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For each first order sentence $\phi$ in the language of groups, define :

$$p_N(\phi)=\frac{\text{number of nonisomorphic groups $G$ of order} \le N\text{ such that } \phi \text{ is valid in } G}{\text{number of nonisomorphic groups of order} \le N}$$

Thus, $p_N(\phi)$ can be regarded as the probability that $\phi$ is valid in a randomly chosen group of order $\le N$.

Now define $$p(\phi)=\lim_{N \to \infty}p_N(\phi)$$ if this limit exists.

We say that the theory of groups fulfills a first order zero-one law if for every sentence $\phi$, $p(\phi)$ exists and equals either $0$ or $1$. I'm asking myself whether this 0-1 law holds indeed in group theory.

Since it is conjectured that "almost every group is a 2-group", statements like $\exists x: x\ne 1 \wedge x^2=1 \wedge \forall y:xy=yx$ (meaning $2|Z(G)$) or $\forall x: x^3=1 \to x=1$ (no element has order 3) should have probability $1$ and I don't see any possibility to construct any sentence with $p\not \in \{0,1\}$. Am I missing an obvious counterexample, or can you show (under the condition that almost every group is indeed a 2-group) that the theory of finite groups fulfills this 0-1 law?

Quinn Culver
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Dominik
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  • Can you elaborate on "first order sentence in the language of groups?" – Alexander Gruber Jun 30 '13 at 21:05
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    $\mathcal L_{\text{grp}}={e,\circ,{}^{-1}}$. What is unclear to you? – Dominik Jun 30 '13 at 21:09
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    I don't deny that this question stems from my own ignorance of logic/formal systems. Would statements like "$G$ is nilpotent" be included in this, for example? – Alexander Gruber Jun 30 '13 at 21:12
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    No since, in first order logic, quantifiers refer only to elements of your group, not to functions or subsets of the group. – Dominik Jun 30 '13 at 21:14
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    @Alexander: there doesn't exist a first-order sentence in the language of groups which is true precisely of the nilpotent groups, although this isn't obvious. The reason is that the class of nilpotent groups isn't closed under ultraproducts (see http://en.wikipedia.org/wiki/Ultraproduct#.C5.81o.C5.9B.27s_theorem). However, there is a first-order sentence which is true precisely of the $k$-step nilpotent groups for a fixed $k$, the reason being that $k$-step nilpotence is equivalent to the vanishing of a finite collection of iterated commutators. – Qiaochu Yuan Jun 30 '13 at 21:21
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    (And my understanding is that standard conjectures imply that almost all groups are $2$-step nilpotent, so I don't think we get a counterexample this way.) – Qiaochu Yuan Jun 30 '13 at 21:21
  • @Dominik: "almost every group is a 2-group" is a folklore, not a conjecture. See e.g. this MSE question, scroll to the answer by m.k. saying that "~99.2% of all groups of order at most 2000 have order 1024". And obviously permutation (1,2,3) has order 3 - unless I misunderstood the assumptions since I am not an expert in logic. – Olexandr Konovalov Jun 30 '13 at 21:24
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    @Alexander K: that link doesn't seem to support your claim. Also, the claim in the OP isn't that the sentence is always true but that it's true with probability $1$. – Qiaochu Yuan Jun 30 '13 at 21:25
  • @QiaochuYuan: thanks, I've edited my comment to point out where to look. – Olexandr Konovalov Jun 30 '13 at 21:35
  • @Dominik: Ok, now I think I understood the question (+1) – Olexandr Konovalov Jun 30 '13 at 22:02
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    @Alexander K: that isn't a proof of the claim, which is an asymptotic claim as the size of the groups you consider goes to infinity. – Qiaochu Yuan Jun 30 '13 at 22:57
  • So if we're assuming that almost all groups are $2$-groups, does this question not reduce to "do $2$-groups have a $0-1$ law?" – Alexander Gruber Jun 30 '13 at 23:16
  • @AlexanderGruber: Yes, it does. – Dominik Jul 01 '13 at 08:09
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    The conjecture should be "almost every finite group is a 2-group". If you want to include infinite groups, then you probably want to restrict yourself to finitely presented groups. Here, it is true that every finitely presented group is hyperbolic, and there is a nice form which one of these "typical" groups takes. See this paper. – user1729 Jul 01 '13 at 09:27
