The probability of generating a finite classical group - CiteSeerX

WILLIAM

M. K A N T O R A N D A L E X A N D E R

THE PROBABILITY

OF GENERATING

CLASSICAL

LUBOTZKY

A FINITE

GROUP

For Jacques Tits on his sixtieth birthday

ABSTRACT. Two randomly chosen elements of a finite simple classical group G are shown to generate G with probability --*1 as IGI ~ co. Extensions of this result are presented, along with applications to profinite groups.

1. I N T R O D U C T I O N If two elements are chosen at random from a finite simple group G, will they probably generate G? Intuition strongly suggests that the answer is 'yes'. The purpose of this paper is to prove that intuition is correct, at least in the case of a classical group G. The case of alternating groups is a beautiful result due to Dixon [11], who also asked whether the corresponding result is true for all finite simple groups. However, whereas his theorem is proved using nineteenth-century results concerning permutation groups, the proof of our result unfortunately uses the classification of finite simple groups. We will prove the following: T H E O R E M . Let G O denote a finite simple classical group, and let Go ~< G-% 7. Let G denote a group satisfying Go ~< G ~< Aut(Go).

GENERATING

A FINITE CLASSICAL GROUP

73

L E M M A 2. I f n is restricted to be at least 21, then P(G) ~ 0 as IGI ~ oo. Proof. Inequality (,) still holds, and by [1] there are still classes CI-C9, but with further conditions on the various subspaces and subgroups involved. For example, subspaces are either totally singular or nonsingular, and many of the direct sum decompositions are orthogonal decompositions, all of which contribute additional cases. If L is in one of these new classes C t - C 8 then it is easy to check that IG:LI >i ½qt~6). In view of the description of the conjugacy classes in CI-C8 found in [I, §1] the following analogue of (**) continues to hold:

(***)

v iL[ = v, [L[ + ~- ]L[ P(6) ~< ,. IGi "~ IGI ~ IGi ~ qM(G°)/2n, where M(Go) is listed in Table I. By [21], it is still true that ILl ~< (q2)3n in the unitary case, but the inequality ILl ~< q3n is no longer true for the remaining classical groups G, because there is a further type of subgroup that must be considered:

E:Sk/2 let Q(H, k) denote the probability that k elements generate H; thus, Q(H, 2) = 1 - P(H) in our previous notation.

GENERATING

A FINITE CLASSICAL

79

GROUP

First we observe that, for our purposes, it suffices to consider the direct product G m of m copies of a group G: L E M M A 5. Let G1 . . . . . Gt be pairwise nonisomorphic simple groups, let ml . . . . . m t be positive integers, and let G = G•' x ... x GT 1. Then (a) A subset of G generates G if and only if its projection into G.~' generates G.~' for each i; and (b) Q(G, k) = 1-I~=, Q(G?', k). Proof Part consequence.

(a) is straightforward,

and

(b) is then

an immediate []

Consequently, we will consider a power G m of a simple group G. Let Dk(G) denote the set of ordered k-tuples generating G. P R O P O S I T I O N 6. Consider a k x m "matrix' (xi~) of km elements of a nonabelian simple group G, with 'rows' r i = ( x i j ) E G m and 'columns' cj = (xij) e G k. Then the r i generate G '~ if and only if the cj lie in different Aut(G)orbits of Ok(G) - where the action of Aut(G) is the diagonal one. (Each such orbit has length IAut(G)l.) Proof Write G ' ~ = G~ x - . - x G m with each G j ~ - G . Clearly H : = ( r 1. . . . . rk~ cannot be all of G m except, perhaps, if the projections of the r i into each factor Gj generate G j, that is, if all the cj lie in Dk(G ). Assume that this is the case. Since H projects onto Gin, H c~ Gm>,2 nonabelian finite simple classical (or alternating) groups. Then the probability that two elements of H ~ ~r~ generate H approaches 1 as the orders of all of the factors of H approach ~ . (In particular, for a fixed m all but finitely many groups in [~. are generated by two elements.) Proof. Lemma 6 allows us to reduce to the case H = G m. For each G we have [Aut(G)l < IGI log IGI. Thus, if G is large then so is ID.(G)l/IAut(G)l, and hence C(dk(G),m) is near 1 (as m is fixed). Since Q(G, 2) ~ 1, Proposition 9 implies the result. [] We give two examples of the uses of Proposition 9 which will be needed in Section 6. EXAMPLE 1. (i) For each prime p >>.5, PSL(2, p)P is generated by two elements. (ii) The probability that two elements generate PSL(2, p)P approaches e- l as the prime p ~ oo. (iii) The probability that three elements generates PSL(2, p)P approaches l as the prime p ~ oo. Proof The subgroup structure of PSL(2, p) has been known for over a century (cf. [10]). Using it, together with the elementary observation (.) in Section 2, it is straightforward to check that P(PSL(2,p))~~ p- 1 _ 2 p - 2 In particular, for each e > 0 we see that l - ( 1 +e)p-~
~ 2 and k, the probability that k elements generate ff l is O. (ii) More generally, the probability that k elements generate an open subgroup (i.e., a closed subgroup of finite index) of Fi is also O. Proof. Since b~2 is a homomorphic image of Fl it suffices to consider the case l = 2. (i) By Example 2 in Section 5, A~ ~/8is a quotient of F2 while the probability of k elements generating A~ !/8 approaches 0 as n ~ oo. Thus, (i) holds. (ii) If H is an open subgroup of Fz then H = F m with m = 1 + [ff~:H[(l- 1) [12, 15.27]. We are only interested in the case in which H is generated by k elements, so that 1 + Ifft:H[(l- 1)~< k. There are only finitely many open subgroups H 1 , . . . , H ~ whose index in Ft is at most (k - 1)/(l- 1) [12, 15.1]. Consequently, Pr(k elements generate a subgroup of finite index in/~,) ~< ~ Pr(k elements generate Hi)= ~ 0 = 0 i=1

i=1

since part (i) can be applied to each of the free profinite groups Hi.

