Simple groups of order ≤ 720

We begin with a few more lemmas to help narrow the cases.

Lemma 5.22 IfHis a group of orderpr_qs_{, wherepandqare primes andr, s≤2thenHis not simple}

Proof: We may assume p > q. If H is simple then it has p-factorization pr_qs _{= p}r _{·ν ·(1 +kp)}

wthk ≥ 1. Since r ≤ 2, the Sylow p-subgroups are abelian, so ν > 1. Hence ν = q, s = 2 and

q= 1 +kp > p, a contradiction

Remark: The restrictions onr ands are unnecessary. Using character theory, Burnside proved that any finite group whose order is divisible by just two primes is not simple.

Lemma 5.23 If|G|=m·pr_{wherepis a prime not dividingm, thenm >7.}

Proof: SinceGis simple, the action on Sylowp-subgroups embedsG ,→ Am, som ∈ {5,6,7}and

m·pr dividesm· · ·3·2. Ifm ≤6thenmpr < 60soGis not simple. Ifm = 7thenp = 5,3,2and

r ≤ 1,2,3respectively. and the7-factorization isν ·(1 + 7k)withν > 1and k > 0. There are no

values ofpandrsatisfying these conditions.

Lemma 5.24 If|G| ≤N thenp <pN/2.

Proof: From Lemma.5.16, we havep·2≤pr_{·ν. Hence}

p·2·(1 +p)≤ |G| ≤N,

sop2 _{< p(1 +p)≤N/2, and the estimate follows.}

Now let G be a simple group of order |G| ≤ 720. Let p be the largest prime dividing |G|, with

p-factorization

|G|=pr·ν·(1 +kp).

We will assumep≥5because groups of order3a2bare not simple, by Burnside’spaqbtheorem (which uses character theory and is not yet in these notes). By Cor. 5.24, we have p < √360 < 19 so

p≤ 17. By Prop. 5.18, we exclude the case (r, ν, k) = (1,2,1). This further excludesp = 17. Since

pr·(1 +p)≤ |G| ≤720we haver= 1forp= 13,11,r≤2forp= 7,5,r ≤4forp= 3. We note also:

1. Ifr= 1then(ν, p−1)>1.

2. Ifν = 2thenkis odd andk ≥3by Cors. 5.13and5.18. 3. Ifk = 1thenr= 1sincep2 _{does not divide(p+ 1)!}

4. Ifr≤2thenν ≥2so1 +pk≤720/2pr_.

5. No prime divisor of1 +kpcan be larger thanp. 6. Ifkis even then either4|νorν is odd.

Sincepr_{·(1 +p)≤ |G| ≤720we haver= 1forp= 13,11,r≤2forp= 7,5,r≤4forp= 3.}

Forp= 13the only surviving case is

|G|= 13·4·(1 + 13) = 728,

which we showed cannot be the order of a simple group, see Example 3 of section5.3. Forp= 11it the only surviving case is

|G|= 11·5·(1 + 11) = 660,

which is the order of the simple groupP SL2(11).

Forp = 7, we haver ≤ 2. Ifr = 2thenν ≥ 2so1 + 7k ≤ (720/2·49) <8, forcingk = 0. Hence

r= 1andν ≥2. And ifk= 1thenν|6, and ifkis even then4|ν. The surviving7-factorizations are

7·3·(1 + 7) = 23·3·7 = 168 7·6·(1 + 7) = 24·3·7 = 336 7·4·(1 + 2·7) = 22·3·5·7 = 420 7·2·(1 + 5·7) = 23·32·7 = 504 7·2·(1 + 7·7) = 22·52·7 = 700

The case|G| = 168 is the order of a simple group PSL2(7), and this is the unique simple group of

order168(see Prop.5.20).

The case|G| = 336hasNG(P)'P oC6, theax+bgroup overZ/7Z. An elementn ∈C6 of order

six fixes a point Q ∈ X − {P} and is free on X − {P, Q}, hence has cycle type [61] so is an odd permutation. HenceG has a subgroup of index two. This occurs in nature: The groupPGL2(7) has

order336and containsPSL2(7)with index two.

The case|G| = 420 has an involutions ∈ CG(P). Suppose there exists Q ∈ X − {P} andx ∈ P

such thats_Q= x_{Q. Thensx ∈N}

G(Q)sox2 ∈ QsoP =Q. Hencesinterchanges the twoP-orbits

inX− {P}and has cycle type[271]onX, which is an odd permutation, soGhas a normal subgroup of index two.

The case|G|= 504is the order of a simple groupPSL2(8).

The case|G|= 700is ruled out by Cor.5.14.

Forp = 5we havek = 1,3,7since5is the largest prime dividing1 + 5k. If k = 1thenr = 1and

ν= 2,4by Lemma5.17. Ifk= 3then5r_{·ν ≤720/16 = 45sor= 1andν≤9is even. Since groups}

of order5·2n_{are not simple, by Lemma5.23we haveν = 6. Ifk = 7then5}r_{·ν ≤720/36 = 20so}

r= 1andν = 2,4.

Thus the surviving5-factorizations are

5·2·(1 + 5) = 2·3·5 = 60 5·4·(1 + 5) = 23·3·5 = 120 5·6·(1 + 3·5) = 25·3·5 = 480 5·2·(1 + 7·5) = 23·32·5 = 360 5·4·(1 + 7·5) = 23·32·5 = 720

The case5·2·(1 + 5) = 60arises from the simple groupA5 =P SL2(5).

In the case5·4·(1 + 5) = 120the normalizer of a Sylow5-subgroup contains a4-cycle inS6, which is

odd. HenceGcontains a normal subgroup of index two. This actually occurs: The groupS5 has order

120and containsA5 of index two.

In the case5·6·(1 + 3·5) = 480, we haveNG(P)of order30, soNG(P) = hg, siwhereg, s have

order15and2respectively, andhg3_{i=P. There are four groups of order30according to the possible}

actions ofs onhgi = C3 ×C5 by inverting one or both or no factors. Note that all four groups occur

inside the normalizer of a15-cycle inS15. By the Transfer Theorem,smust invert theC5-factor (which

isP). Now look at the embedding ofNG(P)inS15via its action onX− {P}. Sinceg3acts with cycle

type[5 5 5], it follows thatg has cycle type[15], sohgiacts freely and transitively onX− {P}. Hence the action ofsonX− {P}is equivalent to its action onC3×C5. Ifsinverts both factors then its cycle

type is[1 27_{]which is odd. Thereforescentralizes theC}

3 factor andNG(P)'C3×D5. 9 Hence the

centralizerCG(Q)of a Sylow3-subgroup has order at least30. In the3-factorization3·ν3·n2we must

haveν3 = 5·2awitha≥2, lestCG(Q) =NG(Q). This forcesn3 = 4, soGis not simple

The case5·2·(1 + 7·5) = 360arises from the simple groupA6.

The case|G| = 720 is harder, but still elementary; in the next section we show there are no simple groups of order720.

9_{Cole (Simple groups from Order 201 to Order 500, Am. J. Math, vol 14, no. 4 1892, 378-88), seems to have missed}

Assuming that result, we have shown that the only possible orders≤720of nonabelian simple groups

Gare60,168,360,504and660. Moreover, there exists a simple group of each of these orders, namely simple groupG |G| A5 'P SL2(5) 60 P SL2(7)'GL3(2) 168 A6 'P SL2(9) 360 P SL2(8) 504 P SL2(11) 660

In fact, there is exactly one simple group having each of these orders, but we have only proved this for

|G|= 60and|G|= 168.