  • (I have just had a though - I think my above comment means that finitely presented groups have a 0-1 law. This is because all torsion-free hyperbolic groups have the same first-order theory, which is a recent result of Sela, and because the typical groups I talk about above are torsion-free hyperbolic. However, my knowledge and understanding of model theory is quite frankly rubbish so I have probably mis-understood something.) – user1729 Jul 01 '13 at 10:54
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    @user1729: Torsion-free hyperbolic groups are not all elemantary equivalent: for example, $\mathbb{Z}$ and $\mathbb{F}_2$ are two torsion-free hyperbolic groups, but they are not elementary equivalent; in fact, they don't have the same universal theory: $\forall x,y ([x,y]=1)$. – Seirios Jul 02 '13 at 07:36
  • @Seirios: Ah, sorry, my mistake (although I really should have said either "non-cyclic" or "one ended" in my previous comment!). The theorem I was thinking of is that it is decidable if two torsion-free hyperbolic groups have the same first-order theory, and looking over the proof quickly I think that it is true that even if we exclude $\mathbb{Z}$ you get torsion-free hyperbolic groups with different first-order theories. It is not unlikely that the "typical" groups, of the type I mention above, all have the same first order theories. I do not know though, and I have not thought about it. – user1729 Jul 02 '13 at 09:24
  • I'm not sure that this makes sense, but would it be possible to take advantage of the 'almost every finite group is a 2-group' folklore to (conjecturally) construct a counterexample by 'asking about' elements or even subgroups of order $2^2$ in some fashion? What's known about the proportion of 2-groups that $V_4$ is a subgroup of? – Steven Stadnicki Sep 08 '13 at 15:49
  • Is it even known whether the $\forall_1$ sentence "G is Abelian" converges to zero, one, or something in between? – HTFB Sep 10 '13 at 13:14
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    @HTFB I believe it's not too hard to show that a vanishingly small proportion of groups are abelian; the fundamental theorem of finite abelian groups gives a very low bound on the number of abelian groups. For instance, the logarithm of the number of abelian groups of order $2^n$ is roughly $c\sqrt{n}$ for some $c$, while the log of the total number of groups of that order is roughly $cn^3$. – Steven Stadnicki Sep 10 '13 at 16:03
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    Since nobody responded to my 500 bounty, I suggest you crosspost this question to MO, if you haven't already. – Alexander Gruber Sep 16 '13 at 23:23
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    @StevenStadnicki By the way, I remember reading a lower bound on the number of $p$-groups of a given order $p^n$ that was pretty high. (EDIT: It's $p^{\tfrac{2}{27} n^2(n-6)}$.) I bet one could show that even just this number is enough to asymptotically exceed the number of abelian groups of order $\leq p^n$. – Alexander Gruber Sep 16 '13 at 23:27
  • Would a positive answer to this not imply that either almost all finite groups are $2$-groups or that hardly any finite groups are $2$-groups, since "is a $2$-group" is a valid sentence (or is a sentence not allowed to assert the existence of an integer with a certain property?) – Tobias Kildetoft Sep 26 '13 at 13:20
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    @AlexanderGruber Has the question been x-posted? – Bruno Joyal Nov 11 '13 at 22:57
  • @BrunoJoyal No, I don't think so. – Alexander Gruber Nov 12 '13 at 22:50
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    The question has now been crossposted at http://mathoverflow.net/questions/150603 . – Emil Jeřábek Dec 05 '13 at 14:14
  • What would be the value of

    $$\phi=\forall x\neq e; (\exists y\neq e\exists z\neq e,; y.z=z.y=x)\vee(\exists y\neq e,; xy=yx)$$

    – Xoff Dec 29 '13 at 10:54
  • @Xoff That sentence is always true: for the second alternative, we can choose y=x. Or did you mean to exclude that? – Erik P. Jan 07 '14 at 18:23
  • @ErikP. yes, $x\neq y$ obviously I wrote it to quickly :) – Xoff Jan 07 '14 at 20:30
  • @AlexanderKonovalov apologies but I don't quite understand your comment above. in what way does either of the answers here prove that almost all finite groups are 2-groups? as far as I can tell all of the discussion there has to do only with groups of order less than 2000, and in fact the accepted answer says that the general statement is indeed a conjecture – Atticus Stonestrom Mar 20 '21 at 01:36

2 Answers2

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Sharing the question within MO has brought some interesting comments - I send you there for reference - but in the last 2 years there has been no new contribution. So I'm wrapping it up for the less technical scope of SE:

As per now, a FO 0-1 law for group theory as defined by Dominik doesn't exist and may never exist.

