We define the upper density $\rho (G)$ of a finite group $G$ as the ratio of the number of finite groups of order $<n$ which contain a subgroup isomorphic to $G$ to the number of groups of order $<n$, in the limit as $n \rightarrow \infty$ (if such a limit exists, but we can always take limsup/liminf). 


For example, $\rho (G) =1$ for $G$ is the trivial group. 


I would like to investigate the value of $\rho (C\_2) $. In general, this seems like a hard problem, so lets simplify things and talk about abelian groups only.


So we define $ \rho (G) = lim\_{n \rightarrow \infty} \frac{ \textrm{number abelian groups order} \ < n \ \textrm{with subgroup iso.} \ G}{ \textrm{number of abelian groups order} < n}$


I tried some rough bounds but I still end up with horrible summations that I can't seem to get much out of.


Some basic facts that will be useful:


* Number of non-isomorphic abelian groups of order $p\_1 ^{\alpha\_1} \cdots p\_k ^{\alpha\_k}$ is $p( \alpha\_1) \cdots p( \alpha\_k)$ where $p(r)$ is the number of partitions of $r$.
* We want to count primes somehow, so maybe a result Bertrand's Postulate would be useful (although will probably only give weak bounds) <https://en.wikipedia.org/wiki/Bertrand%27s_postulate>


	+ The number of abelian groups of order $n$ is a multiplicative function.


Any ideas would be appreciated.



On the Upper Density of $C_2$ in finite groups
Pyber showed that the number of groups of order $n$ is $\leq n^{\frac{2}{27}\nu(n)^3+C\nu(n)^{3/2}}$, where $\nu$ is the highest power of a prime dividing $n$ and $C$ is an absolute constant. On the other hand, Sims showed that for $n$ a power of $p$, this upper bound is achieved. Since for odd $n$ we have $\nu(n)\leq\frac{\log n}{\log 3}$, these results imply that almost all groups contain $C\_2$.


For abelian groups things are simpler. The generating Dirichlet series of the number of abelian groups of order $n$ is $D(s)=\zeta(s)\zeta(2s)\zeta(3s)\dots$, thus the number of abelian groups of order up to $x$ is asymptotically $Cx$, where $C=\zeta(2)\zeta(3)\dots$ is the residuum of $D$ at 1. If you only count groups of odd order, the residuum changes by $\prod\_{k\geq 1}\left(1-\frac{1}{2^k}\right)$, so the density of abelian groups without a subgroup $C\_2$ is $\prod\_{k\geq 1}\left(1-\frac{1}{2^k}\right)$.


