35

I'm asking this question in the same spirit as this other question: Conjectures that have been disproved with extremely large counterexamples?.

What's a nice conjecture relating to finite groups, that first fails for a group of order $N$, with $N$ large? By "nice" I mean a conjecture with a balance between having a simple statement, one that first fails for a large $N$, and something that is not immediately obvious that it is going to fail.

An example might be the conjecture "if $G$ is a finite simple group, then it is the unique simple group of its order", which first fails for $N=20160$.

"Large" in this context is of course undefined, perhaps we'll say $N$ is "large" if the number of groups of order at most $N$ up to isomorphism is "large" in a more generic context. Perhaps I might suggest $N \geq 32$ as large, since $N= 32$ is the smallest number such that there are at least $100$ groups of order at most $N$. The bigger $N$ you can come up with, though, the better!

Robin
  • 3,227
  • 9
    "Every product of commutators is a commutator" works until $N=96$. – Gerry Myerson Dec 26 '23 at 04:57
  • 4
    On the thread https://www.reddit.com/r/math/comments/qy4bwo/conjectures_which_have_very_large_counterexamples/, I found "Let $S(n)$ be the sum of the order of each element in the cyclic group of order $n$. Then, $n$ does not divide $S(n)$ for all $n > 1$." Fails at $N = 614341$. See https://oeis.org/A317480. – The Surgeon of Death Dec 26 '23 at 05:55
  • 1
    Should large be at least 10000? – KCd Dec 26 '23 at 15:20
  • 1
    Some comments here got deleted. Yes, they were answers to the question and should have been posted as such - some of them are in my list below (particularly the commutator subgroup one with $N=96$ came up) but others have been deleted, and I can't remember what they were. Unfortunate. – Robin Dec 27 '23 at 02:50

3 Answers3

40

According to this answer by Chain Markov on the post I linked to, here are a few, ordered from smallest to largest (known) counterexample:

  • Automorphism group of a non-abelian group is non-abelian: counterexample at $N=64$.

  • All products of commutators of any finite group are commutators. Counterexample at $N=96$. (Also mentioned in a comment by Gerry Myerson on this question)

  • Automorphism groups of all finite groups not isomorphic to $\{e \}$ or $C_2$ have even order. Counterexample at $N=2187$ (automorphism group of said group has order $729$)

  • Moreto conjecture (very similar to the one I put in the body of the question): Let $S$ be a finite simple group and $p$ the largest prime divisor of $|S|$. If $G$ is a finite group with the same number of elements of order $p$ as $S$ and $|G| = |S|$, then $G \cong S$. Also fails at $N=20160$ with the simple groups of that order.

  • Any Leinster group has even order. Smallest known counterexample at $N=355433039577$.

  • Any nontrivial complete finite group has even order (a conjecture of S. Rose): smallest known counterexample at $N=788953370457$. (c.f. A341298 in the OEIS.)

  • Hughes conjecture: suppose $G$ is a finite group and $p$ is a prime number. Then $[G : \langle\{g \in G| g^p \neq e\}\rangle] \in \{1, p, |G|\}$. Smallest known counterexample at $N=142108547152020037174224853515625$.

  • Suppose $p$ is a prime. Then, any finite group $G$ with more than $\frac{p-1}{p^2}|G|$ elements of order $p$ has exponent $p$. Smallest known counterexample at $N=142108547152020037174224853515625$ (with $p=5$, same group as in the above one fails).

I'd be interested in any more people have.

Edit: Another answer of mine, kept separate, since it was not on Chain Markov's list.

Robin
  • 3,227
  • 3
    Very nice list Robin! – Nicky Hekster Dec 26 '23 at 14:27
  • 8
    "All commutators of any finite group are commutators". If that can be wrong, what can I still believe in? :) – darij grinberg Dec 26 '23 at 15:09
  • 1
    @darijgrinberg Oops, that meant to say all products of commutators..., thanks. – Robin Dec 26 '23 at 16:12
  • Or even sharper, the set of commutators equals the commutator subgroup $G’$. – Nicky Hekster Dec 26 '23 at 19:03
  • 1
    @NickyHekster True, thank you for that and your kind words! :) P.S. for everyone, I just found this question, according to the answers, there are two non-isomorphic groups of order $96$ with this property. The answers there give some descriptions for them. – Robin Dec 26 '23 at 20:38
  • There are five entries for $142108547152020037174224853515625$ at the Online Encyclopedia of Integer Sequences, none mentioning these group theory examples. – Gerry Myerson Dec 28 '23 at 22:37
  • @GerryMyerson The groups of order that number are the smallest known counterexamples, but are not proven to be the the smallest counterexamples as far as I know. Hence no OEIS entry, as they would want to list all the possible values in ascending order. – Robin Dec 29 '23 at 02:53
  • 1
    Understood, Robin, but I was surprised to see so many entries for a number that large. – Gerry Myerson Dec 29 '23 at 03:34
  • We can also add this from this question: non isomorphic groups of order $125000$ with isomorphic group algebras. – Jianing Song Jan 08 '24 at 04:33
10

Another answer from me, separate from my other answer since it was not on the list that I cited there:

Suppose $G$ is the unique group of order $N$ (that is, $G$ is cyclic). Then $N$ has at most $2$ prime factors. Fails first for $N=255 = 3 \times 5 \times 17$.

Interestingly, it'll also fail for all Carmichael numbers: Korselt's criterion for Carmichael numbers happens to be a stricter criterion than the classification of cyclicity-forcing numbers, which can be seen quite directly.

(Less interestingly, $1$ group of order $N$ $\implies N$ prime fails first for $N=15$.)

What about at most $3$ prime factors? Fails first for $N=5865=3 \times 5 \times 17 \times 23$.

At most $4$ prime factors? $N=146965=5 \times 7 \times 13 \times 17 \times 19$.

At most $5$ prime factors? $N=3380195 = 5 \times 7 \times 13 \times 17 \times 19 \times 23$, ...

Robin
  • 3,227
  • 1
    I think the quoted values are incorrect after 3 factors and the correct value for e.g. 4 prime factors should be $N=146965=5 \times 7 \times 13 \times 17 \times 19$: see https://oeis.org/A264907 . – Steven Stadnicki Jan 11 '24 at 23:00
  • @StevenStadnicki That's right, thank you. I'll edit this now. – Robin Jan 11 '24 at 23:04
3

Yet another answer from me. Not nearly as big, failing at $N=24$ with $2$ groups. For context, there are $59$ groups of order $\leq 23$, $74$ groups of order $\leq 24$.

"The commutator subgroup of a finite group is abelian" fails for two groups of order $24$ (they are $S_4$ and $SL_2(\mathbb{F}_3)$. $[S_4, S_4] = A_4$, $[SL_2(\mathbb{F}_3), SL_2(\mathbb{F}_3)] \cong Q_8$) and no smaller groups.

Robin
  • 3,227