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Given a finite group $G$, let $p(G)$ denote the number of prime factors of the order of $G$ (counting multiplicities).

Does there exist a function $f: \mathbb{N} \rightarrow \mathbb{N}$ which grows faster than $O(1)$ and such that for every finite group $G$ and every maximal subgroup $M$ of $G$ we have $p(M) > f(p(G))$?

If the answer is yes, how fast can $f$ grow at most?

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    $\begingroup$ Probably not. If there is a sequence of primes powers $(p_n)$ such that $p_n-1$ is divisible by boundedly many primes, but not $p_n+1$, then $\mathrm{GL}_2(p_n)$ would do the job, with $M$ being the upper triangular subgroup. (I guess such $(p_n)$ is conjectured to exist.) $\endgroup$
    – YCor
    Commented Mar 10 at 10:27
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    $\begingroup$ $A_5$ is a maximal subgroup of $\operatorname{PSL}(2,p)$ when $p$ is $\pm 1$ mod $10$. $\endgroup$
    – Dave Benson
    Commented Mar 10 at 10:58
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    $\begingroup$ @DaveBenson Ah, indeed! -- That should answer the question. -- Thank you! $\endgroup$
    – Stefan Kohl ♦
    Commented Mar 10 at 11:08
  • $\begingroup$ Yes, I think you just need to combine that with Erdős's work on the number of prime factors of $p-1$. $\endgroup$
    – Dave Benson
    Commented Mar 10 at 11:10
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    $\begingroup$ @DaveBenson Erdös' work is not even needed here. -- Just Dirichlet's prime number theorem (there exist primes congruent to 1 modulo $10 \cdot 2^k$ for every $k$). $\endgroup$
    – Stefan Kohl ♦
    Commented Mar 10 at 11:12

2 Answers 2

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$A_5$ is a maximal subgroup of $\operatorname{\rm PSL}(2,p)$ when $p$ is $\pm 1$ mod $10$. Combining this with Dirichlet's theorem for primes congruent to one modulo $10$ times a highly composite number proves that there is no such bound.

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    $\begingroup$ A reference for the statement about $A_5$ is Suzuki's "Group Theory I" §3.6. $\endgroup$
    – Dave Benson
    Commented Mar 10 at 11:26
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    $\begingroup$ Even better (or worse?) $A_4$ is a maximal subgroup of ${\rm PSL}(2,p)$ whenever $p \equiv \pm3, \pm 13 \bmod 40$. $\endgroup$
    – Derek Holt
    Commented Mar 10 at 12:14
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    $\begingroup$ By way of contrast, using a theorem of Feit and Thompson ( not the odd order theorem), it follows that if G is a finite group which has a maximal subgroup isomorphic to $S_{3}$, then $|G|$ has at most four prime divisors, (including multiplicities). $\endgroup$
    – Geoff Robinson
    Commented Mar 11 at 5:16
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    $\begingroup$ I wonder whether it's possible to classify the groups $H$ with the property that there exists $n>0$ such that if $H$ is a maximal subgroup of a finite group $G$ then $G$ has at most $n$ prime divisors (there are two versions of this wonderment, depending on whether you want to count multiplicities). $\endgroup$
    – Dave Benson
    Commented Mar 13 at 14:54
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I claim that there are finite groups whose order has arbitrarily many repeating prime factors but which contain a cyclic maximal subgroup of prime order.

Let $F$ be a finite field of order $q=p^m$. Let $a\in F$ and suppose that the order of $a$ is $n$. Define a group automorphism $\phi:F\rightarrow F$ by setting $\phi(b)=ab$. Then we can use this automorphism to obtain a semi-direct product $G$ of $\mathbb{Z}_n$ with $F$.

We observe that $\mathbb{Z}_n$ is a subgroup of $G$. If $a$ is not contained in any proper subfield of $F$, then $\mathbb{Z}_n$ is a maximal subgroup of $G$. We have $p(G)=m+\Omega(n)$ where $\Omega(n)$ denotes the number of prime factors of $n$ ($p(\mathbb{Z}_n)=\Omega(n)$). I claim that we can make $n$ prime while $m$ is arbitrarily large so that $\Omega(n)=1$ and $p(G)=m+1$ while $p(\mathbb{Z}_n)=1$

Suppose that $p,n$ are primes with $n>p$. Then as a consequence of the base $p$ expansion of $1/n$, there is some $m$ where $n$ is a factor of $p^m-1$. Let $m$ be the least positive integer where $n$ is a factor of $m$. In this case, we can find an element $a\in F_q$ with order $m$ and here $a$ is not contained in any proper subfield.

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    $\begingroup$ I think you can also see this by Zsygmondy's theorem. Let n be a positive integer, say n > 6, and let q be a prime. Then there is a prime p such that p divides q^n -1, but p does not divide q^i - 1 for 0 < I < n. Let P be a cyclic group of order p. Then there is a semidirect product PV, where V is elementary Abelian of order q^n, and P acts irreducibly on V. Then P is a maximal subgroup of prime order of PV, but the order of PV has n+1 prime divisors $\endgroup$
    – Geoff Robinson
    Commented Mar 14 at 1:42

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