Dear all,

Does someone know of any paper/method that enables us counting/estimating the number of normal subgroups of some p-group of order $p ^n $ ($ n$ is some natural number ? ) .

Is there anyway we can count the maximal subgroups it has (i.e.- the groups of order $p^{n-1} $ ? ) ?

Thanks in advance

  • $\begingroup$ en.wikipedia.org/wiki/Hall_algebra In abelian groups count of subgroups with fixed factor is related to Hall-Littlewood polynoms. What happens for non-abelian - I asked MO: mathoverflow.net/questions/107537/… with no reply $\endgroup$ – Alexander Chervov Oct 2 '12 at 5:18
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    $\begingroup$ Same question simultaneously asked at m.se, math.stackexchange.com/questions/205681/… --- coincidence? $\endgroup$ – Gerry Myerson Oct 2 '12 at 5:51
  • $\begingroup$ @Alexander Chevov: Thanks ! I had no idea about the Hall Algebra notion... But I'm still skeptic about it... Have you got any paper the gives some more details about it? Thanks again! $\endgroup$ – Jason Mraz Oct 2 '12 at 12:38
  • $\begingroup$ I think MacDonald's book "Symmetric functions ... " discuss this... I am not sure - I can send you file of the book, if you need. There have been modern developments about Hall algebras, which go very very far from p-groups - they are surveyed in arxiv.org/abs/math/0611617 $\endgroup$ – Alexander Chervov Oct 3 '12 at 11:00
  • $\begingroup$ That's excatly the thing... I only need kind of "simple" estimates and bounds on the number of subgroups... I'll try to go over the lecture notes you sent and I might find something useful in them... Thanks a lot ! (I'll try to look for the book you mentioned) $\endgroup$ – Jason Mraz Oct 3 '12 at 19:03

For a $p$-group $P$, the number of maximal subgroups is $\sum_{k=0}^r p^k$ where $r$ is the minimum size of a generating set for $P$. You can see this from looking at the maximal subgroups of $P/\Phi(P)$, which is elementary abelian of order $p^r$.

What I can tell you is that there is at least one normal subgroup for every power of $p$ up to the order of the group. Sylow theory style orbit counting gives us that the number of normal subgroups of each order $p^k$ is going to be congruent to $1 \mod{p}$, so the total number of normal subgroups in a $p$-group of order $p^n$ will then be congruent to $n+1 \mod{p}$.

EDIT: I thought of a bound.

$n+1$ is the lower bound, attained by the cyclic group of order $p^n$. There must be at least one normal subgroup for every prime power divisor, so this is the lowest it can go.

On the other hand, I claim that elementary abelian groups $E_{p^n}$ contain the largest number of normal subgroups. This is because it has the maximum rank of all groups of order $p^n$. Thinking of $E_{p^n}$ as an $\mathbb{F_p}$-vector space, we obtain the number of subspaces by $$\mathcal{N}(E_{p^n})=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{p^n-p^k}{p^m-p^k}.$$ Here we count the number of ordered combinations of $m$ linearly independent vectors in $\mathbb{F_p}^n$, then divide by the number of possible bases of an $m$-dimensional subspace. Summing over $m$ we have the total number of normal subgroups in $E_{p^n}$.

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    $\begingroup$ Forgive my ignorance: how do Sylow theorems give you information about subgroups of a p-group? $\endgroup$ – Nick Gill Oct 2 '12 at 8:37
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    $\begingroup$ Indeed, in a cyclic group G of order p, there are two normal subgroups: G and {1}. And 2 is not 1 mod p... Do you mean the number of normal subgroups OF A GIVEN ORDER will be 1 mod p? I could maybe believe that but I'd have to think about how I'd prove it. $\endgroup$ – Nick Gill Oct 2 '12 at 8:41
  • $\begingroup$ From Sylow Theorem, we see that the number of subgroups of a given order in a finite $p$-group is congruent to 1 mod $p$. Maybe Alexander means this. $\endgroup$ – Wei Zhou Oct 2 '12 at 9:48
  • $\begingroup$ @Wei Zhou, I've never seen Sylow theorems applied to subgroups of a $p$-group. So, while I can believe the result you state, I'm not sure how you use Sylow to prove it. Maybe it's just that the method by which we prove the $1\mod p$ part of the Sylow theorems can be applied here (?). $\endgroup$ – Nick Gill Oct 2 '12 at 13:35
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    $\begingroup$ @Nick: Let $G$ be a $p$-group of order $p^n$, and S the set of all subgroups of order $p^m$. Let $P \in S$. Then $P$ can acts on $S$ by conjugation. By counting the orbits of this action, we see $|S|$ congruent to 1. This trick is use to prove Sylow theorem by someone. So in some book I can not find, I think this is also called Sylow theorem. $\endgroup$ – Wei Zhou Oct 2 '12 at 15:50

