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A note on involution centralizers in black box groups

  • Veronica Kelsey EMAIL logo and Peter Rowley
Published/Copyright: November 20, 2019

Abstract

Here we note a minor variation on the method in [J. N. Bray, An improved method for generating the centralizer of an involution, Arch. Math. (Basel) 74 2000, 4, 241–245] which enables calculations of CH(t) for H a subgroup of a black box group G and t an involution of G. We then give some applications of this, the first to investigate the action of the Lyons sporadic group via conjugation on its involutions, a number of calculations in E6(2) and E8(2), then finally to carry out some calculations in GL670(2).

1 Introduction

In [7], Bray revealed a method for calculating centralizers of involutions in black box groups with an order oracle. This method extended one introduced earlier by R. Parker (see [20]). In recent times, the Bray method has had many ramifications in computational group theory (for a fraction of these, consult [8, 2, 6, 11, 12, 13, 14, 19, 26, 25, 20]). The purpose of this short note is to observe a further twist to this story. Although this perturbation to [7] is very minor, it results in some substantial gains. In Section 2, we make use of it to determine the suborbits of the Lyons sporadic simple group in its action by conjugation on its involutions. A further demonstration of the efficacy of this variation, in Section 2.4, are some calculations in GL670(2). Also in Section 2, we include statistics of some sample calculations in E6(2) and E8(2).

Suppose G is a black box group with an order oracle. Assume t is an involution of G. In [7], the elements 𝒦(t,g) of G are key. For gG and letting n be the order of [t,g], we define

𝒦(t,g)={[t,g]mifn=2m,g[t,g]mifn=2m+1.

These elements 𝒦(t,g) (and also 𝒦(t,g-1) if n is even) supply elements in CG(t). Those 𝒦(t,g) obtained when n is odd have the property observed by R. Parker that they are uniformly distributed throughout CG(t).

For HG, set 𝒪H={hhHand[t,h]is of odd order}.

Lemma 1.1.

Suppose t is an involution in G, HG, and let cCH(t). Then |{hOHK(t,h)=c}| is independent of c.

Proof.

Since cCH(t), [t,h]=[t,ch] for all hH, so we only need prove the lemma for each right coset CH(t)h (hH) for which [t,h] has odd order. For ehCH(t)h, we have

𝒦(t,eh)=eh[t,eh]m=eh[t,h]m=e𝒦(t,h),

where [t,h] has order 2m+1. Hence each such coset contributes 1 to

|{h𝒪H𝒦(t,h)=c}|,

so giving the lemma. ∎

Theorem 3.1 of [7] is the case H=G, and its proof is virtually identical to that for Lemma 1.1. The point is that t does not need to be in H. So to compute CH(t), we may proceed as follows.

  1. Initialize S:={}.

  2. Choose a random element hH.

  3. Compute 𝒦(t,h).

  4. Check whether 𝒦(t,h)H; if yes, then add 𝒦(t,h) to S.

  5. Go to step (2).

Then S will be a subgroup of CH(t). All the analysis and caveats discussed in [7] will apply here. Lemma 1.1 shows that the set of elements passing test (4) will be uniformly distributed in CH(t). Also, the membership problem raises its head in step (4), and the exact nature of H may help in resolving this. For example, we may have H=CG(X) (XG) in which case step (4) can be settled by checking whether 𝒦(t,h) commutes with the set X. The other conundrum facing us is what is the probability that the elements 𝒦(t,h) are actually in H. Without further specific information about H, there does not seem much that can be said in general. Indeed, we could have that CH(t)=1. In this connection, in Section 2, we gather statistics of some actual calculations.

Suppose H=CG(s), where s is an involution of G (in fact, the situation that sparked this note); experimentally, the following works well. In place of (2)–(5), do the following.

  1. Compute 𝒦(s,g), where g is a random element of G (so applying the Bray method for CG(s)).

  2. Compute 𝒦(t,𝒦(s,g)).

  3. Check whether s and 𝒦(t,𝒦(s,g)) commute; if yes, then add 𝒦(t,𝒦(s,g)) to S.

  4. Go to step (2).

Observe that Lemma 1.1 applied twice shows that these will be uniformly distributed in CG(t)CG(s). Of course, we could determine generating sets for CG(t) and CG(s) using the Bray method, and then we could attempt to compute CG(t)CG(s). In computationally hard groups, the latter step may prove impossible. To illustrate this with an example, take G=E6(2), as given in the electronic Atlas [24] in its 27-dimension GF(2) representation. There G=a,b, where a has order 2, b has order 3 with ab of order 62. Taking t=a and s=(ab)31, using the method discussed here (that is, steps (1), (2), (3), (4), (5)) with 10,000 random elements gG, we get S=CG(s)CG(t) with |S|=21237. This is done in the blink of an eye, whereas first calculating CG(s) and CG(t) (which is quick) and then CG(s)CG(t) takes forever (1552 seconds on a 16×1248 MHz machine running Magma [9] version 222-10). This disparity will be even greater for larger groups.

