# Generating series - Finite groups of order $n$

I am wondering if something of interest can be said about one of the two series

$$G_1(x)=\sum_{n=1}^{+\infty}{\mathcal{G}(n)z^n}$$ $$G_2(s)=\sum_{n=1}^{+\infty}{\frac{\mathcal{G}(n)}{n^s}}$$

where $\mathcal{G}(n)$ is the number of finite groups of order $n$. (I don't even know if those series are converging for any $z$ or $s$...)

Thanks a lot !

• I doubt much can be said. Not much is known about $\mathcal{G}(n)$. – Alexander Gruber Sep 1 '14 at 17:35

Some things can be said about this, you should take a look at the book "$\underline{\text{Enumeration of finite groups}}$" by Blackburn, Neumann and Venkataraman. In this book, if for any $n$, $\mu(n)$ denotes $max(\alpha_i)$ where :

$$n=\prod_{i=1}^rp_i^{\alpha_i}$$

Then you have the following upper bound proven by Laszlo Piber proven in 1991 :

$$\mathcal{G}(n)\leq n^{\frac{2}{27}\mu(n)^2+O(\mu(n)^{\frac{5}{3}})}$$

Now, if you want to talk convergence of $G_1$ I think it is possible, take $n$ and $k$ such that :

$$2^k\leq n<2^{k+1}$$

It is somehow clear that $\mu(n)<k+1$ (otherwise $n$ would be divisible by $p^{k+1}\geq 2^{k+1}$), on the other hand :

$$k\leq log_2(n)<k+1$$

Hence we have that :

$$\mu(n)<log_2(n)+1$$

Now I would like to find a good majoration for the upper bound. You have that :

$$(n^{\frac{2}{27}\mu(n)^2+O(\mu(n)^{\frac{5}{3}})})^{\frac{1}{(log_2(n)+1)^2}}\leq n^K$$

For some positive constant $K$. Hence :

$$\mathcal{G}(n)\leq n^{K(log_2(n)+1)^2}$$

Now you have a criterion for convergence of such series (I don't remember the name) if you positive terms $a_n$ for a serie with $R$ its radius of convergence then if :

$$\sqrt[n]{a_n} \text{ has a limit }L$$

Then $L=\frac{1}{R}$ but if $a_n= n^{K(log_2(n)+1)^2}$ then :

$$\sqrt[n]{a_n}=exp(K(log_2(n)+1)^2log(n)\frac{1}{n})$$

This is easily seen to be convergent to $1$ hence the radius for the serie associated to $(n^{K(log_2(n)+1)^2})$ is $1$ and for the serie $G_1$ it is greater than $1$ (by comparison theorem). On the other hand it is more than clear that $\mathcal{G}(n)$ being $\geq 1$ and hence that the radius of convergence for $G_1$ will be $\leq 1$.

Hence the radius of convergence of $G_1$ is $1$.

For $G_2$ I think it is nowhere convergent. I will actually show that there are no real numbers where it is convergent. First, because $\mathcal{G}(n)\geq 1$ you have that if $G_2(s)$ converge then $s>1$. Now take $s>0$, one thing you know about lower bound of $\mathcal{G}(n)$ is the following :

$$\mathcal{G}(p^k)\geq p^{\frac{2}{27}k^3-O(k^2)}$$

Now if $N>p^m$ then :

$$\sum_{n=1}^N\frac{\mathcal{G}(n)}{n^s}\geq \sum_{k=1}^m\frac{\mathcal{G}(p^k)}{p^{ks}} \geq \sum_{k=1}^m p^{\frac{2}{27}k^3-O(k^2)-ks}$$

Now $O(k^2)-ks=O(k^2)$ hence :

$$\sum_{n=1}^N\frac{\mathcal{G}(n)}{n^s}\geq\sum_{k=1}^m p^{\frac{2}{27}k^3-O(k^2)}\geq \sum_{k=1}^m p^{\frac{2}{27}k^3-C_sk^2}$$

Which is easily seen to be converging (when $m$ goes to infinity) to infinity. Hence the sum cannot converge when $N$ goes to infinity hence there cannot be any convergence...