[]

Note that the preceding proof did not use the classification of finite simple groups: the proof of (i) used only elementary properties of A,. The preceding result should be compared to the situation in the abelian case (see [12, 16.15] for the case 1 = 1): P R O P O S I T I O N 12. Let A t ~-~t be the free profinite abelian group on l generators. Then, for k >1 l, (i) Pr(k elements generate/it) =

0 Flk=k_l+ l ((0_ 1 > 0

/fk = l /fk > l

where ~ is the usual zeta function; and (ii) Pr(k elements generate a subgroup of finite index in 4,) = ifk>l.

84

WILLIAM

M. K A N T O R A N D A L E X A N D E R

LUBOTZKY

Proof. For each prime p let En be the set of k-tuples in /i~ whose projections to Z~ generate Z~, (or equivalently, whose projections to ~:~, generate F~). Then {Ep I P prime} consists of independent sets, and/~(Ep) is the probability that k vectors in H:~span that space. The latter probability is the same as the probability that I vectors in U:~are linearly independent (using the equality of row and column rank), and hence pk __ 1 pk _ p

~Ep) =

p~

p~

ptt _ p l - 1

p~

= (1 - p-kXl - p-{k-1})... (1 - p-{k-I+ l}). By independence, ~ Ep) = Hz/z(Ep), which = H~=k_t+ 1 ((i) -~ if k > l and = 0 if k = I. This proves (i). Moreover, this shows that, if/~p denotes the complement of Ep, then E #(/~p) converges if k > I and is ~ if k = I. By the Borel-Cantelli lemma 1-12, 16.7], it follows that Pr(x E/i~ is in Ep for almost all p) is 1 if k > I and is 0 if k = i. This completes the case k = ! since x can generate a subgroup of finite index in ,'it only if x is in Ep for almost all p. Now let k > I. We first claim that almost every k-tuple of elements of Z~, generates a subgroup of finite index. Indeed, for every subgroup H of index p~ in Ztn consider H k 0, P(An) < n- l +~ for all sufficiently large n.) (ii) The first part follows from an examination of the estimates occurring in the proof of the theorem (see the remarks at the end of Sections 2 and 3). For the second part proceed as in Section 5, Example 1. [] We will consider (unrestricted) products P = I-I6~. G of subsets ad of , ~ . Each such product P is a compact group, with Haar measure # satisfying #((So)a) = [Ic~oalSol/lGI whenever the subsets So ~-G satisfy So = G for almost all G ~ q/. Throughout the remainder of this section, when we consider the case q / = ~-a we will always assume: ($) For every finite simple group G of Lie type defined over ~:q, P( G) = O(l(G)a(log q)2q -tto~). By Proposition 13, ($) is already known to be true in the case of classical groups; and in fact it is also known for all the cases discussed in Section 4. We will consider the probability of generating a dense subgroup of P using two or three elements of G. P R O P O S I T I O N 14. Let P be the product of all the members of ~ c, or of ,~a assuming ($). Then (i) Pr(g, h ~ P generate a dense subgroup of P) = 0; and (ii) Pr(g, h, k ~ P generate a dense subgroup of P) ~ O, 1.

86

W I L L I A M M. K A N T O R AND A L E X A N D E R L U B O T Z K Y

P R O P O S I T I O N 15. Let P be the product of all the members of ~ c, or of ~a assumino ($), but excludin9 the alternating groups and those of the form PSL(2, p) with p prime. Then Pr(9, h • P 9enerate a dense subgroup of P) ~ O, 1. Proof of Propositions 14 and 15. Consider 9, h • P. It is easy to check that (g, h> is dense if and only if 9 and h project onto a pair of generators of each factor in the definition of P. Thus, if Te -~ P x P is the set of all such pairs (g,h), then Te = H a TG where T~ is defined similarly for G • ~ A. Then #(Te) = l-I~ (1 - P(G)), and so #(Te) = 0 if and only if E GP(G) diverges. Now Proposition 13 shows that, in Proposition 14(i), E~ P(G) has a subsum ~>E n - 1 and hence diverges; whereas in Proposition 15, E6P(G)converges (by Proposition 13, since we are avoiding part (i) there, while EG n3(log q)2q-,6~ converges provided that G is restricted so that q is not a prime when n = 2/(G) = 2). The proof of Proposition 14(ii) is similar: it is only necessary to check that Pr(g, h, k • PSL(2, q) do not generate PSL(2, q)) = O(q- 2) and Pr(g, h, k • A, do not generate An) = O(n-3/2). In fact, since Pr(g, h, k • G do not generate G)~< P(G) 2, Proposition 13(ii) implies the first of these inequalities; while the second follows either from Proposition 13(i), or from the less difficult result from [6-1 mentioned in the proof of Proposition 13(i) (namely, use e = ¼). []

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Authors' addresses:

William M. Kantor, Department of Mathematics, University of Oregon, Eugene, OR 97403, U.S.A. Alexander Lubotzky, Department of Mathematics, The Hebrew University, Jerusalem,

Israel. (Received, November 14, 1989)