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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.

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    $\begingroup$ Conjecturally almost all finite groups are $2$-groups, and this would imply $\rho(C_2)=1$ in the non-abelian case. $\endgroup$ – Julian Rosen Oct 23 '17 at 0:19
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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 $\prod_{k\geq 1}\left(1-\frac{1}{2^k}\right)$.

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