On products of Sylow subgroups in finite groups

D.F.Holt and P. Rowley

November 18, 1997

Let G be a finite group and let A_{i}(1 ≤ i ≤ r) be an ordered sequence of subsets of G. The product of the A_{i}, ΠA_{i}, is defined by

ΠA_{i} = {a_{1}a_{2} … a_{r}|a_{i}ϵA_{i}}.

In this note, we are concerned with the following question.

(∗) Is it possible to find a Sylow p_{i}-subgroup P_{i} of G, one for each prime p_{i} dividing the order of G, such that, in some order, G = ΠP_{i}?

If G is a soluble group then, by a theorem of P. Hall (Theorem 6.4.3 of [1]), we can choose the P_{i} such that P_{i}P_{j} = P_{j}P_{i} for all i and j, and it follows immediately that the answer to (∗) is yes, and the product can be taken in any order. More generally, (∗) has a positive answer whenever G = AB and A and B are soluble Hall-subgroups of G. So, for example, if G = A_{5} then we have G = HP_{5} with H = A_{4}, and so we can find P_{2} and P_{3} such that G = P_{2}P_{3}P_{5} = P_{3}P_{2}P_{5}. However, it is easily checked that there is no choice of the P_{i} for which G = P_{2}P_{5}P_{3}, and in fact all products of this form have order 48. (Many of the calculations described here were carried out on a computer using the CAYLEY package.) Moving on to G = L_{3}(2) ≅ L_{2}(7), we find that this group has Hall π-subgroups for both π = {2, 3} and π = {3, 7}, and so the product can be taken in any order in this example.

Now let G = P GL(2, q) for any prime power q. Then G = H_{1}H_{2}H_{3} = H_{1}H_{3}H_{2}, where H_{1} is cyclic of order q + 1, H_{2} is abelian of order q, and H_{3} is cyclic of order q − 1. Furthermore, H_{3} normalizes H_{2}. Since 2 is the only possible common prime factor of |H_{1}| and |H_{3}|, we can choose Sylow 2-subgroups of H_{1} anmd H_{3} which together generate a Sylow 2-subgroup of G, and (∗) holds. It then follows easily that (∗) is also true in the groups P SL(2, q), which have index 1 or 2 in G. A similar argument works in P GL(3, q). Here, we have G = H_{1}H_{2}H_{3}H_{4}, where H_{1}, H_{2} and H_{3} are as in P GL(2, q) (except that H_{2} has order q^{3} and H_{3} is abelian of order (q − 1)^{2}), and H_{4} is cyclic of order q^{2} + q + 1. The only common possible common prime factor of |H_{3}| and |H_{4}| is 3, and so we move the Sylow 2-subgroup of H_{3} past the H_{2} to bring it next to that of H_{1} and leave the Sylow 3-subgroup of H_{3} next to that of H_{4}.

Arguments like those above work in a number of other examples, such as A_{7} and A_{8}, but they become more difficult as the the number of primes dividing |G| increases, and we cannot always bring the different parts of all of the Sylow subgroup together.

The smallest nonabelian simple group for which this type of argument apparently does not work is the unitary group G = U_{3}(3), and (∗) does indeed turn out to have a negative answer in this case. We have |G| = 6048 = 2^{5}.3^{3}.7. To verify the failure of (∗), we proceed as follows:

(i) select a Sylow 2-subgroup P_{2} of G; (ii) choose orbit representatives P_{7}^{j}(1 ≤ j ≤ 9) for the action of P_{2} by conjugation on

Syl_{7}(G);

(iii) determine the orders of P_{2}P_{3}P_{7}^{j}, P_{3}P_{2}P_{7}^{j} and P_{2}P_{7}^{j}P_{3}, for 1 ≤ j ≤ 9 and each P_{3}ϵSyl_{3}(G).

The orders of the 756 products arising in Step (iii) were determined using CAYLEY. The orders that actually arose were 2592, 3024, 3360, 3456, 3584, 3780, 4032, 4256, 4320, 4536, 4704, 4914, 5184 and 5292. Since |G| = 6048, the answer to (∗) is no for G = U_{3}(3).

In order to give the reader the opportunity to repeat the above calculations, we give generators for P_{2} and the P_{7}^{j} in a 28 point permutation representation of G. Any P_{2} and P_{7} will together generate G. P_{2} is generated by the two permutations

(1,17,3,2)(4,15,13,19)(5,12)(6,14,9,11)(7,24,23,22)(8,18,25,28)(10,27,26,21)(16,20) and (2,15,17,19)(5,16,20,12)(6,25,11,21)(7,18,22,10)(8,14,27,9)(23,28,24,26),

whereas the P_{7}^{j} are respectively generated by

(1,6,15,25,13,8,9)(2,22,26,17,4,20,3)(5,7,23,11,21,18,28)(10,24,27,12,19,16,14), (1,6,21,13,8,12,14)(2,4,20,7,26,5,3)(9,10,28,17,22,23,27)(11,15,16,19,25,18,24), (1,9,13,2,16,20,15)(3,4,24,8,6,27,22)(5,7,23,12,28,11,10)(14,17,18,25,21,26,19), (1,6,22,8,12,18,15)(2,28,25,14,16,27,20)(3,7,26,10,4,9,5)(11,24,13,23,19,17,21), (1,14,8,3,11,26,21)(2,5,7,19,12,6,25)(4,24,15,20,23,13,22)(9,10,28,18,27,17,16), (1,24,18,22,26,3,8)(2,21,25,20,28,27,23)(4,13,16,14,11,9,15)(5,6,17,7,19,12,10), (1,4,11,19,9,7,25)(2,13,10,12,21,6,24)(3,20,8,16,23,27,26)(5,22,15,28,18,14,17), (1,2,19,23,8,9,13)(3,26,17,18,22,14,28)(4,25,15,20,21,27,5)(6,16,12,7,24,10,11), and (1,8,19,26,3,9,23)(2,17,16,28,13,27,20)(4,14,18,25,10,21,24)(5,12,22,6,15,7,11).

The fact that the groups P_{7}^{j} really do lie in distinct orbits of the conjugation action of P_{2} on Syl_{7}(G) can be checked by hand in the following manner. The orbits of P_{2} on the set Ω = {1, 2, …, 28} are O_{1} = {1, 17, 3, 2, 15, 19, 4, 13}, O_{2} = {5, 12, 16, 20} and O_{3} = Ω \ (O_{1} ∪ O_{2}). The sizes of the intersections of O_{k}(1 ≤ k ≤ 3) with the cycles of the generators of the P_{7}^{j} are invariant under conjugation by P_{2}, and this enables us to check that no two of the P_{7}^{j}(1 ≤ j ≤ 9) are conjugate under P_{2}.

Finally, Jon Hall has observed that the question (∗) has a representation-theoretical interpretation. Let K be a field and, for H ≤ G, let e_{H} = ∑_{hϵH} h (ϵKG). Then (∗) is equivalent to the following question. Can we find Sylow subgroups P_{i} of G, one for each prime divisor of |G|, such that, in some order, Πe_{P_{i}} = e_{G}?

References.

[1] D. Gorenstein, “Finite Groups”, Harper and Row (1968).