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The orders of the simple groups (ignoring the matrix groups for which the problem is solved) all seem to be a lot like this:

2^46 3^20 5^9 7^6 11^2 13^3 17 19 23 29 31 41 47 59 71

starts with a very high power of 2, then the powers decrease and you get a tail - it's something like exponential decay.

Why does this happen? I want to understand this phenomenon better.

I wanted to find counter-examples, e.g. a simple group of order something like

2^4 3^2 11^5 13^9

but it seems like they do not exist (unless it slipped past me!).

We have the following bound $|G| \le \left(\frac{|G|}{p^k}\right)!$ which allows $3^2 11^4$ but rules out orders like $3^2 11^5$, $3^2 11^6$, .. while this does give a finite bound it is extremely weak when you have more than two primes, it really doesn't explain the pattern but a much stronger bound of the same type might?

I also considered that it might be related to multiple transitivity, a group that is $t$-transitive has to have order a multiple of $t!$, and e.g. 20! =

2^18 3^8 5^4 7^2 11 13 17 19

which has exactly the same pattern, for reasons we do understand. But are these groups really transitive enough to explain the pattern?

Alexander Gruber
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    For one, any finite simple -non-abelian group must be divisible by at least three different primes (Burnside), one of them must be 2 (Feit-Thompson) , and the group's 2-rank must be higher than 1, from which it follows that the group's order must be divisible at least by 4. In fact, I thin I once read somewhere that the group's order must be divisible by 12...anyway, there can'be be a simple non-abelian group of order the second number you wrote down. – DonAntonio Mar 14 '13 at 23:24
  • ok good point, I've improved the example from it now. I'm sure there's no simple group of that order. –  Mar 14 '13 at 23:29
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    I'm not sure about yet, but my guess is that a group with such a distribution of orders would somehow necessarily contain too many elements which centralize a Sylow $2$-subgroup $P$ (or a suitably sized subgroup) - in particular, $|P||C_G(P)|\geq |G|$. I will look into this further and let you know if I come up with anything. – Alexander Gruber Mar 14 '13 at 23:59
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    Apparently there are non-abelian finite simple groups who's order is not divisible by 3: http://brauer.maths.qmul.ac.uk/Atlas/v3/group/Sz8/ – Josh B. Mar 15 '13 at 00:12
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    @DonAntonio One can show using Burnside's transfer theorem that the order of a simple group is divisible by either $8$ or $12$. – Tobias Kildetoft Mar 15 '13 at 00:53
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    I think the idea is that the Sylow subgroups cannot be too large. We know that $p$-groups are very far from simple, so it seems vaguely plausible that if a Sylow subgroup gets too large then that ruins simplicity somehow. – Qiaochu Yuan Mar 15 '13 at 01:27
  • Thanks, @TobiasKildetoft. That together with what Josh wrote would imply, apparently, that a simple group must be divisible either by 12 or 24. – DonAntonio Mar 15 '13 at 02:36
  • Well @jug, there you are. Didn't know that. Thanks. – DonAntonio Mar 15 '13 at 12:42
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    Well @jug, the fact that 3 divides both 12 and 24 I did learn, after making a huge effort, a little after beginning my PhD studies...No, the "didn't know that" means that I didn't know the Suzuki groups' order are not divisible by $,3,$ . And I misread what Josh wrote. I thought it was that "apparently there are no non-abelian finite simple...". I was, as usual, being honest and direct in my thanking you and in my acknowledging my ignorance, just as I am being about my stupidity for not reading correctly Josh's comment. – DonAntonio Mar 15 '13 at 13:39
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    Multiply-transitive groups are severely limited, so this will not help. I am also a little confused by the question. For example, the order of $PSL(3,17)$ is $2^93^217^3307$. –  Mar 19 '13 at 19:24
  • @SteveD: The Lie-type groups are probably excluded by the remark "ignoring the matrix groups for which the problem is solved". (I think, the exponent of $p$ in $|PSL_n(p)|$ grows quadratically for $n \to \infty$, but only linear for all other primes.) – j.p. Mar 21 '13 at 17:17
  • I feel that number falls into the same type of pattern that I've been trying to describe. –  Mar 21 '13 at 17:17
  • @caveman: What exactly did you mean with your remark "ignoring the matrix groups for which the problem is solved"? – j.p. Mar 21 '13 at 17:34
  • @jug, I have formulas for their orders in terms of a factorial like product, this explains why the exponents decay over the primes for the simple matrix groups. –  Mar 21 '13 at 18:15
  • @caveman: I think it would be nice if you could add your formulas to the question. – j.p. Mar 21 '13 at 18:19
  • @jug, they are all here http://en.wikipedia.org/wiki/List_of_finite_simple_groups just look at the order row of the tables for the matrix groups. –  Mar 21 '13 at 18:25
  • Now that I look at it again.. it seems the steinberg groups break this pattern.. –  Mar 21 '13 at 18:29
  • @caveman: Yes, and I think there are more groups that do (unless I have misunderstood your pattern). But can you tell me if ${}^2A_5(79^2)$ satisfies your pattern or not? –  Mar 21 '13 at 20:26
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    Sorry, the order of ${}^2A_5(79^2)$ is $2^{23}\cdot 3^4\cdot 5^6\cdot 7^2\cdot 11^1\cdot 13^3\cdot 43^1\cdot 79^{15}\cdot 641^1\cdot 1091^1\cdot 3121^1\cdot 6163^2$. –  Mar 21 '13 at 20:28
  • @SteveD, this is definitely a counter-example due to the $79^{15}$. That answers my question if you would like to post it. Thank you –  Mar 21 '13 at 20:54

