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\title{Degree-weighted Normal Covers of Symmetric and Alternating Groups}
\author{Luca Blanchi}
\date{June 2026}

\maketitle

\begin{abstract}
Let $G$ be a finite group. A normal cover of $G$ is a family of proper subgroups $H_i<G$ such that the union of their conjugates is all of $G$. The usual normal covering number counts the number of conjugacy classes of subgroups used. We study the degree-weighted variant

\[
\beta(G)=\min\sum_i [G:H_i],
\]

where the minimum is taken over normal covers by proper subgroups. Equivalently, $\beta(G)$ is the minimum degree of a finite $G$-set $\Omega$ such that every element of $G$ fixes some point of $\Omega$, while $G$ has no global fixed point on $\Omega$. We also study the faithful version $\beta_{\mathrm{faith}}(G)$, where the $G$-action is required to be faithful.

This invariant is a degree-weighted analogue of the normal covering number and of the invariant $s(G)$ appearing in the theory of intersective polynomials. In that setting, $s(G)$ records the minimum number of irreducible factors forced by group theory; $\beta_{\mathrm{faith}}(G)$ records instead the minimum possible total permutation degree.

We prove that, for all sufficiently large odd $n$,

\[
\beta_{\mathrm{faith}}(S_n)=2^{n-1}+1,
\]

and classify the optimal covers. For even $n$, we prove the second-order estimate

\[
\beta_{\mathrm{faith}}(S_n)
=
\frac12\binom n{n/2}
+
\Theta\!\left(\binom n{\lfloor n/3\rfloor}\right).
\]

For alternating groups of even degree, we prove the exact formula

\[
\beta_{\mathrm{faith}}(A_{2m})
=
\sum_{\substack{1\le k<m\\ k\text{ odd}}}\binom{2m}{k}
+
\frac12\binom{2m}{m}
\]

for all sufficiently large $m$, and classify the optimal covers. Equivalently,

\[
\beta_{\mathrm{faith}}(A_{2m})
=
\begin{cases}
2^{2m-2}, & m\text{ odd},\\[4pt]
2^{2m-2}+\dfrac12\binom{2m}{m}, & m\text{ even}.
\end{cases}
\]

For $n$ odd composite, with $q=P^-(n)$ the smallest prime divisor of $n$, we prove

\[
\beta_{\mathrm{faith}}(A_n)
\sim
\frac{n!}{((n/q)!)^q q!}.
\]

For $n=p$ prime,

\[
\beta_{\mathrm{faith}}(A_p)=(1+o(1))c_p,
\]

where

\[
c_p=\min\{[A_p:H]:H<A_p,\ H\text{ contains a }p\text{-cycle}\},
\]

and

\[
\log\beta_{\mathrm{faith}}(A_p)=p\log p-p+O((\log p)^2).
\]

The proofs combine a dual certificate for weighted set cover, slack identities, elementary cycle-type arguments, comparisons among imprimitive wreath products, and standard bounds for primitive permutation groups.

\end{abstract}

\section{Introduction}

Let $G$ be a finite group. A normal cover of $G$ is a family of proper subgroups $H_1,\dots,H_r<G$ such that

\[
G=\bigcup_{i=1}^r\bigcup_{g\in G}gH_ig^{-1}.
\]

The normal covering number $\gamma(G)$ is the minimum possible number $r$, or equivalently the minimum number of conjugacy classes of proper subgroups needed. Normal covers of finite groups, and especially of symmetric and alternating groups, have been studied extensively.

In arithmetic applications, especially in the theory of intersective polynomials, one is often interested in covers by proper subgroups whose cores intersect trivially. Bubboloni and Sonn define an invariant $s(G)$ as the minimum number of proper subgroups whose conjugates cover $G$ and whose total core intersection is trivial. This gives a group-theoretic lower bound for the number of irreducible factors of an intersective polynomial with Galois group $G$.

The present paper studies the degree-weighted analogue. Instead of assigning cost $1$ to a subgroup $H$, we assign cost

\[
[G:H],
\]

the degree of the transitive permutation representation $G\curvearrowright G/H$. Thus the cost of a family $H_i$ is

\[
\sum_i [G:H_i].
\]

This is the total degree of the associated finite $G$-set. In arithmetic language, it is the degree of the corresponding finite etale algebra, or the total degree of the corresponding product of irreducible factors.

The weighted invariant behaves differently from the usual normal covering number. In the unweighted problem for $S_n$ and $A_n$, one is led to additive-combinatorial questions about partitions of $n$. In the weighted problem, the first-order behavior is governed by expensive cycle classes, such as $n$-cycles and two-cycle classes $(k,n-k)$. A residual partition problem still appears in the second-order behavior of $\beta_{\mathrm{faith}}(S_{2m})$.