The consensus seems to be that an accepted and complete view of what group theoretic concepts correspond to in a classical logic framework is needed to address the problem but - as per now - it is not available.

The actual methods available to cross-translate group theoretic concepts into logic ones seem to suggest that such a law cannot exist for the whole group theory or even for finite group theory. Limiting laws have infact been identified such that group theoretical entities can be partitioned into different classes exhibiting different logical behaviours (i.e. quantification over functions which is ubiquitous in group theory; but also abelianity appears to be relevant). While yet to be proven, some of these classes can be expected to obey a 0-1 law for FOL while some others can be expected to not obey it.

Lorenzo
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Sentence proposals:

  1. The cardinality of $G$ is even.

    • I strongly suspect the limit diverges by having a zero $\lim\inf$ and a unit $\lim \sup$. Via supermultiplicity, the number of groups of order $2^k n$ is bounded below by the number of order $2^k$ times the number of order $n$, so is infrequently "small" ($n$ prime) and frequently large. Additionally, the number of groups of a given order is upper bounded (see http://www.jstor.org/stable/2946623 ) by something not too stupendously rapidly growing, so no $N$ is going to miraculously overwhelm the running average.
  2. Let $P(n)$ be the cardinality of the set of isomorphism classes of groups of order $n$. For all positive integers $n$, $P(n)>0$ since there is at least a cyclic group of order $n$. Let $C(n) = \sum_{i=1}^n P(n)$, which is clearly positive and monotonically increasing on the positive integers. Set $a_0=1$ and, for $j>0$, set $a_j = \min_i\{i \mid C(i)-a_{j-1} > 2^{2j}\}$, such an $i$ exists because $C$ is monotonically increasing. Let $T = \bigcup_{j=0}^\infty [a_{2j}, a_{2j+1})$ and let $\phi$ be the predicate $|G| \in T$. By construction, $p_N(\phi)$ swings by a factor of two through each interval $[a_j, a_{j+1})$ (from something < $2^{-j}$ to something $> 1- 2^{-j}$ and vice versa). Therefore, the limit as $N$ goes to infinity does not exist.

