# Does the sequence (Number of groups of even order $\le n$) / (Number of groups of order $\leq n$) converge? If not, what are its cluster points?

I recently gave an undergraduate course on group theory (which is not entirely my field of expertise, so the following questions might have a well-known answer of which I am simply unaware). As I was explaining the concept of solvability, I digressed a little and told the class about the odd-order theorem, also known as the Feit-Thompson theorem, which states that every finite group of odd order is solvable. I made the remark: Among finite groups, solvability is the rule rather than the exception, because solvability is at least as likely as oddity. One of my students asked: "So if I take an arbitrary finite group, how likely is it then that this group is of odd order?" To which I knew no reply.

So I would like to ask the following series of related questions:

(1.) If $$\begin{equation*}x_{n}=\frac{\#\text{Isomorphy classes of groups of even order \leq n}}{\#\text{Isomorphy classes of groups of order \leq n}}\end{equation*}$$ does the series $$x_{n}$$ converge? If not, what are its cluster points?

(2.) If $$m\in\mathbb{N}$$ and $$\begin{equation*}y_{n}=\frac{\#\text{Isomorphy classes of groups of order \leq n, divisible by m}}{\#\text{Isomorphy classes of groups of order \leq n}}\end{equation*}$$ does the series $$y_{n}$$ converge? If not, what are its cluster points?

(3.) If $$\begin{equation*}z_{n}=\frac{\#\text{Isomorphy classes of solvable groups of order \leq n}}{\#\text{Isomorphy classes of groups of order \leq n}}\end{equation*}$$ does the series $$z_{n}$$ converge? If not, what are its cluster points?

My simple intuition is that in all three cases, the answer should be "yes, it converges", and it should converge to $$\frac{1}{m}$$ in case (2.), and to a value $$\geq\frac{1}{2}$$ in case 3.

I beg your forgiveness in advance if the answers are well-known, I am not an expert on group theory.

• "$\#$ Number of" sounds redundant, since $\#$ means "number of", so it sounds like "number of number of". – YCor 2 days ago
• There is widespread belief among specialists in the area that almost all finite groups (meaning isomorphism classes) have order a power of two, but it remains unproven, and the current techniques do not appear to be strong enough to prove it. So that would imply that the if we let $t_n$ be the number of isomorphism classes of groups of order a power of two less than $n$ divided by the number of all groups od order less than $n$, then $t_n \to 1$ as $n \to \infty$. – Derek Holt 2 days ago
• There are even stronger conjectures that almost all finite groups are $2$-groups $G$ of nilpotency class $2$ in which $Z(G)=[G,G]$ and $G/Z(G)$ and $Z(G)$ are both elementary abelian. The known lower bounds are derived from counting groups of this type. – Derek Holt 2 days ago
• If you do a search for "almost all groups are 2-groups" you will get plenty of hits and information on known results. – Derek Holt 2 days ago
• Re: almost all groups are 2-groups, you might find this old StackExchange answer interesting. – Carl-Fredrik Nyberg Brodda 2 days ago

As mentioned in the comments, conjecturally almost all finite groups are $$2$$-step nilpotent $$2$$-groups, so conjecturally the answers to 1) and 3) are that the limits both exist and both equal $$1$$; that is, almost all finite groups have even order and almost all finite groups are solvable (even nilpotent). As numerical evidence for this, almost all of the first $$50$$ billion groups have order $$1024$$. The conjectural answer to 2) is then that if $$m$$ is a power of $$2$$ then the limit is equal to $$1$$ and otherwise if $$m$$ has a nontrivial odd divisor then the limit is equal to $$0$$.
It's worth knowing as context here that a result due to Higman and Sims states that asymptotically the number of $$p$$-groups of order $$p^n$$ is $$p^{ \frac{2}{27} n^3 + O \left( n^{8/3} \right)}$$. The lower bound comes from counting $$2$$-step nilpotent $$p$$-groups; you can see an analogous argument for nilpotent Lie algebras here. Thinking of this count as a function of the order $$p^n$$ it's not hard to check that it's maximized, if $$p^n$$ is bounded by some reasonably large $$N$$, by making $$p$$ as small as possible (equivalently, by making $$n$$ as large as possible), which is what singles out $$p = 2$$. It should be possible to write down a similar heuristic argument showing that the count of nilpotent groups (which are products of their Sylow subgroups) is dominated by groups of order $$2^n$$ also.
• In this context it is worth mentioning Pyber's upper bound ${\rm gnu}(n) \le n^{(2/27)\mu(n)^2 + O(\mu(n)^{3/2})}$ on the number of groups of order $n$, where $\mu(n) \le \log_n(2)$ is the highest power of any prime dividing $n$. So the gap between proven upper and lower bounds is, in some sense, not large, it unfortunately it occurs in the exponent, rather than in the number itself. – Derek Holt yesterday