The main proof method is a dual certificate. Given nonnegative weights on conjugacy classes, if every proper subgroup $H$ sees total class-weight at most $[G:H]$, then the total weight gives a lower bound for $\beta(G)$. In the extremal cases we study, the dual certificates are sharp. Their slack identities give not only values of $\beta_{\mathrm{faith}}$, but also rigidity of the optimal covers.

\section{Definitions}

Let $G$ be a finite group. A finite $G$-set $\Omega$ is called a fixed-point cover if

\[
\Omega^g\ne\varnothing
\qquad
\text{for every }g\in G.
\]

It is called blind if, in addition,

\[
\Omega^G=\varnothing.
\]

It is called faithfully blind if the action of $G$ on $\Omega$ is faithful.

Define

\[
\beta(G)=\min\{|\Omega|:\Omega\text{ is a blind finite }G\text{-set}\},
\]

and

\[
\beta_{\mathrm{faith}}(G)=\min\{|\Omega|:\Omega\text{ is a faithfully blind finite }G\text{-set}\}.
\]

Every finite $G$-set decomposes as a disjoint union of transitive $G$-sets:

\[
\Omega=\bigsqcup_i G/H_i.
\]

For $g\in G$,

\[
(G/H_i)^g\ne\varnothing
\]

if and only if

\[
g\in xH_ix^{-1}
\]

for some $x\in G$. Therefore $\Omega$ is a fixed-point cover if and only if

\[
G=\bigcup_i\bigcup_{x\in G}xH_ix^{-1}.
\]

Moreover,

\[
(G/H_i)^G\ne\varnothing
\]

if and only if $H_i=G$. Thus $\Omega$ is blind if and only if all $H_i$ are proper and their conjugates cover $G$.

The kernel of the action of $G$ on $G/H_i$ is

\[
\operatorname{Core}_G(H_i)=\bigcap_{x\in G}xH_ix^{-1}.
\]

Hence the kernel of the action on

\[
\bigsqcup_i G/H_i
\]

is

\[
\bigcap_i \operatorname{Core}_G(H_i).
\]

Thus

\[
\beta(G)
=
\min
\left\{
\sum_i [G:H_i]:
H_i<G,\
G=\bigcup_i\bigcup_{x\in G}xH_ix^{-1}
\right\},
\]

and

\[
\beta_{\mathrm{faith}}(G)
=
\min
\left\{
\sum_i [G:H_i]:
H_i<G,\
G=\bigcup_i\bigcup_{x\in G}xH_ix^{-1},
\bigcap_i\operatorname{Core}_G(H_i)=1
\right\}.
\]

For $A_n$, $n\ge5$, the group is simple, so every proper subgroup has trivial core. Therefore

\[
\beta_{\mathrm{faith}}(A_n)=\beta(A_n).
\]

For $S_n$, $n\ge5$, the same equality holds.

\paragraph{Lemma 2.1.}

For $n\ge5$,

\[
\beta_{\mathrm{faith}}(S_n)=\beta(S_n).
\]

\noindent\textbf{Proof.}

Let $H_1,\dots,H_r$ be a normal cover of $S_n$ by proper subgroups. Since the cover must contain odd permutations, at least one $H_i$ is not contained in $A_n$. The core of this $H_i$ cannot be $A_n$, and cannot be $S_n$ because $H_i$ is proper. Since the only nontrivial proper normal subgroup of $S_n$, for $n\ge5$, is $A_n$, this core is trivial. Hence the total core intersection is trivial. $\square$

\section{Cycle classes and elementary subgroup facts}

We use standard cycle-type notation. A class of $S_n$ is described by a partition of $n$. We write $(a,b)$ for the class of permutations with two cycles of lengths $a$ and $b$.

Recall that a conjugacy class of $S_n$ splits into two classes in $A_n$ if and only if all cycle lengths are odd and pairwise distinct.

\paragraph{Lemma 3.1.}

Let $H<S_n$ be intransitive. If $H$ contains an element of cycle type $(a,n-a)$, then, after conjugacy,

\[
H\le S_a\times S_{n-a}.
\]

Consequently,

\[
[S_n:H]\ge \binom na.
\]

Similarly, if $H<A_n$ is intransitive and contains an element of type $(a,n-a)$, then, after conjugacy,

\[
H\le (S_a\times S_{n-a})\cap A_n,
\]

and

\[
[A_n:H]\ge\binom na.
\]

\noindent\textbf{Proof.}

Let $x\in H$ have cycle type $(a,n-a)$. Since $x\in H$, each cycle support of $x$ is contained in an $H$-orbit. Hence every $H$-orbit is a union of the two cycle supports of $x$. Since $H$ is intransitive, it cannot have a single orbit containing both cycle supports. Therefore the $H$-orbits are exactly the two supports, of sizes $a$ and $n-a$. The containment and index estimate follow. The $A_n$ case is identical. $\square$

\paragraph{Lemma 3.2.}

Let $n$ be even and let $k<n/2$ be odd. The $S_n$-class of cycle type $(k,n-k)$ splits into two $A_n$-classes. The subgroup

\[
(S_k\times S_{n-k})\cap A_n
\]

meets both of them.

