I posted this question on Math.SE (link), but it didn't get any answers so I'm going to ask here. This is an edited version of the question.

Let $p$ be a prime and $n \geq 1$ some integer. Furthermore, let $G$ be a finite group where $p$-Sylow subgroups have order $p^n$. Denote by $n_p(G)$ the number of Sylow $p$-subgroups of $G$. Denote the number of elements in the union of all Sylow $p$-subgroups of $G$ by $f_p(G)$. I am interested in finding lower bounds for $f_p(G)$ that do not depend on the group $G$, but only on $p$, $n$ and $n_p(G)$.

By Sylow's theorem, we know that $n_p(G) = kp + 1$ for some integer $k \geq 0$. What I know so far:

  • If $k = 0$, then $f_p(G) = p^n$.
  • If $k = 1$, then $f_p(G) = p^{n+1}$.
  • If $k \geq 2$, then $f_p(G) \geq 2p^{n+1} - p^n$.

This is a theorem due to G. A. Miller, see also this question from Math.SE. To prove the inequality in the case $k \geq 2$, you first prove that then $f_p(G) > p^{n+1}$. Then observe that $f_p(G) - 1$ is divisible by $p-1$, then the inequality follows from Frobenius theorem (*). Details are in a book of Miller, Blichfeldt and Dickson (Theory and Applications of Finite Groups) and a paper of Miller ("Some deductions from Frobenius Theorem").

My main question is the following:

What is a better lower bound for the case $k > 2$?

The case $n = 1$ is easy, because then we know the value of $f_p(G)$ precisely. If $n = 1$, then $f_p(G) = n_p(G)(p-1)+1$. What about when $n > 1$? Answers regarding particular $n$ or particular $k$ are also welcome.

If the Sylow $p$-subgroups are cyclic, then we have $f_p(G) \geq n_p(G)\varphi(p^n) + p^{n-1}$ and this bound is okay. But most $p$-groups are not cyclic..

I think the following example shows that $f_p(G)$ gets arbitrarily large values for fixed $p$ and $n$ (not surprising). By Dirichlet's theorem, there exist arbitrarily large primes $q$ such that $q \equiv 1 \mod{p}$. Then in a direct product $G = C_{p^{n-1}} \times H$, where $H$ is a non-abelian group of order $pq$, the Sylow subgroups of $G$ have $C_{p^{n-1}}$ as their common intersection. There are exactly $q$ Sylow $p$-subgroups, because otherwise $G$ would be nilpotent but its subgroup $H$ is not. Therefore the number of elements in the $p$-Sylow subgroups is $f_p(G) = q(p^{n} - p^{n-1}) + p^{n-1}$, and this goes to infinity as $q$ goes to infinity. Thus there exist groups $G$ with Sylow $p$-subgroups of order $p^n$ such that $f_p(G)$ is arbitrarily large.

Also, $f_p(G) \rightarrow \infty$ as $k \rightarrow \infty$. This is seen by noticing that $f_p(G)^{p^n} \geq n_p(G)$, so

$$f_p(G) \geq (kp + 1)^{p^{-n}}$$

which goes to infinity as $k \rightarrow \infty$.

One more observation: not all integers $\equiv 1 \mod{p}$ are possible amounts of Sylow $p$-subgroups. For example, there does not exist a group with exactly $22$ Sylow $3$-subgroups, although $22 \equiv 1 \mod{3}$. I don't know if this complicates things.

(*) Frobenius Theorem says that when $G$ is a finite group with order divisible by $s$, the number of solutions to $x^s = 1$ in $G$ is a multiple of $s$. We know that $f_p(G)$ is the number of solutions to $x^{p^n} = 1$ in $G$.

