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
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 $E_{p^n}$.
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...
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.
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Commented
Oct 2, 2012 at 13:16
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.
