# Are there overwhelmingly more finite posets than finite groups? [closed]

A function $$f:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$$ overwhelms $$g:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$$ if for any $$k\in \mathbb{Z}_{\geq 1}$$ the inequality $$f(n)\leq g(n+k)$$ holds only for finitely many $$n\in\mathbb{Z}_{\geq 1}$$.

For example $$n\to n^2$$ overwhelms $$n\to n$$.

Does the number of non-isomorphic posets of cardinality $$n$$ overwhelm the number of non-isomorphic groups of cardinality $$n$$?

• This is an interesting question, but you give no evidence that it’s research level. Partial information about these functions is not hard to come by. – Kevin Arlin May 2 at 12:45
• In your deleted question today (where you initially compared groups vs topologies), I wrote the following comment, which seems still relevant here: Is there a reason to compare these two seemingly unrelated numbers? the second is highly sensitive to the prime decomposition of $n$ (high for powers of $2$, $=1$ for primes), the first increases with $n$. – YCor May 2 at 13:05
• I’m voting to close this question: see @YCor's comment above and @‍Wojowu's comment on your very similar question. Especially notice that these questions were asked at nearly the same time. – LSpice May 2 at 14:36

On one hand, the number of groups of order $$n$$ is at most $$2^{O((\log n)^3)}$$ (see here). On the other hand, by considering posets which are disjoint unions of total orders, the number of posets of order $$n$$ is at least equal to the number $$p(n)$$ of partitions of $$n$$. Since $$p(n)\gg 2^{\sqrt{n}}$$, we get the desired result.
• @firn You're right, I missed the $n$ in the base of the exponential. – Wojowu May 2 at 13:09
• Not relevant to the question, but the number of posets of order $n$ is $2^{\frac{n^2}{4}+o(n^2)}$. – Richard Stanley May 2 at 14:45
• Either way, since $\log n! = o(\log 2^{n^2/4})$. – Richard Stanley May 3 at 13:23