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In the comments to this question, it was remarked that the question of whether a finite group of order $p(p+1)$ is necessarily not simple becomes fairly interesting if we do not assume that $p$ is a prime— if it is a prime, that question's answer and the linked duplicate's answers (and comments) answers in the affirmative.

The following basic GAP code

for n in [2..10^5] do 
  iter:=SimpleGroupsIterator(n*(n+1),n*(n+1)); 
  for i in iter do 
    Print(n, "->", i, "\n"); 
    break; 
  od; 
od;

Shows that no simple groups of such order exist for $n\leq 10^5$ (so no examples with $|G|\leq 10^5(10^5+1)$). The iterator function quickly starts hitting issues soon after this point, especially once the order meets or exceeds $|A_{15}|$, which occurs before $n=10^6$, limiting how far one can search with this method. It takes a few minutes to check up through $n=10^5$. The following variant finds no simple groups of suitable order before $A_{15}$:

iter := SimpleGroupsIterator(3,Factorial(15)/2-1);; 
for G in iter do 
  if (Tau(4*Order(G)+1) mod 2) = 1 then 
    Print(G, "  ", Order(G), "\n"); 
    break; 
  fi; 
od;

I would expect one should be able to hammer out a proof that all such groups are not simple (assuming this is a true statement) by using the classification of finite simple groups and doing a brute-force check on the sporadic and infinite family orders. But is there a less brutal and more clever way, or does such a simple group actually exist?

EDIT: A few hours later, and the first bit of code finished checking up through $n=10^6$ and found no examples, as was claimed by Derek Holt in the aforementioned comments.

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    $\begingroup$ Going by the en.wikipedia.org/wiki/List_of_finite_simple_groups, clearly the Cyclic ones do not satisfy this. I checked the alternating groups from $A_n$ for $n \in \{5,...,99\}$ and found nothing, nor did I find anything in the sporadic simple groups. EDIT: perhaps I should mention that, rather than using GAP to go through all groups, I'm simply using the formulas given there to go through all possible orders. $\endgroup$
    – sTertooy
    May 14, 2016 at 17:02
  • $\begingroup$ @SteamyRoot Yes, that's what I meant that you should be able to brute force the answer via the classification theorem. For the record, I would accept, or at least up vote and appreciate, a conclusive answer that used the classification. Though I would prefer an answer that doesn't use one, if possible. $\endgroup$ May 14, 2016 at 19:05
  • $\begingroup$ I agree that a proof not using the classification would be nicer, but currently I'm just trying to brute-force for some counterexamples. Anyway, any non-abelian finite simple group has even order, and a necessary requirement to be of the form $n(n+1)$ is that $4|G|+1$ is a perfect square. I'm not sure if that would be useful, but who knows... $\endgroup$
    – sTertooy
    May 14, 2016 at 19:30
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    $\begingroup$ Ah, I hadn't really looked at that. Anyway, the list on madore.org/~david/math/simplegroups.html contains no orders such that $4|G|+1$ is a perfect square. $\endgroup$
    – sTertooy
    May 14, 2016 at 19:53
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    $\begingroup$ Well, I upgraded the script quite a bit and managed to check all orders up to $10^{23}$, with no counterexamples found. The update code can be found at pastebin.com/HyKGVbWR. I probably won't bother looking any further, though. $\endgroup$
    – sTertooy
    May 15, 2016 at 18:34

1 Answer 1

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The alternating groups $A_m$ are ruled out for sufficiently large $m$ by the abc conjecture. If $n(n+1) = \frac{m!}{2}$, then by the abc conjecture we have (starting from the identity $n + 1 = (n+1)$) for any $\varepsilon > 0$ a constant $K_{\varepsilon}$ such that

$$n < K_{\varepsilon} \text{rad}(n(n+1))^{1 + \varepsilon}$$

for all $n$, but $\text{rad}(n(n+1)) = \text{rad} \left( \frac{m!}{2} \right)$ is the product of all the primes less than or equal to $m$. This is asymptotically about $e^m$, whereas $n$ is asymptotically about $\left( \frac{m}{e} \right)^{\frac{m}{2}}$ by Stirling's approximation.

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    $\begingroup$ Is there any way to decide how large is "sufficiently large" here? $\endgroup$ May 14, 2016 at 19:28
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    $\begingroup$ Not without proving the abc conjecture... $\endgroup$ May 14, 2016 at 19:57

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