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Inspired by the answers given by these three questions (here, here, and here), what is the general solution for constructing a group with a specific number of Sylow subgroups?

That is, given a prime $p$ and a positive integer $n\equiv1\pmod p$, is it always possible to construct a group $G$ with exactly $n$ subgroups of order $p$?

For example, is it possible to construct a group with exactly $15$ subgroups of order $7$, or exactly $35$ subgroups of order $17$?

From Hölder's Theorem (Theorem 19 here), if there is at least one prime $q \mid n$ such that $q \neq 1\pmod p$, then we can conclude that $g=|G|$ cannot be squarefree.

At the same time, it is not a necessary condition that every prime $q\mid n$ is congruent to $1\pmod p$.

In our example with $n=15$ and $p=7$, we know that $g$ cannot be square-free. From the Sylow Theorems, $np \mid g$, so $g$ is a multiple of $105$. However, $g$ is at least $315$ because of the requirement that $g$ must not be square-free. That is as far as I can get with constructing such groups.

Edit:

Derek's answer suggests that the answer to my original question is false in general. That is, it is not always possible to construct a group with $n$ subgroups of order $p$, even when $n$ is restricted to $1\pmod p$. How would one prove this?

Secondly, under what conditions would there exist a group with exactly $n$ $p$-Sylow subgroups?

user26857
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J. Linne
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2 Answers2

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No, there is no finite group with exactly $15$ Sylow $7$-subgroups.

Let $G$ be the image of such a group in $S_{15}$ under the conjugation action of the group on its Sylow $7$-subgroups.

Then $G$ would be a transitive group of degree $15$ with order divisible by $7$.

There are databases of the transitive groups of degree $n$ for small $n$ available in GAP and Magma. From these, we find that there are $104$ transitive groups of degree $15$ (up to conjugation in $S_{15}$), but only four of these have order divisible by $7$ (numbers 47, 72, 103, and 104), and you can check by computer that none of these group have exactly $15$ Sylow $7$- subgroups - in fact these numbers are 120, 960, 370656000, and 370656000.

I expect this could be proved without using computers, but I will leave that to someone else!

Derek Holt
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For a prime $p$ and finite group $G$, let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$.

By Sylow's theorem we know that $n_p(G) \equiv 1 \pmod{p}$. Your question is whether for any integer $n \equiv 1 \pmod{p}$ there exists a finite group $G$ such that $n_p(G) = n$.

For $p = 2$ the answer is yes, since for any odd integer $n$ the dihedral group $G = D_{2n}$ has $n_2(G) = n$.

But for every prime $p > 2$ there exist an integer $n \equiv 1 \pmod{p}$ such that there is no finite group $G$ with $n_p(G) = n$. This is a result of Marshall Hall:

Marshall Hall, Jr., On the number of Sylow subgroups in a finite group, J. Algebra 7 (1967), 363-371. DOI

For example:

  • There is no finite group $G$ with $n_3(G) = 22$.
  • There is no finite group $G$ with $n_5(G) = 21$.
  • For every prime $p > 5$, there is no finite group $G$ with $n_p(G) = 1+3p$.

Hall also describes integers $n \equiv 1 \pmod{p}$ such that there exists a finite group $G$ with $n_p(G) = n$. By Theorem 2.2 from his paper:

Let $G$ be a finite group. Then $$n_p(G) = n_p(X_1) \cdots n_p(X_t) q_1 \cdots q_s$$ where $X_i$ is a finite simple group for all $1 \leq i \leq t$, and $q_i$ is a prime power $\equiv 1 \pmod{p}$ for all $1 \leq i \leq s$.

Note that conversely for $n = n_p(X_1) \cdots n_p(X_t) q_1 \cdots q_s$, for the group $$G = X_1 \times \cdots \times X_t \times \operatorname{AGL}(1,q_1) \times \cdots \times \operatorname{AGL}(1,q_s)$$ we have $n_p(G) = n$.

So to understand integers $n$ of the form $n = n_p(G)$, we are reduced to the case where $G$ is simple. With the classification of finite simple groups, in principle you could analyze the values of $n_p(G)$ further, but I am not sure if there is much more to be said in general.

EDIT: Some related references:

  • Marcel Herzog, Counting group elements of order $p$ modulo $p^2$. Proc. Amer. Math. Soc. 66 (1977), no. 2, 247–250. DOI

  • Benjamin Sambale, Pseudo Frobenius numbers. Expo. Math. 37 (2019), no. 2, 200-206. DOI

  • Benjamin Sambale, Pseudo Sylow numbers. Amer. Math. Monthly 126 (2019), no. 1, 60-65. DOI

  • Masafumi Murai, On the number of p-subgroups of a finite group. J. Math. Kyoto Univ. 42 (2002), no. 1, 161–174. link

There is also a related question by Jack Schmidt.

Mikko Korhonen
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  • Interesting! Thanks for the references. In the same paper you referenced in your answer, it appears that (3.1) is an almost necessary condition for there to be a group with $n$ $p$-Sylow Subgroups. I am not sure if it also a sufficient condition? If so, then that would imply the existence of groups that contain exactly $50, 85$ and $120$ subgroups of order $7$. – J. Linne Mar 16 '22 at 04:35
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    I doubt it is sufficient. $\operatorname{PSL}(2,7^2)$ has $50$ Sylow $7$-subgroups, and $A_7$ has $120$ Sylow $7$-subgroups. But it seems to me there is no finite simple group with $85$ Sylow $7$-subgroups, hence by thm 2.2 no finite group with $85$ Sylow $7$-subgroups. – Mikko Korhonen Mar 16 '22 at 05:18
  • I notice now that you also ask about a slightly different question. Given a prime $p$ and integer $n \equiv 1 \mod{p}$, when does there exist a group $G$ with $n$ subgroups of order $p$? Well, if $p^2 \nmid |G|$, then this is a question about Sylow $p$-subgroups as before. If $p^2 \mid |G|$, then by a theorem of Kulakoff and Hall, the number of subgroups of order $p$ is congruent to $1$ or $1+p$ modulo $p^2$. See also the paper by Herzog. Now $85 \not\equiv 1,8 \mod{49}$, so I think we can conclude there is no finite group $G$ with $85$ subgroups of order $7$. – Mikko Korhonen Mar 16 '22 at 05:59
  • It would be nice to have some way of trivially proving the existence or nonexistence of a group with n subgroups of order p, for any n. – J. Linne Mar 16 '22 at 06:53
  • In the paper, the proof that there are no groups with $22$ subgroups of order $3$ and $21$ subgroups of order $5$ aren't trivial as they require more theory about the possible group structures, which are independent of any trivial conditions, like those given in 3.1 or 2.2. – J. Linne Mar 16 '22 at 06:54
  • @J.Linne: The subject of the AMM paper of Sambale is an elementary proof that there is no finite group $G$ with $n_{17}(G) = 35$. He also notes in the end that then by Hall/Kulakoff, for all $k \geq 1$ there is no finite group with $35$ subgroups of order $17^k$. – Mikko Korhonen Mar 16 '22 at 08:14
  • @J.Linne: $G$ does not have to be simple, but if $n_7(G) = 85$, then there exists a finite simple group $X$ with $n_7(X) = 85$. There is probably a way to rule this out as in the paper by Hall, or with a computer calculation as in the answer by Derek (which is what I had in mind in my first comment to this answer). – Mikko Korhonen Mar 16 '22 at 08:26