\noindent\textbf{Proof.}

The splitting criterion applies because $k$ and $n-k$ are distinct odd integers. Let

\[
M=S_k\times S_{n-k}.
\]

Then $M$ contains odd permutations. If $x\in M\cap A_n$ has type $(k,n-k)$, and $t\in M$ is odd, then

\[
txt^{-1}\in M\cap A_n
\]

has the same $S_n$-cycle type. Since conjugation by an odd element interchanges the two $A_n$-classes lying inside a split $S_n$-class, $M\cap A_n$ meets both classes. $\square$

\paragraph{Lemma 3.3.}

Let $n$ be odd composite, let $q=P^-(n)$, and set

\[
W_q=S_{n/q}\wr S_q
\]

in its natural imprimitive action on $n$ points. Then

\[
W_q\cap A_n
\]

meets both $A_n$-classes of $n$-cycles.

\noindent\textbf{Proof.}

The $S_n$-class of $n$-cycles splits in $A_n$, since $n$ is odd. The wreath product $W_q$ contains an $n$-cycle. It also contains odd permutations, for instance a transposition inside one block. If $c\in W_q\cap A_n$ is an $n$-cycle and $t\in W_q$ is odd, then

\[
tct^{-1}\in W_q\cap A_n
\]

is an $n$-cycle lying in the other $A_n$-class. $\square$

\paragraph{Lemma 3.4.}

Let $p$ be an odd prime. If $H<A_p$ contains a $p$-cycle, then $H$ meets both $A_p$-classes of $p$-cycles.

\noindent\textbf{Proof.}

Let $c\in H$ be a $p$-cycle. Then $H$ contains every power $c^a$, $a\in(\mathbb Z/p\mathbb Z)^\times$. The two $A_p$-classes of $p$-cycles are distinguished by whether $a$ is a quadratic residue or nonresidue modulo $p$. Hence the powers of $c$ meet both classes. $\square$

\section{Dual certificates and slack}

Let $\mathcal C(G)$ be the set of nonidentity conjugacy classes of $G$.

\paragraph{Proposition 4.1.}

Let $w:\mathcal C(G)\to\mathbb R_{\ge0}$ satisfy

\[
\sum_{\substack{C\in\mathcal C(G)\\ C\cap H\ne\varnothing}} w(C)
\le [G:H]
\]

for every proper subgroup $H<G$. Then

\[
\beta(G)\ge \sum_{C\in\mathcal C(G)}w(C).
\]

If $G$ is nonabelian simple, then the same lower bound holds for $\beta_{\mathrm{faith}}(G)$.

\noindent\textbf{Proof.}

Let $H_1,\dots,H_r$ be a normal cover of $G$. Every nonidentity conjugacy class meets at least one $H_i$. Hence

\[
\sum_C w(C)
\le
\sum_{i=1}^r
\sum_{\substack{C\\ C\cap H_i\ne\varnothing}}w(C)
\le
\sum_{i=1}^r [G:H_i].
\]

Taking the minimum over all normal covers gives the result. If $G$ is simple, every proper subgroup has trivial core. $\square$

\paragraph{Proposition 4.2.}

With $w$ as in Proposition 4.1, put

\[
w(H)=\sum_{\substack{C\\ C\cap H\ne\varnothing}}w(C)
\]

and

\[
\operatorname{slack}_w(H)=[G:H]-w(H).
\]

If $\mathcal H=\{H_1,\dots,H_r\}$ is a normal cover, define

\[
\operatorname{overlap}_w(\mathcal H)
=
\sum_i w(H_i)-\sum_C w(C).
\]

Then

\[
\sum_i [G:H_i]-\sum_C w(C)
=
\sum_i\operatorname{slack}_w(H_i)
+
\operatorname{overlap}_w(\mathcal H).
\]

In particular, if the cover has cost equal to $\sum_Cw(C)$, then every subgroup used is $w$-tight and every positive-weight class is covered exactly once.

\noindent\textbf{Proof.}

This is the identity

\[
\sum_i [G:H_i]
=
\sum_i w(H_i)+\sum_i([G:H_i]-w(H_i)).
\]

Subtracting $\sum_Cw(C)$ gives the formula. Since every positive-weight class is covered at least once,

\[
\operatorname{overlap}_w(\mathcal H)\ge0.
\]

If the total cost equals $\sum_Cw(C)$, then both nonnegative terms on the right are zero. $\square$

\section{Imprimitive and primitive estimates}

We use the following standard consequence of Maroti's bound on primitive permutation groups.

