# Conjectures for finite groups that fail with large counterexamples

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!

• "Every product of commutators is a commutator" works until $N=96$. Commented Dec 26, 2023 at 4:57
• On the thread reddit.com/r/math/comments/qy4bwo/…, 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 oeis.org/A317480. Commented Dec 26, 2023 at 5:55
• Should large be at least 10000?
– KCd
Commented Dec 26, 2023 at 15:20
• 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. Commented Dec 27, 2023 at 2:50

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.

• Very nice list Robin! Commented Dec 26, 2023 at 14:27
• "All commutators of any finite group are commutators". If that can be wrong, what can I still believe in? :) Commented Dec 26, 2023 at 15:09
• @darijgrinberg Oops, that meant to say all products of commutators..., thanks. Commented Dec 26, 2023 at 16:12
• @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. Commented Dec 26, 2023 at 20:38
• Understood, Robin, but I was surprised to see so many entries for a number that large. Commented Dec 29, 2023 at 3:34

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? $$N=3380195 = 5 \times 7 \times 13 \times 17 \times 19 \times 23$$, ...

• 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 oeis.org/A264907 . Commented Jan 11 at 23:00
• @StevenStadnicki That's right, thank you. I'll edit this now. Commented Jan 11 at 23:04

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.