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While I was thinking about solvable groups and how the Feit-Thompson theorem states that every finite group of odd order is solvable, I wondered how strong this result really is or how many groups it covers. I know that there isn't an explicit formula for the number of isomorphism classes of groups with order less or equal to some given integer $n$, but the following question came to my mind:

Let $G(n)$ denote the number of isomorphism classes of groups with order less or equal to a given integer $n$ and let denote $S(n)$ the number of isomorphism classes of groups with odd order less or equal to $n$.

Question: What is known about: $$ \lim_{n\to\infty}\frac{S(n)}{G(n)} $$ Does this limit even exist? If yes, what is it? If not, what about the $\limsup$ and $\liminf$?

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    $\begingroup$ it goes to zero $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '16 at 21:16
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    $\begingroup$ @JorgeFernándezHidalgo Are you aware of a reference, where this is proven? What is the main idea? $\endgroup$ – Severin Schraven Dec 25 '16 at 21:23
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    $\begingroup$ The groups of order a power of $2$ are the "most". $\endgroup$ – Dietrich Burde Dec 25 '16 at 21:24
  • $\begingroup$ I don't, sorry. It just sounded true, but I wasn't able to find or prove it later. $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '16 at 21:36
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    $\begingroup$ It is conjectured that almost all finite groups are 2-groups, in the sense that $\lim\frac{T(n)}{G(n)}=1$ if $T(n)$ is the number of 2-groups of order $\leq n$. So your limit is almost certainly $0$, though I don't know if that's actually been proven. Basically, the intuition is that in order to have lots of different groups of a given order, it needs to have repeated prime factors. Powers of $2$ grow much slower than powers of any other prime, so by far the most efficient way to get lots of different groups of small order is to take 2-groups. $\endgroup$ – Eric Wofsey Dec 25 '16 at 21:56
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Looking at the book enumeration of finite groups,

We obtain that the number of groups of order $p^k$ is at least: $$\frac{p^{\frac{2m^3}{27}}}{p^{\frac{2}{3}m^2}}$$

We also obtain that the number of groups of order $n$ is:

$$n^{\frac{2\mu^2}{27}}n^{\mathcal O(\mu^{3/2})}$$

where $\mu$ is the largest exponent for a prime power dividing $N$.

So take a natural $N$, and suppose that $2^m$ is the largest power of $2$ not exceeding $N$, we obtain that the number of groups of order $2^m$ is at least:

$$\frac{2^{\frac{2m^3}{27}}}{2^{\frac{2}{3}m^2}}=\frac{(2^m)^{\frac{2}{27}m^2}}{(2^m)^{\frac{2}{3}m}}\geq\frac{N^{\frac{2m^2}{27}}}{N^{\frac{2}{3}m}2^{\frac{2m^2}{27}}}\geq \frac{N^{\frac{2m^2}{27}}}{N^m}$$

On the other hand, if $n$ is an odd integer less than $N$, clearly its $\mu$ is at most $m\log_3(2)+1$. For large values of $m$ we can just bound this by $\alpha m$, for some pre-selected $\alpha<1$.

So the number of groups of odd order is :

$$N(N^{\frac{2\alpha^2 m^2}{27}}N^{\mathcal O(m^{3/2})})$$.

So the fraction between the number of groups of order $2^m$ and the number of groups of odd order less than $N$ for large $N$ is at least:

$$\frac{N^{\frac{2m^2}{27}}}{N^mN(N^{\frac{2\alpha m^2}{27}}N^{\mathcal O(m^{3/2})})}=N^{\frac{2(1-\alpha^2)m^2}{27}-\mathcal O(m^{3/2})}$$

Which clearly goes to infinity.

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