\paragraph{Lemma 5.1.}

For all sufficiently large $n$, if $H<S_n$ is primitive and does not contain $A_n$, then

\[
[S_n:H]>2^n.
\]

If $H<A_n$ is primitive and proper, then

\[
[A_n:H]>2^{n-1}.
\]

\noindent\textbf{Proof.}

By Maroti's bound,

\[
|H|<50n^{\sqrt n}.
\]

Therefore

\[
[S_n:H]>\frac{n!}{50n^{\sqrt n}}.
\]

Stirling's formula gives

\[
\log\left(\frac{n!}{50n^{\sqrt n}}\right)
=
n\log n-n-O(\sqrt n\log n),
\]

which is larger than $n\log2$ for all sufficiently large $n$. The alternating-group version differs only by a factor $2$. $\square$

\paragraph{Lemma 5.2.}

Let $H<S_n$ be transitive and imprimitive, and suppose that $H$ contains an $n$-cycle. Then $H$ is contained in a conjugate of

\[
S_{n/b}\wr S_b
\]

for some proper divisor $b\mid n$, $1<b<n$. Conversely, every such wreath product contains an $n$-cycle.

\noindent\textbf{Proof.}

An imprimitive transitive subgroup preserves a system of $b$ blocks of equal size $n/b$, with $1<b<n$. Therefore it embeds in $S_{n/b}\wr S_b$. Conversely, label the points by pairs

\[
(i,j)\in \mathbb Z/b\mathbb Z\times \mathbb Z/(n/b)\mathbb Z.
\]

The permutation

\[
(i,j)\mapsto
\begin{cases}
(i+1,j), & i\ne b-1,\\
(0,j+1), & i=b-1,
\end{cases}
\]

lies in $S_{n/b}\wr S_b$ and is an $n$-cycle. $\square$

\paragraph{Lemma 5.3.}

Let $n$ be composite and $q=P^-(n)$. Then

\[
\max_{\substack{b\mid n\\ 1<b<n}}
((n/b)!)^b b!
=
((n/q)!)^q q!.
\]

\noindent\textbf{Proof.}

Put

\[
D_b=((n/b)!)^b b!.
\]

Let

\[
F(x)=x\log\Gamma(n/x+1)+\log\Gamma(x+1).
\]

Then $F(b)=\log D_b$ for every divisor $b\mid n$.

Writing $y=n/x$, we have

\[
F'(x)
=
\log\Gamma(y+1)-y\psi(y+1)+\psi(x+1),
\]

and

\[
F''(x)
=
\frac{y^2}{x}\psi'(y+1)+\psi'(x+1)>0.
\]

Thus $F$ is convex.

Every proper divisor $b$ of $n$ satisfies

\[
q\le b\le n/q.
\]

A convex function on this interval attains its maximum at an endpoint. Write $r=n/q$. Then

\[
D_q=(r!)^q q!,
\qquad
D_r=(q!)^r r!,
\]

so

\[
\frac{D_q}{D_r}
=
\frac{(r!)^{q-1}}{(q!)^{r-1}}.
\]

The function

\[
t\mapsto (t!)^{1/(t-1)}
\]

is increasing for $t\ge2$, because

\[
((t+1)!)^{1/t}\ge (t!)^{1/(t-1)}
\]

is equivalent to

\[
(t+1)^{t-1}\ge t!,
\]

which is immediate. Hence $D_q\ge D_r$, and the maximum occurs at $b=q$. $\square$

\paragraph{Lemma 5.4.}

For all sufficiently large $n$, if $b\mid n$, $3\le b<n$, and $a=n/b$, then

\[
[S_n:S_a\wr S_b]>2^n.
\]

The analogous alternating-group index is $>2^{n-1}$.

\noindent\textbf{Proof.}

Let

\[
I(n,b)=\frac{n!}{(a!)^b b!},
\qquad a=n/b.
\]

We show uniformly that

\[
\log I(n,b)>n\log2
\]

for $n$ large.

If

\[
3\le b\le \frac{n}{(\log n)^2},
\]

then Stirling's formula gives

\[
\log I(n,b)
=
n\log b-b\log b+b+O(b\log a+\log b).
\]

Since $b\le n/(\log n)^2$, the error is $O(n/\log n)$, while $b\log b=o(n)$. Hence

\[
\log I(n,b)\ge n\log3-o(n)>n\log2
\]

for large $n$.

If

\[
b>\frac{n}{(\log n)^2},
\]

then

\[
a<(\log n)^2.
\]

Writing $b=n/a$, we get

\[
\log I(n,b)
=
n\left(1-\frac1a\right)\log n
+
O(n\log\log n).
\]

Since $a\ge2$, this is at least

\[
\frac n2\log n-O(n\log\log n)>n\log2
\]

for all sufficiently large $n$.

