# Abundancy index and non-solvable finite groups

Let $$\sigma$$ be the sum-of-divisors function. A number $$n$$ is called abundant if $$\sigma(n)>2n$$. Note that the natural density of the abundant numbers is about $$25 \%$$. The abundancy index of $$n$$ is $$\sigma(n)/n$$. The following picture displays the abundancy index for the $$10000$$ first orders of non-solvable groups (see A056866). Observe that for $$G$$ non-solvable with $$|G| \le 446040$$ then $$|G|$$ is abundant, with minimal abundancy index $$\frac{910}{333} \simeq 2.73$$.

Question 1: Are the non-solvable groups of abundant order?

Note that the number of integers $$n \le 446040$$ with $$\sigma(n)/n \ge 910/333$$ is exactly $$19591$$, so of density less than $$5 \%$$ with more than half of them being the order of a non-solvable group. Among those which are not the order of a non-solvable group, the maximal abundancy index is $$512/143 \simeq 3.58$$, realized by $$n=270270$$, whereas there are exactly $$896$$ numbers $$n \le 446040$$ with $$\sigma(n)/n > 512/143$$, which then are all the order of a non-solvable group.

Question 2: Is a number of abundancy index greater than $$512/143$$ the order of a non-solvable group?
Weaker version 1: Is there $$\alpha >3$$ such that a number of abundancy index greater than $$\alpha$$ must be the order of a non-solvable group?
Weaker version 2: Is there $$\beta < 1$$ such that a number $$n$$ of abundancy index greater than $$\beta e^{\gamma} \log \log n$$ must be the order of a non-solvable group?

Recall that $$\limsup \frac{\sigma(n)}{n \log \log n} = e^{\gamma}$$ with $$\gamma$$ the Euler-Mascheroni constant.

Finally, there are non-solvable finite groups $$G$$ with $$|G| \gg 446040$$ and abundancy index less than $$\frac{910}{333}$$. The non-abelian simple groups $$G$$ with $$|G|=n \le 749186071932$$ and $$\sigma(|H|)/|H|>\sigma(n)/n$$ for all non-abelian simple groups $$H$$ of order less than $$n$$ are exactly the 39 the simple groups $$\mathrm{PSL}(2,p)$$ with $$p$$ prime in {5, 37, 107, 157, 173, 277, 283, 317, 563, 653, 787, 907, 1237, 1283, 1307, 1523, 1867, 2083, 2693, 2803, 3413, 3643, 3677, 4253, 4363, 4723, 5443, 5717, 6197, 6547, 6653, 8563, 8573, 9067, 9187, 9403, 9643, 10733, 11443}. Let $$n_p:=|\mathrm{PSL}(2,p)| = p(p^2-1)/2$$.

It follows that for $$G$$ non-abelian simple with $$|G| \le 749186071932$$ then $$|G|$$ is abundant, with minimal abundancy index $$\sigma(n_{11443})/n_{11443} = 50966496/21821801 \simeq 2.33.$$

The following picture displays the adundancy index of $$n_p$$ for $$p$$ prime and $$5 \le p \le 10^6$$. The minimal is $$579859520520/248508481289 \simeq 2.3333 \simeq 7/3$$, given by $$p=997013$$.

Question 3: Is it true that $$\inf_{p \ge 5, \text{ prime}} \sigma(n_p)/n_p = 7/3$$?

Question 4: Is the abundancy index of the order of a non-solvable group greater than $$7/3$$?

Fun fact: the smallest integer $$n$$ such that there exists two non-isomorphic simple groups of order $$n$$ is $$20160$$, whereas the biggest integer that is not the sum of two abundant numbers is $$20161$$ (see A048242). Any explanation?

