Answer by Asinomás for Known bounds for the number of groups of a given order.
We can obtain an even better lower bound by looking at the groups of exponent $p$ and nilpotency class $2$.Apparently this tells us that there are at least $p^{\frac{2}{27}m^3-\frac{2}{3}m^2}$ groups...
View ArticleAnswer by Jack Schmidt for Known bounds for the number of groups of a given...
Geoff's answer is exactly correct, but I wanted to give the specifics.If you only want bounds that are easy to compute without being able to prove them yourself, then this answer should be just fine....
View ArticleAnswer by Geoff Robinson for Known bounds for the number of groups of a given...
The number of $p$-groups of order $p^{n}$ is (asymptotically) around $p^{ \frac{2 n^{3}}{27}}$. This suggests that one can't expect to do better than $n^{c\log(n)^{2}}$ for some constant $c$, for the...
View ArticleKnown bounds for the number of groups of a given order.
The number of nonisomorphic groups of order $n$ is usually called $\nu(n)$. I found a very good survey about the values. $\nu(n)$ is completely known absolutely up to $n=2047$, and for many other...
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