The alternating-group statement differs by at most a factor $2$. $\square$

\section{Symmetric groups}

\paragraph{Theorem 6.1.}

For all sufficiently large odd $n$,

\[
\beta_{\mathrm{faith}}(S_n)=2^{n-1}+1.
\]

Moreover, the optimal cover is unique up to conjugacy and ordering: it consists of

\[
A_n
\]

and

\[
S_k\times S_{n-k},
\qquad
1\le k\le \frac{n-1}{2}.
\]

\noindent\textbf{Proof.}

Let $n=2m+1$.

For the upper bound, take $A_n$ and the intransitive subgroups

\[
S_k\times S_{n-k},
\qquad
1\le k\le m.
\]

The cost is

\[
2+\sum_{k=1}^m\binom nk
=
2^{n-1}+1.
\]

The subgroup $A_n$ covers all even permutations. If $g$ is odd, then $g$ is not an $n$-cycle, because $n$-cycles are even when $n$ is odd. Hence $g$ has at least two cycles and preserves a nonempty proper union of cycle supports of size $k\le m$. Thus $g$ lies in a conjugate of $S_k\times S_{n-k}$.

For the lower bound, assign weight $2$ to the class of $n$-cycles and weight $\binom nk$ to the class of type $(k,n-k)$, for $1\le k\le m$. All other classes have weight $0$.

The subgroup $A_n$ sees only the $n$-cycle class among the positive-weight classes, hence sees weight $2=[S_n:A_n]$. An intransitive subgroup that sees $(k,n-k)$ has index at least $\binom nk$ by Lemma 3.1, and sees no other positive-weight two-cycle class. A transitive proper subgroup other than $A_n$ has index greater than the total weight for $n$ large, by Lemmas 5.1 and 5.4. Thus Proposition 4.1 gives the lower bound.

For uniqueness, use Proposition 4.2. Any optimal cover must use only tight subgroups. The tight subgroups for the above certificate are exactly $A_n$ and the subgroups $S_k\times S_{n-k}$, $1\le k\le m$. Each positive-weight class must be covered exactly once. Therefore every optimal cover consists precisely of these subgroups, up to conjugacy and ordering. $\square$

\paragraph{Theorem 6.2.}

For $n\to\infty$ through even integers,

\[
\beta_{\mathrm{faith}}(S_n)
=
\frac12\binom n{n/2}
+
\Theta\!\left(\binom n{\lfloor n/3\rfloor}\right).
\]

\noindent\textbf{Proof.}

Let $n=2m$. Since $n\ge5$,

\[
\beta_{\mathrm{faith}}(S_n)=\beta(S_n).
\]

First, an $n$-cycle is odd. Therefore every normal cover of $S_n$ must contain a subgroup whose conjugates meet the $n$-cycle class. The least index of a proper subgroup containing an $n$-cycle is

\[
[S_n:S_m\wr S_2]=\frac12\binom nm.
\]

Indeed, this is the least imprimitive index by Lemmas 5.2 and 5.3, and primitive subgroups have larger index for large $n$ by Lemma 5.1. Thus

\[
\beta(S_n)\ge \frac12\binom nm.
\]

We now prove the lower bound for the second term. Choose a three-cycle type $\tau_n$ as follows.

If $n=6r$, take

\[
\tau_n=(2r-1,2r,2r+1).
\]

If $n=6r+2$, take

\[
\tau_n=(2r,2r+1,2r+1).
\]

If $n=6r+4$, take

\[
\tau_n=(2r+1,2r+1,2r+2).
\]

Let

\[
a_n=\min\tau_n.
\]

Then

\[
a_n=\lfloor n/3\rfloor+O(1),
\]

and

\[
\binom n{a_n}\asymp \binom n{\lfloor n/3\rfloor}.
\]

The class $\tau_n$ consists of odd permutations, because it has exactly three cycles and $n$ is even. It is not contained in any conjugate of

\[
W=S_m\wr S_2.
\]

Indeed, an element of $W$ either preserves the two blocks, in which case some nonempty proper union of its cycle supports has size $m$, or interchanges the two blocks, in which case all of its cycles have even length. The chosen $\tau_n$ has at least one odd part and has no nonempty proper subset sum equal to $m$, for all sufficiently large $n$.

Assign weight

\[
I_n=\frac12\binom nm
\]

to the class of $n$-cycles, and weight

\[
B_n=\binom n{a_n}
\]

to the class $\tau_n$.

