Are there more even or odd groups? 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 $\liminf$?
 A: 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.
