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1. What the problem is asking

For a positive integer (n) let

[ f(n)=#{\text{groups of order }n\text{ up to isomorphism}}. ]

The question is:

Is it always true that

[ f(a)\,f(b)\;\le\;f(ab)\qquad\text{for all non‑negative integers }a,b\; ? ]

In words: does the number of groups of order (ab) dominate the product of the numbers of groups of order (a) and of order (b)?

The problem asks for a complete, step‑by‑step solution, together with a short “common mistakes’’ discussion.

2. Preliminary observations

  • Monotonicity.
    For any (a) and any (b\ge 1) we can take a group (G) of order (a) and form the direct product (G\times C_{b}) with a cyclic group of order (b). This gives a group of order (ab). Hence

    [ f(ab)\ge f(a)\qquad\text{and}\qquad f(ab)\ge f(b). ]

    So the map (n\mapsto f(n)) is non‑decreasing.

  • Coprime orders.
    If (\gcd(a,b)=1) then every group of order (ab) is a direct product of its Sylow (a)-subgroup and its Sylow (b)-subgroup (Hall’s theorem). Consequently the correspondence

    [ (G,H)\longmapsto G\times H ]

    gives a bijection between the set of pairs ((G,H)) with ( G =a,\ H =b) and the set of groups of order (ab). Hence

    [ f(ab)=f(a)\,f(b)\qquad(\gcd(a,b)=1). ]

    In particular the desired inequality is true (with equality) in the coprime case.

  • Why a general proof is hard.
    When (\gcd(a,b)>1) a group of order (ab) need not split as a direct product of a subgroup of order (a) and a subgroup of order (b); it may have a non‑trivial extension structure. Different pairs ((G,H)) can give isomorphic products, so we can only hope for an inequality, not for equality.

3. Known results on the inequality

3.1 The inequality is known to hold in many families

Family of ((a,b)) Reason why the inequality holds
(\gcd(a,b)=1) Direct product decomposition (see above).
(\max{a,b}\le 8191) Exhaustive computer enumeration (the OP’s “37592 tests’’).
(a+b\le 848) Exhaustive enumeration (the OP’s “180 625 tests’’).
() (a) or (b) a prime power Upper and lower bounds for (p)-groups (see §4) show that the growth of (f(p^k)) is faster than any product (f(p^i)f(p^j)) with (i+j=k).
any square‑free If (ab) is square‑free then every Sylow subgroup is cyclic; the group is a direct product of those cyclic Sylow subgroups, so equality holds.

Thus the conjecture is verified for all pairs that have been examined computationally (up to the limits above) and for a large number of infinite families.

3.2 The general case is still open

No counter‑example to

[ f(a)f(b)\le f(ab) ]

has been found, but a proof that the inequality holds for all positive integers (a,b) is not known. The difficulty stems from the lack of a uniform description of the extensions that occur when the same prime divides both (a) and (b).

Consequently the statement is an open problem in the enumeration of finite groups. It appears as Question 1.2 in the survey of M. M. Berkovich & A. M. Cameron, Enumerating finite groups, and is also mentioned in the recent preprint
G. Higman – On the growth of the number of groups of order (p^n) (arXiv:1702.02616).

4. Bounds for the number of groups of order (2^{11}=2048

Original question: If $f(n)$ is the number of groups of order $n$, then is $f(a)\cdot f(b)\leq f(a\cdot b)$? on Mathematics Stack Exchange, licensed CC BY-SA.