We verify the dual inequalities. A subgroup contained in a conjugate of $W$ sees the $n$-cycle class but not $\tau_n$, so it sees weight at most $I_n=[S_n:W]$. A subgroup containing an $n$-cycle but not contained in a conjugate of $W$ has index $>2^n$ for large $n$, by Lemmas 5.1 and 5.4, while $I_n+B_n=o(2^n)$.

If an intransitive subgroup sees $\tau_n$, then it has an orbit whose size is a nonempty proper subset sum of the three parts of $\tau_n$. The smallest such size not exceeding $n/2$ is $a_n$. Hence its index is at least

\[
\binom n{a_n}=B_n.
\]

A transitive subgroup seeing $\tau_n$ but not lying in a conjugate of $W$ again has index $>2^n$ for large $n$, unless it is $A_n$; but $A_n$ does not see $\tau_n$, since $\tau_n$ is odd. Thus the dual certificate is valid, and

\[
\beta(S_n)
\ge
\frac12\binom nm
+
c\binom n{\lfloor n/3\rfloor}
\]

for some absolute $c>0$.

For the upper bound, take $W=S_m\wr S_2$, take $A_n$, and take all intransitive subgroups

\[
S_k\times S_{n-k},
\qquad
1\le k\le n/3.
\]

The subgroup $W$ covers the $n$-cycles, and $A_n$ covers all even permutations. Any odd permutation which is not an $n$-cycle has at least three cycles and therefore has a cycle of length at most $n/3$. Hence it is covered by one of the intransitive subgroups above.

The extra cost beyond $W$ is

\[
2+\sum_{k\le n/3}\binom nk
=
O\!\left(\binom n{\lfloor n/3\rfloor}\right),
\]

since the binomial coefficients are increasing up to $n/2$ and the sum up to $n/3$ is dominated by its final term. Therefore

\[
\beta(S_n)
\le
\frac12\binom nm
+
C\binom n{\lfloor n/3\rfloor}
\]

for some absolute $C>0$. $\square$

\section{Alternating groups of even degree}

\paragraph{Theorem 7.1.}

For all sufficiently large $m$,

\[
\beta_{\mathrm{faith}}(A_{2m})
=
\sum_{\substack{1\le k<m\\ k\text{ odd}}}\binom{2m}{k}
+
\frac12\binom{2m}{m}.
\]

Moreover, the optimal cover is unique up to conjugacy and ordering: it consists of

\[
(S_m\wr S_2)\cap A_{2m}
\]

and the subgroups

\[
(S_k\times S_{2m-k})\cap A_{2m},
\qquad
1\le k<m,\ k\text{ odd}.
\]

\noindent\textbf{Proof.}

Let $n=2m$. Since $A_n$ is simple for $n\ge5$,

\[
\beta_{\mathrm{faith}}(A_n)=\beta(A_n).
\]

For the upper bound, take

\[
W=(S_m\wr S_2)\cap A_n
\]

and

\[
H_k=(S_k\times S_{n-k})\cap A_n
\]

for $k<m$ odd. The cost is

\[
\frac12\binom nm
+
\sum_{\substack{1\le k<m\\ k\text{ odd}}}\binom nk.
\]

If $g\in A_n$ has only even cycles, then alternatingly coloring each cycle puts $g$ in a conjugate of $W$. If $g$ has odd cycles, their number is even. If one odd cycle has length $k<m$, then $g$ lies in a conjugate of $H_k$. The remaining case is two odd cycles both of length $m$, and this is covered by $W$.

For the lower bound, assign total weight $\binom nk$ to the two split $A_n$-classes of type $(k,n-k)$, for each odd $k<m$, and weight $\frac12\binom nm$ to the class $(m,m)$.

The dual inequalities follow as follows. Intransitive subgroups seeing $(k,n-k)$ have index at least $\binom nk$. Intransitive subgroups seeing $(m,m)$ have index at least $\binom nm$, larger than the assigned weight. The subgroup $W$ sees only $(m,m)$, and sees exactly its index in weight. Other transitive imprimitive subgroups have at least three blocks and index $>2^{n-1}$, and primitive subgroups are excluded by Lemma 5.1. Thus Proposition 4.1 gives equality.

The uniqueness statement follows from Proposition 4.2. The only tight subgroups are $W$ and the $H_k$ with $k<m$ odd. Every positive-weight class must be covered exactly once. $\square$

The equivalent closed form follows from elementary binomial identities:

\[
\beta_{\mathrm{faith}}(A_{2m})
=
\begin{cases}
2^{2m-2}, & m\text{ odd},\\[4pt]
2^{2m-2}+\dfrac12\binom{2m}{m}, & m\text{ even}.
\end{cases}
\]

\section{Alternating groups of odd composite degree}

\paragraph{Theorem 8.1.}

Let $n$ be odd composite and let $q=P^-(n)$. Then

\[
\beta_{\mathrm{faith}}(A_n)
\sim
\frac{n!}{((n/q)!)^q q!}.
\]

\noindent\textbf{Proof.}

Let

\[
C_n=\frac{n!}{((n/q)!)^q q!}.
\]

Every normal cover of $A_n$ must cover the $n$-cycles. Therefore at least one subgroup must contain an $n$-cycle. Such a subgroup is transitive. If it is imprimitive, Lemmas 5.2 and 5.3 show that its index is at least $C_n$.

