/16$ for every sufficiently large n. Thus we have

\begin{align*} k(G) &\leq e_K \cdot n\left(H, {{\operatorname{Irr}}}(X^{n})\right) + \frac{k^{n}}{16}\\ & \lt e_K \cdot n\left(H/K, {{\operatorname{Irr}}}(Y^{r})\right) + \frac{e_K \cdot k^{n}}{|H|} + \frac{e_K \cdot n k^{n/2}}{|H|} + \frac{k^{n}}{16}, \end{align*}

where e_K denotes the maximum of the numbers of conjugacy classes of those inertia subgroups of H which intersect with K trivially. Note that e_K is at most the maximum of the numbers of classes of subgroups of $H/K$.

Recall that we are done when $H/K$ is not a large base group, and so we assume in the remainder of the proof that $H/K$ is a large base group. We shall follow the notation in Definition 4.1 and Notation 4.2, with Y and r in place of X and n, respectively. In particular, $(\mathsf{A}_m)^t\leq H/K\leq \mathsf{S}_m \wr \mathsf{S}_t$ for some $t\geq 1$ and $m\geq 5$. Also, $r={m\choose\ell}^t$.

Since H is not abelian, we have $e_K \leq (5/8) |H|$ by [Reference Gustafson11]. It follows that

\begin{equation*}\frac{e_K \cdot k^{n}}{|H|} + \frac{e_K \cdot n k^{n/2}}{|H|} + \frac{k^{n}}{16} \leq \frac{3}{4} k^{n}\end{equation*}

for every sufficiently large n. Note that $H/K$ can be viewed as a subgroup of $\mathsf{S}_{mt}$, and so $e_K \leq 5^{mt}$ by [Reference Kovács and Robinson16, Theorem 1.2]. To establish $k(G)\leq k^n$ for sufficiently large n or k, it is now sufficient to show that

(6.1)\begin{equation} n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq \frac{k^{n}}{4\cdot 5^{mt}} \end{equation}

for every sufficiently large r, or equivalently, every sufficiently large m or t.

Let $(t,\ell) = (1,1)$. In this case we may replace the above bound $e_K \leq 5^{mt}$ by $e_K \leq 5^{mt/3}$ as discussed above. We have

\begin{equation*}n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq 2 \cdot \binom{r + k^{n/r} - 1}{r} \leq 2 \cdot 3^{r} \Big(\frac{r + k^{n/r} -1}{r}\Big)^{r} \lt 2 \cdot 3^{r} \Big(\frac{k^{n/r}}{r} + 1 \Big)^{r}.\end{equation*}

If $k^{n/r}/r \geq 1$, then

\begin{equation*}n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq 2 \cdot 6^{r} \Big( \frac{k^{n}}{r^r} \Big) \lt k^{n}/(4\cdot 5^{mt/3})\end{equation*}

for every sufficiently large r = m. Let $k^{n/r} \leq r$. Then $n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq 4^{r}$. This is less than $k^{n}/(4\cdot 5^{r/3})$ for every sufficiently large r, unless $n = 2r$ and k = 2. In the exceptional case $n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq (r+3)(r+2)(r+1)/3$, which is again less than $k^{n}/(4\cdot 5^{r/3})$ for every sufficiently large r. Let $(t,\ell) \not= (1,1)$.

First, arguing as in the proof of Proposition 5.2 and using Lemma 5.3, we have

\begin{equation*} n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right)\leq 2^tn\left((\mathsf{S}_m)^t,B_t\right)+k(Y)^{2r/3}=2^tn\left(\mathsf{S}_m,B_1\right)^t+k(Y)^{2r/3}. \end{equation*}

When $t\to\infty$, one may use the obvious bound $n\left(\mathsf{S}_m,B_1\right)\leq |B_1|=k(Y)^{m\choose\ell}$ to achieve (6.1). So we assume that t is bounded.

Next, using Proposition 4.6, we deduce that

\begin{align*} n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq 4^t k(Y)^{\frac{7}{8}{m\choose\ell}t} + 4^t(m!)^{-0.58t} k(Y)^{{m\choose\ell}t} + k(Y)^{2r/3}. \end{align*}

for every sufficiently large m. As $k(Y)=k^{n/r}$, it follows that

\begin{equation*} n(H/K, {{\operatorname{Irr}}}\left(Y^{r})\right) \leq 4^t k^{\frac{7}{8r}{m\choose\ell}tn} + 4^t(m!)^{-0.58t} k^{\frac{1}{r}{m\choose\ell}tn} + k^{2n/3}. \end{equation*}

With $n\geq 2r=2{m\choose\ell}^t$, $m\to\infty$, and t being bounded, it is straightforward to verify that this sum is less than the right-hand side of (6.1), and the proof is complete.