Does the percentage of groups of order at most $n$ of even order aproach $1$? Let $E_n$ be the number of isomorphism classes of groups of even order at most $n$, let $G_n$ be the number of isomorphism classes of groups of order at most $n$ and $T_n$ be the number of isomorphism classes of $2$-groups of order less than $n$.
I think that it is an open problem if $\frac{T_n}{G_n}$ approaches $1$, so I figured that it is probably not so hard to prove that $\frac{E_n}{G_n}$ approaches $1$, although I was not able to do so.
 A: I think the following works:
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.