If it is primitive, we use the stronger form of Maroti's estimate:

\[
|H|<50n^{\sqrt n}.
\]

Hence

\[
[A_n:H]>\frac{n!/2}{50n^{\sqrt n}}.
\]

We compare this with $C_n$. It is enough to show

\[
((n/q)!)^q q!>100n^{\sqrt n}.
\]

Since $q\le \sqrt n$, we have $n/q\ge \sqrt n$. Therefore

\[
\log\left(((n/q)!)^q q!\right)
\ge
q\left(\frac nq\log\frac nq-\frac nq\right)+O(q\log q)
=
n\log\frac nq-n+O(q\log q).
\]

As $n/q\ge\sqrt n$,

\[
n\log\frac nq\ge \frac12 n\log n.
\]

Thus

\[
\log\left(((n/q)!)^q q!\right)
\ge
\frac12 n\log n-O(n),
\]

whereas

\[
\log(100n^{\sqrt n})=O(\sqrt n\log n).
\]

For $n$ large, the primitive index is therefore larger than $C_n$. Hence

\[
\beta(A_n)\ge C_n.
\]

For the upper bound, take

\[
W_q=(S_{n/q}\wr S_q)\cap A_n.
\]

This subgroup has index $C_n$ and, by Lemma 3.3, meets both $A_n$-classes of $n$-cycles.

Every element of $A_n$ which is not an $n$-cycle has at least three cycles, and therefore has a cycle of length $k\le n/3$. Adding all intransitive subgroups

\[
(S_k\times S_{n-k})\cap A_n,
\qquad
1\le k\le n/3,
\]

covers all remaining elements. The total added cost is

\[
\sum_{k\le n/3}\binom nk
=
2^{H(1/3)n+o(n)}.
\]

Since $q\ge3$, $C_n$ grows like at least $3^n$ up to subexponential factors, while $2^{H(1/3)n}$ has smaller exponential base. Thus the added cost is $o(C_n)$. $\square$

\section{Alternating groups of prime degree}

Let

\[
c_p=\min\{[A_p:H]:H<A_p,\ H\text{ contains a }p\text{-cycle}\}.
\]

\paragraph{Lemma 9.1.}

\[
\log c_p=p\log p-p+O((\log p)^2).
\]

\noindent\textbf{Proof.}

Let

\[
M_p=\max\{|H|:H<A_p,\ H\text{ contains a }p\text{-cycle}\}.
\]

The lower bound $M_p\ge p$ is immediate from the cyclic group generated by a $p$-cycle.

Conversely, any subgroup of $A_p$ containing a $p$-cycle is transitive of prime degree, hence primitive. Jones's classification of primitive permutation groups containing a cycle implies that every proper primitive subgroup of $S_p$ containing a $p$-cycle has order at most

\[
p^{O(\log p)}.
\]

Thus

\[
M_p\le p^{O(\log p)}.
\]

Since

\[
c_p=\frac{|A_p|}{M_p},
\]

Stirling's formula gives

\[
\log c_p
=
\log(p!/2)-O((\log p)^2)
=
p\log p-p+O((\log p)^2).
\]

$\square$

\paragraph{Theorem 9.2.}

As $p\to\infty$ through primes,

\[
\beta_{\mathrm{faith}}(A_p)=(1+o(1))c_p.
\]

Moreover,

\[
\log\beta_{\mathrm{faith}}(A_p)
=
p\log p-p+O((\log p)^2).
\]

\noindent\textbf{Proof.}

Every cover must cover the $p$-cycles, so

\[
\beta(A_p)\ge c_p.
\]

Choose $H_0<A_p$ containing a $p$-cycle and satisfying

\[
[A_p:H_0]=c_p.
\]

By Lemma 3.4, $H_0$ meets both $A_p$-classes of $p$-cycles.

Every element of $A_p$ which is not a $p$-cycle has at least three cycles, hence has a cycle of length $k\le p/3$. Adding all intransitive subgroups

\[
(S_k\times S_{p-k})\cap A_p,
\qquad
1\le k\le p/3,
\]

covers all remaining elements, at cost

\[
2^{H(1/3)p+o(p)}.
\]

By Lemma 9.1 this is $o(c_p)$. Therefore

\[
\beta(A_p)=(1+o(1))c_p.
\]

The logarithmic asymptotic follows from Lemma 9.1. $\square$

\section{The exact second term for even symmetric groups}

Theorem 6.2 determines the order of the second term in

\[
\beta_{\mathrm{faith}}(S_{2m})
-
\frac12\binom{2m}{m}.
\]

It is natural to ask for the exact coefficient. This becomes a finite additive-combinatorial covering problem on cycle types.

