5

Denote the number of groups of order $n$ by $gnu(n)$.

A natural number $n\ge 1$ is called group-abundant, if $gnu(n)>n$, group-perfect, if $gnu(n)=n$ and group-deficient, if $gnu(n)<n$.

I wonder, which prime powers $p^k$ ($p$ prime , $k\ge 1$) are group-abundant, group-perfect and group-deficient.

I could solve the case $k\le 7$ completely using the higman's PORC-functions.

The only group-abundant prime powers $p^k$ with $k\le 7$ are $2^5,2^6,2^7$ and $3^7$, all the other prime powers are group-deficient (there are no group-perfect prime powers for $k\le 7$):

The numbers $2^8$ and $3^8$ are group-abundant as well as $2^9,2^{10}$ and $2^{11}$.

Is it known whether all prime powers $p^k$ with $k\ge 8$ are group-abundant ?

Peter
  • 86,576

1 Answers1

7

According to this math.SE question the number of groups of order $p^k$ satisfies $$ gnu(p^k) \geq p^{\frac{2}{27}k^2(k-6)}.$$ Filling in $k \geq 8$ gives $$ gnu(p^k) \geq p^{\frac{2}{27}k\cdot 8\cdot 2} > p^{k},$$ so all prime powers $p^k$ with $k \geq 8$ are group-abundant.

Community
  • 1
Marc
  • 3,067
  • 1
    Thank you. A superb answer. I hoped that the lower limits would give a value $k$, for which it can be shown that $p^k$ is group-abundant, but I did not expect that it is already the case for $k=8$. – Peter Jan 24 '16 at 20:41