  • $\begingroup$ Nice question. You say you want a formula for $f_p(G)$ depending only on $n_p(G)$ but you seem to be happy to have $p$ and $n$ in the formula as well, right? (Well, there'd be no hope if not, so I hope this is OK!) Perhaps you should adjust the question to make this clear. $\endgroup$
    – Nick Gill
    Jan 21, 2013 at 10:49
  • $\begingroup$ @Nick Gill: Yes, the lower bound will of course depend on $p$ and $n$, just like the lower bound given by Miller's thm does. Basically given $p$ and $n$, we're looking at functions $g$ such that $f_p(G) \geq g(k)$ for any finite group $G$ with Sylow subgroups of order $p^n$ and $n_p(G) = kp + 1$. Is this what you meant? $\endgroup$ Jan 21, 2013 at 12:33
  • $\begingroup$ Perhaps one way to start with this is that given $n$, $p$ and $k$, find the smallest possible value for $f_p(G)$. I don't know if the bound $2p^{n+1}−p^n$ is sharp for the case $k=2$. It is sharp for all primes $p$ such that $2p+1$ is prime, which can be seen by the construction in my question. $\endgroup$ Jan 21, 2013 at 13:21
  • $\begingroup$ @Mikko, no that's not what I meant. I'm suggesting that you replace "a formula for $f_p(G)$ depending only on $n_p(G)$" by "a formula for $f_p(G)$ depending only on $p$, $n$ and $n_p(G)$". $\endgroup$
    – Nick Gill
    Jan 23, 2013 at 9:43
  • 1
    $\begingroup$ @Nick: Ok, I'll edit the post. What I mean by the construction is this. Fix a prime $p$ and integer $n \geq 1$. The construction shows that we can find a group $G$ with Sylow $p$-subgroups of order $p^n$ (not just some arbitrary power of $p$ like in your comment) such that $f_p(G)$ is arbitrarily large. $\endgroup$ Jan 23, 2013 at 11:02

1 Answer 1


This is just a small partial result and some comments. I think the following should settle the case $k = 2$.

Suppose that $G$ is a group with Sylow $p$-subgroups of order $p^n$ and that $n_p(G) = 2p + 1$. According to a theorem of Marshall Hall (see theorem 3.1 in [*]), this can only happen if $2p + 1 = q^t$ is a power of a prime, so let's assume that this is the case.

First of all, the lower bound $p^n(2p - 1)$ given in the question is attained. Let

$$G = C_{p^{n-1}} \times AGL(1, q^t),$$

where $AGL(1, q^t)$ is the group of invertible affine transformations $x \mapsto ax + b$ of the field of order $q^t$. Here $G$ has exactly $2p + 1$ Sylow $p$-subgroups, the Sylow $p$-subgroups have order $p^n$ and $f_p(G) = p^n(2p - 1)$.

Now by Frobenius theorem $f_p(G) = tp^n$ where $t$ is an integer. Since $2p - 1 \leq t < 2p + 1$, we see that $t = 2p - 1$ or $t = 2p$.

If $p \neq 2$, then $t = 2p - 1$ because $t-1$ must be a multiple of $p - 1$ [**]. Therefore in this case $f_p(G) = p^n(2p - 1)$.

If $p = 2$ and $n \geq 2$, then $f_2(G) = 2^{n+2}$ is attained by a semidirect product $G = C_{2^n} \ltimes_\theta C_5$ (I think, I'll check this later).

I have not made much progress for the cases where $n_p(G) = kp + 1$ and $k > 2$. In the case where $n_p(G) = 3p + 1$, a theorem of Marshall Hall (see theorem 3.2 in [*]) shows that $p = 2$, $p = 3$ or $p = 5$. It seems that things get messy from now on with this approach, perhaps it's best to disregard "impossible values" like $n_3(G) = 22$. Or we could start with the case $p = 2$ where there are no impossible values.

[*] M. Hall, On the number of Sylow subgroups in a finite group (1967) DOI link

[**] Proof: Now $f_p(G) - 1$ is the number of elements of order $p^k$, where $1 \leq k \leq n$. Since the number of elements of order $s$ is always a multiple of $\varphi(s)$, we get that $f_p(G) - 1$ must be a multiple of $p-1$. Thus $t-1$ is also a multiple of $p-1$.


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