Erdős Problem #1160

Erdős Problem 1160 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Let $g(n)$ count the isomorphism classes of groups of order $n$. Must \[ g(n)\leq g(2^m) \] hold for every positive integer $n\leq2^m$?

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

ATTACK PLAN AND WORK

A useful comparison is available at odd orders. If $n$ is odd, the assignment $G\mapsto G\times C_2$ embeds isomorphism classes of groups of order $n$ into those of order $2n$. The subgroup of all odd-order elements in this direct product is exactly $G\times\{1\}$: the second coordinate is nontrivial precisely when the element has even order. It is characteristic, so an isomorphism of the products recovers an isomorphism of the original groups. Thus \[g(n)\le g(2n)\qquad(n\text{ odd}).\tag{1}\] This does not compare $n$ with a nearby power of two.

One can also exhibit many different groups at the target order without invoking classification. For $0\le j\le\lfloor m/2\rfloor$, the groups $C_{2^{m-j}}\times C_{2^j}$ have order $2^m$ and distinct exponents $2^{m-j}$, hence \[g(2^m)\ge\lfloor m/2\rfloor+1.\tag{2}\] In particular the desired comparison holds for $n$ prime, where a nonidentity element generates the whole group and $g(n)=1$. It also holds for prime-square orders when $m\ge2$: a group of order $p^2$ is abelian (its nontrivial center and the fact that a cyclic central quotient forces commutativity prove this), and the only two possibilities are $C_{p^2}$ and $C_p\times C_p$.

VERIFICATION AND REMAINING GAP

Control the number of groups at arbitrary mixed-prime orders below $2^m$. The recoverable-product injection changes the order in the wrong way for the full comparison, and the explicit abelian family is only a lower bound.

SOURCES

https://www.erdosproblems.com/1160

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$5\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.