2 Answers2

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I am posting the following counterexample to the question, as requested by caveman in the comments.

The Steinberg group ${}^2A_5(79^2)$ has order $$ 2^{23}\cdot 3^4\cdot 5^6\cdot 7^2\cdot 11^1\cdot 13^3\cdot 43^1\cdot 79^{15}\cdot 641^1\cdot 1091^1\cdot 3121^1\cdot 6163^2.$$

There are other counterexamples, too. For example ${}^2A_9(47^2)$ has order $$ 2^{43}\cdot 3^{13}\cdot 5^2\cdot 7^3\cdot 11^1\cdot 13^2\cdot 17^2\cdot 23^5\cdot 31^1\cdot 37^1\cdot 47^{45}\cdot 61^1\cdot 97^1\cdot 103^3\cdot 3691^1\cdot 5881^1\cdot 14621^1\cdot 25153^1\cdot 973459^1\cdot 1794703^1\cdot 4778021^2.$$

I would guess there are infinite counterexamples, but the numbers (of course) get very very large!

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    It seems that in general for $G={} ^2!A_5(p^n)$, $v_p!(|G|)=10n$, which can indeed exceed $v_2!(|G|)$. – Alexander Gruber Mar 21 '13 at 23:07
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    @AlexanderGruber: Indeed, and increasing that $5$ also gives bigger and bigger discrepancies. –  Mar 21 '13 at 23:12
  • Ah, looks like we can find similar examples - "exponential decay except for one prime" - in the families $^2!D_n$, $^2!E_6$, $E_n$ ($n=6,7,8$), and $F_4$. A particularly striking counterexample is the Ree groups $^2G_2$ in which $v_2=3$ and $v_3$ can be arbitrarily high. – Alexander Gruber Mar 21 '13 at 23:36
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    So in the case of groups of Lie type, I think if we exclude the characteristic of the underlying field from the factorization, caveman's observation still holds. – Alexander Gruber Mar 21 '13 at 23:44
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Emil Artin showed in his 1955 papers$^1{}^2$ that groups of Lie type of characteristic $p$ have a large Sylow $p$-subgroup. That explains why the families $^2D_n,^2E_6,F_4,^2G_2$ and $E_n$ ($n=6,7,8$) have a "spike" at exactly their characteristic, and why the $2$-power in the orders of other groups of Lie Type of characteristic $2$ is much higher. It doesn't explain the underlying exponential pattern.

Orders of the alternating groups are easy to explain, so that leaves the sporadics. Of these, everything adheres to your observations except for the O'Nan group, which has has a little bump at $7$ (the order is $2^9 \cdot 3^4 \cdot 5 \cdot 7^3 \cdot 11 \cdot 19 \cdot 31$) and the same for $J_{11}$ (the order is $2^{21} \cdot 3^3 \cdot 5 \cdot 7 \cdot 11^3 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 43$). These don't violate your predictions by very much, though, so I suspect the extra $7$'s and $11$'s are just a number theoretic coincidence.

We observe that in Sylow subgroups of other groups of Lie type, the characteristic is $2$, so their order has a peak at the $2$-power. It makes sense that the orders of these groups look like especially sharp exponential decay. Several of the sporadic groups (sporadics of characteristic type $2$, for which $F^\star(Y)$ is a $2$-group for every $2$-local subgroup $Y$) have been described$^3$ to me as sporadics "trying to have characteristic $2$." In particular, they have large Sylow $2$-subgroups just like like groups of Lie type of characteristic $2$, so they also have a peak at $2$. The rest of the sporadic groups, I have no intuitive explanation.

  1. Emil Artin. "The orders of the classical simple groups." Communications in Pure and Applied Mathematics 8 (1955), 455–472.

  2. Emil Artin. "The orders of the linear groups." Communications in Pure and Applied Mathematics 8 (1955). 355–365.

  3. Thanks to G. Glauberman for this phrasing and his conversation on this topic.

Alexander Gruber
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