# Is there an efficient method to decide whether $gnu(n)<n$ , $gnu(n)=n$ or $gnu(n)>n$?

Denote : $gnu(n)$ = number of groups of order $n$

It is much easier to decide whether a natural number $n$ is group-deficient ($gnu(n)<n$) , group-perfect ($gnu(n)=n$) or group-abundant ($gnu(n)>n$) than to actually calculate $gnu(n)$.

For example $gnu(2048$) is unknown, but it is known that $gnu(2048)>2048$, so $2048$ is group-abundant.

Squarefree numbers are always group-deficient ($gnu(n)<n$) and it seems that all numbers $n$ , for which there is no prime $p$ with $p^4|n$ are group-deficient.

I am not sure whether it is known for all prime powers $p^k$ , whether $p^k$ is group-deficient, group-perfect or group-abundant.

The cubefree numbers upto $50,000$ are group-deficient and the odd numbers upto $9,000$ except $3^7$ and $3^8$ are also group-deficient. (Perhaps, someone can doublecheck it ?)

There are no group-perfect numbers (except $1$) upto $2499$. (Again, can someone please doublecheck it ?)

Is there an efficient method for the decision problem (lower or upper bounds would do the job in most cases) ?

Is it true, that $m|n$ implies $gnu(m)\le gnu(n)$ ?

Is it true that $p<q$ implies $gnu(p^m)<gnu(q^m)$ for $m>4$?

• The answer to the second question should be yes because we can take direct products. How can I prove that $G_1\times H$ isomorphic to $G_2\times H$ implies $G_1$ isomorphic to $G_2$ ? – Peter Jan 5 '16 at 17:41
• I don't know about the first question, but the answers to the second and third questions are certainly yes. Proofs are left as exercises! The second is very easy - I thought you might have asked that before. – Derek Holt Jan 5 '16 at 17:44
• For third question, use the cyclic group of order $p^n$ as an extra example to give the strict inequality. – Derek Holt Jan 5 '16 at 17:47
• I feel the answer to your third question should be yes, but I am not sure it is known. By math.auckland.ac.nz/~obrien/research/gnu.pdf (which I link to in almost all of your question), the answer is yes for $m\leq 7$. While finding exact formulas of this type for every $m$ is probably not possible, some decent estimates might be enough to establish what you ask for. – verret Jan 6 '16 at 2:03
• As for your first question, the answer is almost certainly that there is no efficient method known at the moment. – verret Jan 6 '16 at 2:04