The wikipedia article on p-groups reminded me that

Every normal subgroup of a finite p-group intersects the center nontrivially.

This implies immediately that minimal normal subgroups of a p-group $G$ will be central. This fact can be used to prove the statement that Wei Zhou made:

A $p$-group of maximal class and size $p^n$ has the least number of normal subgroups of all groups of order $p^n$.

(If I'm thinking straight this number is $n+1$ and the bound is also achieved by the cyclic group of order $p^n$.)

It seems to me that one might be able to prove something a little stronger using an inductive argument: counting the minimal normal subgroups in the center $Z$, and then counting the normal subgroups in $G/Z$, and then putting these two numbers together... It's that last bit that's going to be tricky though. If the center is cyclic, then everything is fine but when it's not cyclic, eek...

  • $\begingroup$ Dear @Nick: Thanks a lot ! I 'll be glad if you'll be able to tell me what do you mean by a "group of maximal class" ... After verifying this little detail, I'll reread your answer in order to check again that I understand it... Thanks again! $\endgroup$ – Jason Mraz Oct 2 '12 at 12:43
  • $\begingroup$ By class', I mean nilpotency class' i.e. the length of the upper (or lower) central series. A group $G$ of order $p^n$ (with $n>2$ )is of maximal class if this length is $n-1$; in this case $G$ has center $Z(G)$ cyclic of order $p$, then $G/Z(G)$ has center cyclic of order $p$. This pattern continues until you get to a normal group of index $p^2$ in $G$ at which point one has an abelian quotient. Note that, a priori, there may be more than one isomorphism class of group of order $p^n$ of class $n-1$ - not all of them will necessarily have the minimal number of normal subgroups. $\endgroup$ – Nick Gill Oct 2 '12 at 13:16
  • $\begingroup$ I think that the $p$-groups of maximal class have the least number of normal subgroups except for the cyclic groups. If I'm not mistaken $p$-groups of maximal class and order $p^n$ will have one normal subgroup for each $p$th power up to $p^{n-2}$, then a few normal subgroups of order $p^{n-1}$, as opposed to cyclic groups which of course have unique normal subgroups for every power of $p$. $\endgroup$ – Alexander Gruber Oct 2 '12 at 22:58
  • $\begingroup$ Dear @Nick Gill and @Alexander: Where can I find proofs for the facts you mention? I can't see this straight away... Can you give me some reference for the proof of these facts? Thanks ! $\endgroup$ – Jason Mraz Oct 3 '12 at 9:16
  • $\begingroup$ @Alexander, if $G/ G_1$ is cyclic of order $p^2$ (where $G_1$ is the first term in the lower central series), then I think one gets $n+1$ normal subgroups. HOWEVER I do not know enough about $p$-groups of maximal class to be sure that this can happen! If all $p$-groups of maximal class have $G/G_1$ elementary abelian, then I agree with you. $\endgroup$ – Nick Gill Oct 3 '12 at 13:09

As I know, for the p-group of maximal class, the number of normal subgroups are known. And the number of normal subgroups in p-group of maximal class the the smallest.

  • $\begingroup$ Reference please $\endgroup$ – Alexander Chervov Oct 2 '12 at 15:38
  • $\begingroup$ About the minimality of normal subgroup, you can get this fact mentioned in the comment of Alexander Gruber following the answer of Nick Gill, from the paper by N. Blackburn, On a special class of p-groups. By the way, the above paper is an important paper for the theory of p-group $\endgroup$ – Wei Zhou Oct 3 '12 at 1:19
  • $\begingroup$ Thank you ! PS as a remark may say that adding more details to the answers would be more valuable in general and more easy convince the readers to press on +1 button:) $\endgroup$ – Alexander Chervov Oct 3 '12 at 10:50

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