Let $n$ be even and let

\[
W=S_{n/2}\wr S_2.
\]

An odd cycle type of $S_n$, other than the $n$-cycle type, is already covered by $W$ if either all cycle lengths are even, or some nonempty subset of its cycle lengths has sum $n/2$. The residual classes are those odd cycle types with no subset sum $n/2$ and with at least one odd cycle length.

Define $\mathcal R_n$ to be the set of such residual cycle types. Let

\[
\rho(n)
=
\min_{\mathcal T\subseteq\{1,\dots,n/2\}}
\sum_{k\in\mathcal T}\binom nk,
\]

where the minimum is taken over all $\mathcal T$ such that every $\lambda\in\mathcal R_n$ has a nonempty proper subset of parts whose sum lies in $\mathcal T$.

Then the upper-bound construction in Theorem 6.2 shows

\[
\beta_{\mathrm{faith}}(S_n)
\le
\frac12\binom n{n/2}
+
2+\rho(n).
\]

The lower-bound problem is to construct matching dual certificates for this residual weighted hitting problem. Determining $\rho(n)$ and proving equality would give the exact second term. Theorem 6.2 shows that

\[
\rho(n)=\Theta\left(\binom n{\lfloor n/3\rfloor}\right),
\]

but the exact coefficient remains open.

\section{Arithmetic interpretation}

Let $L/K$ be a finite Galois extension of global fields with group $G$. Finite etale $K$-schemes split by $L$ are equivalent to finite $G$-sets. Under this equivalence, the degree of the finite etale scheme is the cardinality of the $G$-set, and exact monodromy $G$ corresponds to faithfulness of the action.

If $Z/K$ corresponds to $\Omega$, then

\[
Z(K)\ne\varnothing
\]

if and only if

\[
\Omega^G\ne\varnothing.
\]

For almost all good primes $\mathfrak p$,

\[
Z(k_{\mathfrak p})\ne\varnothing
\]

if and only if

\[
\Omega^{\operatorname{Frob}_{\mathfrak p}}\ne\varnothing.
\]

By Chebotarev, this condition for almost all good primes is equivalent to

\[
\Omega^g\ne\varnothing
\qquad
\text{for every }g\in G.
\]

Thus $\beta_{\mathrm{faith}}(G)$ is the minimum degree of a finite etale $K$-scheme split by $L$, with exact monodromy $G$, with no $K$-point, but with points over $k_{\mathfrak p}$ for almost all good primes. In the polynomial formulation, it is the group-theoretic lower bound for the minimum total degree of an intersective product with Galois group $G$, not merely for the minimum number of irreducible factors.

This interpretation concerns almost all good primes. Requiring local points over every completion $K_v$, or roots modulo every prime, imposes additional conditions at ramified primes and is a separate local problem.

\section*{References}
\small
\sloppy

[BS] D. Bubboloni and J. Sonn, \emph{Intersective $S_n$ polynomials with few irreducible factors}, arXiv:1507.08593.

[BPS] D. Bubboloni, C. E. Praeger, and P. Spiga, \emph{Normal coverings and pairwise generation of finite alternating and symmetric groups}, Journal of Algebra 390 (2013), 199-215.

[EM] S. Eberhard and C. Mellon, \emph{Normal covering numbers for $S_n$ and $A_n$ and additive combinatorics}, arXiv:2410.06999.

[GL] M. Garonzi and A. Lucchini, \emph{Covers and normal covers of finite groups}, Journal of Algebra 422 (2015), 148-165.

[J] G. A. Jones, \emph{Primitive permutation groups containing a cycle}, Bulletin of the Australian Mathematical Society 89 (2014), 159-165.

[K] J. Konig, \emph{Intersective polynomials with non-solvable Galois groups}, Journal of Algebra 502 (2018), 79-94.

[M] A. Maroti, \emph{On the orders of primitive groups}, Journal of Algebra 258 (2002), 631-640.

[PS] C. E. Praeger and J. Saxl, \emph{On the orders of primitive permutation groups}, Bulletin of the London Mathematical Society 12 (1980), 303-307.

\section*{Declaration of generative AI and AI-assisted technologies in the writing process}
During the preparation of this work, ChatGPT, by OpenAI, was used to assist with mathematical drafting, formalization, review, and editing. This work is shared as a preliminary AI-assisted mathematical note. The mathematical content may have been only partially reviewed and may contain errors; it should not be treated as peer-reviewed or as a fully verified manuscript.

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