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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.

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  • $\begingroup$ "$\#$ Number of" sounds redundant, since $\#$ means "number of", so it sounds like "number of number of". $\endgroup$ – YCor 2 days ago
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    $\begingroup$ 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 $n \to \infty$. $\endgroup$ – Derek Holt 2 days ago
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    $\begingroup$ 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. $\endgroup$ – Derek Holt 2 days ago
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    $\begingroup$ If you do a search for "almost all groups are 2-groups" you will get plenty of hits and information on known results. $\endgroup$ – Derek Holt 2 days ago
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    $\begingroup$ Re: almost all groups are 2-groups, you might find this old StackExchange answer interesting. $\endgroup$ – Carl-Fredrik Nyberg Brodda 2 days ago
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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.

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    $\begingroup$ Is this an answer or some ideas based on the above comments? $\endgroup$ – Alireza Abdollahi yesterday
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    $\begingroup$ 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. $\endgroup$ – Derek Holt yesterday

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