• Question 3 seems hard. It has something to do with the factorizations of $p-1$ and $p+1$. – Thomas Browning 2 days ago
• If there are infinitely many primes $p$ such that $p-1$ is 4 times a prime and $p+1$ is 6 times a prime, then question 3 has a positive answer. 997013 is one such prime. We may also rearrange the factors of 2 and 3 to get other sufficient conditions. Any of these follow from well-known conjectures, like en.wikipedia.org/wiki/Dickson%27s_conjecture but they are still open. – S. Carnahan 2 days ago
• One question that fits into this theme but you didn't ask is: Among the numbers $n$ such that all groups of order $n$ are solvable, is the abundancy index of $n$ bounded? The answer is no, because the abundancy index of odd numbers is unbounded, and the Odd Order Theorem asserts that any odd natural number is a satisfactory value for $n$. – S. Carnahan 2 days ago
• $29$ is a smaller prime one more than four times a prime and one less than six times a prime. – Gerry Myerson 2 days ago
• @GerryMyerson That's great, but I mentioned 997013 because it is briefly considered just before question 3, as the prime $p$ satisfying $5 \leq p < 10^6$ that yields the smallest abundancy index for $|PSL(2,p)|$. – S. Carnahan 2 days ago

I can answer Questions 1 and 4.

Make sure you look at S. Carnahan's answer. It deals with Questions 2 and 3.

Questions 1 and 4: If a finite group $$G$$ is not solvable then $$|G|$$ is divisible by $$|G_0|$$ for some finite simple group $$|G_0|$$. By the CFSG, either $$12\bigm||G_0|$$ or $$G_0$$ is a Suzuki group. If $$12\bigm||G_0|$$ then $$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(12)}{12}=\frac{7}{3}.$$ If $$G_0$$ is a Suzuki group then $$320\bigm||G_0|$$ and $$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(320)}{320}>\frac{7}{3}.$$ Thus, every non-solvable group has abundancy index larger than $$\frac{7}{3}$$.

• Question 2 had typos, what I wanted to ask (leading by previous paragraph) is: Is a number of abundancy index greater than 512/143 the order of a non-solvable group? (same correction for the weaker versions). Fixed! – Sebastien Palcoux 2 days ago
• What is the reference for: if $G$ non-abelian finite simple group then its order is a multiple of $12$, or it is Suzuki? – Sebastien Palcoux 2 days ago
• By a theorem of Burnside, dating back to the early 20th (or maybe late 19th) century, if $G$ is a finite non-Abelian simple group, then $|G|$ is divisible either by $12$ or the cube of its smallest prime divisor. Suzuki groups were proved by Thompson to be the only simple groups of order prime to $3$ (actually before the full classification), and no non-Abelian simple group has odd order, by Feit-Thompson- more generally, no non-Abelian simple group has a cyclic Sylow $2$-subgroup. Hence every non-Abelian simple group other than a Suzuki group has order divisible by $12$. – Geoff Robinson 2 days ago
• @ThomasBrowning: Question 2 (or its weaker versions) asks about a sufficient condition on an integer to be the order of a non-solvable group. It is not about a necessary condition. – Sebastien Palcoux 2 days ago
• @SebastienPalcoux Then doesn't Question 2 have a negative answer by S. Carnahan's comment? Take n to be odd with large abundancy index. – Thomas Browning yesterday

As I mentioned in a comment, Question 2 (in its revised form) has a negative answer, because odd natural numbers have unbounded abundancy index, while the Odd Order Theorem implies all groups of odd order are solvable.

Weaker version 2 has a positive answer: If $$\beta$$ is sufficiently close to 1, then any $$n > 1$$ whose abundancy index is greater than $$\beta e^\gamma \log \log n$$ is a multiple of 60, so there is a group of order $$n$$ that is unsolvable.

As I mentioned in a different comment, Question 3 is true subject to well-known open conjectures, such as [Dickson's conjecture]. In particular, it suffices to show that there are infinitely many primes $$p$$ such that $$p-1$$ is 4 times a prime and $$p+1$$ is 6 times a prime. : https://en.wikipedia.org/wiki/Dickson%27s_conjecture

• That’s right! One way for Question 2 (or weaker version 1) to survive would be to exclude the multiple of odd abundant numbers. The least odd number of abundancy index greater than $3$ is $1018976683725$. – Sebastien Palcoux yesterday