    • It's not clear to me how this sort of see-saw argument fails in any set that has something like a positive semidefinite inverse norm with infinite support. ($P(n)$ is that inverse norm here, taking cardinalities (a usual norm) to the number of isomorphism classes.)
Eric Towers
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  • @PedroTamaroff: For 6 month old questions, it's time to start proposing things and see if anything sticks. – Eric Towers Feb 12 '14 at 08:15
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    You can comment. – Pedro Feb 12 '14 at 08:19
  • @PedroTamaroff: Because being lost in a sea of 30 comments is productive? – Eric Towers Feb 12 '14 at 08:19
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    Maybe make the answer community Wiki then, and people can add sentences? – Pedro Feb 12 '14 at 08:21
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    Heuristically "almost all groups are $2$-groups," so we can confidently predict that "the cardinality of $G$ is even" will correspond to a probability of zero. – anon Feb 12 '14 at 08:22
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    @anon Shouldn't that be probability $1$ instead? – EuYu Feb 12 '14 at 08:28
  • Errr, yes. Derp. – anon Feb 12 '14 at 08:29
  • @PedroTamaroff: Wikified. – Eric Towers Feb 12 '14 at 08:31
  • In addition, is your second sentence even possible to write in the language of groups? –  Feb 12 '14 at 17:26
  • @Mike: And how many are of order $3 \cdot 2^k$, at least as many as there are of order 3 times those of order $2^k$. Q.v. superlinearity w.r.t the order of groups. The odd numbers are about as dense as the even numbers, so after some initial bias this effect should settle out... – Eric Towers Feb 12 '14 at 17:27
  • @Mike: Sentence 2: The sentence asks whether the cardinality of a group is in a list. This is a first order sentence. – Eric Towers Feb 12 '14 at 17:29
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    @EricTowers The point is that before you get to the groups of order $3\cdot 2^k$ you hit the groups of order $2\cdot 2^k = 2^{k+1}$ and there are so vastly many of those that they dwarf the groups of $3\cdot 2^k$. The bias doesn't 'settle out' because 2 is the smallest number, so among the numbers with prime multiplicity $k$ $2^k$ is always smallest. – Steven Stadnicki Feb 12 '14 at 17:29
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    How do you propose to write the cardinality of a group in the language of groups? In particular, see Emil Jeřábek's comments here –  Feb 12 '14 at 17:31
  • How many groups of order 2^k are there? How many of order 3^k, 5^k, 7^k, 9^k, 11^k,... Which of these lists is denser and made up only of odd order groups? – Eric Towers Feb 12 '14 at 17:34
  • @Mike: The language of groups includes ZFC by reference. – Eric Towers Feb 12 '14 at 17:37
  • Empirically (and there are probably some conceptual arguments), the number of groups of order $2^k$ will overwhelm all of the groups between $2^k$ and $2^{k+1}$ (collectively), which is why we should expect groups to be even order (indeed, to be $2$-groups) with probability $1$. I don't see how your argument in #1 connects any blow to this fact. – anon Feb 12 '14 at 19:29
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    In #2: would I be correct in saying "cardinality $n$" (for fixed $n$) can be expressible as a simple combination of $\exists,\not\exists,=,\ne$s? However, $\phi$ seems to implicitly quantify over an infinite number of possible cardinalities $-$ is this an accurate impression? $-$ so I don't see why $\phi$ would be expressible in the first-order language of groups. – anon Feb 12 '14 at 19:30
  • @EricTowers Which of the numbers $2^k, 3^k, 5^k,$ etc. is going to be smaller than $N$ for a given $N$? Remember that we're looking at an asymptotic density, with ratio given by group-size. Yes, there are more groups of size $3^k$ than of size $2^k$ - but there are even more groups of size $2^{k+1}$ than of size $3^k$, and $2^{k+1}\lt 3^k$ for all sufficiently large $k$. – Steven Stadnicki Feb 12 '14 at 21:36
  • @Steven: Which diverges, the number of powers of $2$ less than $2^k$ (i.e. $k$) or the number of powers of odds less than $2^k$: the sum $k\ln(2) \left( \frac{1}{\ln 3} + \frac{1}{\ln 5} + \frac{1}{\ln 7} + \frac{1}{\ln 9} + \cdots \right)$. – Eric Towers Feb 13 '14 at 01:17
  • @EricTowers Yes, but none of those powers has exponent $k$, which is the driving factor in the number of groups that exist. It's not proven, but it's very strongly conjectured that 'almost all groups are 2-groups', and there's good reason both theoretical and empirical; I encourage you to look at the table through $n=400$ at http://mathworld.wolfram.com/FiniteGroup.html to try and understand this phenomenon. – Steven Stadnicki Feb 13 '14 at 01:29
  • @StevenStadnicki: I have done so. And worked up through 2047. And worked up through a few tens of thousands using various bounds for the currently un-precisely-known values. What I see is (obvious) jumps for even cardinalities and two opposing effects: gradual majority of odds to evens between the powers of two and increasing densities of middling-sized jumps at odd powers. The whole thread of "even must win" sounds like small number bias. – Eric Towers Feb 24 '14 at 07:04
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    @EricTowers More than 99% of all the groups of order less than 2047 have order 1024. And it's known that there are at least 20,000 times as many groups of order 2048 as of order 1024. I'm not sure where your 'gradual majority of odds to evens' is coming from. – Steven Stadnicki Feb 24 '14 at 07:30
  • @StevenStadnicki: Actual plotting of odds/total. Between large negative excursions (at powers of two) are gradually increasing plateaus containing positive excursions (at powers of odds). – Eric Towers Feb 25 '14 at 05:30
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    @EricTowers Can you point to a single $n\gt 256$ where the majority of groups of order $\leq n$ are not of order a power of 2? – Steven Stadnicki Feb 25 '14 at 07:14
  • @StevenStadnicki: Can you cite a single $x$ where $\pi(x) < \mathrm{li}(x)$? Experience with trifling small number (say $<10^{400}$) is inadequate to inform of asymptotic behaviour. – Eric Towers Feb 28 '14 at 06:58