Finally, we observe that the above process for the centralizer of two involutions may be iterated so as to find CG(H), where H is generated by involutions – see Section 2.4 for an example of this.

2 Some applications

2.1 Involutions in the Lyons group

The sporadic simple group Ly, first pin-pointed by Lyons in [15], has order 283756711313767. Its smallest non-trivial permutation degree is 8,835,156 – indeed, this representation was built in the 1970s by Sims [22] in the course of showing the existence of Ly. Despite considerable advances in computer algebra, such a permutation representation is not viable for probing the structure of Ly. More tractable is the 111-dimensional representation over GF(5), first displayed in [17], which we employ here. From [24], we obtain this representation where Ly is generated by the standard generators a,b with a2A, b5A and ab14A (using Atlas [10] names for the conjugacy classes of Ly).

For many of the sporadic simple groups G, the action (by conjugation) of G on a G-conjugacy class of involutions has been analyzed. See [4, 5, 21, 23], where the suborbits are determined. This type of information is valuable when studying such topics as commuting involution graphs [3], C-strings and polytopes [16].

Here we investigate the action of GLy upon X=2A. Before stating our main conclusion, we introduce two pieces of notation. Let tX. For C a G-conjugacy class, XC={xXtxC}, and for yG,

CG*(y)={gGyg=yoryg=y-1}.

Observe that XC will either be empty or the union of CG(t)-orbits (under conjugation) and, of course, X=˙XC, as C runs through all the G-conjugacy classes. As is well known, |XC| equals a particular structure constant which may be calculated using the complex irreducible character table of G. Also, we have [CG*(y):CG(y)]=1 or 2.

The following easy lemma will be used in the proof of Theorem 2.2.

Lemma 2.1.

For t,xX, put z=tx. Then

CCG(t)(x)=CG(t)CG(x)=CCG(z)(t)=CCG(z)(x).

So, for z=txC, where xX if we know the structure of CG(z), we may be able to calculate CG(t)CG(x). In particular, note that [CG*(z):CG(z)]=2 and, as G=Ly has just one conjugacy class of involutions, CCG(z)(y) as y ranges through all involutions in CG*(z)\CG(z) gives all the possible structures for

CG(t)CG(y),yXC,

that must occur (but not the possible multiplicities).

As we shall see, a number of the XC may be easily broken into CG(t)-orbits. However, for three G-conjugacy classes, it is not so straightforward, and we make use of the routine (1), (2), (3), (4), (5) given in Section 1.

Theorem 2.2.

Suppose that G=Ly, X=2A and tX. Then the permutation rank of G on X is 78 and the suborbit (that is, CG(t)-orbit) sizes are as follows.

C|XC|CG(t)-orbit sizes
1A11
2A34,65034,650
3A5,0405,040
3B677,600554,400 + 123,200
4A1,663,200831,600 + 831,600
5A199,584299,584
5B5,322,2401,330,560 + 3,991,680
6A831,600831,600
6B6,098,4001,108,800 + 4,989,600
6C13,305,6003,326,400 + 3,326,400 + 3,326,400 + 3,326,400
7A19,958,40019,958,400
8A979,2004,989,600 + 4,989,600
8B19,958,4009,979,200 + 9,979,200
9A26,611,20019,958,400 + 6,652,800
10A4,989,6004,989,600
10B39,916,80019,958,400 + 19,958,400
12A14,968,8004,989,600 + 9,979,200
12B26,611,2006,652,800 + 19,958,400
14A19,958,40019,958,400
15A7,983,3601,995,840 + 1,995,840 + 1,995,840 + 1,995,840
15B39,916,80019,958,400 + 19,958,400
15C39,916,80039,916,800
18A39,916,80019,958,400 + 19,958,400
20A39,916,80019,958,400 + 19,958,400
21A39,916,80019,958,400 + 19,958,400
21B39,916,80019,958,400 + 19,958,400
24A39,916,80019,958,400 + 19,958,400
24B39,916,80019,958,400 + 19,958,400
24C39,916,80019,958,400 + 19,958,400
25A39,916,80039,916,800
28A39,916,80019,958,400 + 19,958,400
30A39,916,80019,958,400 + 19,958,400
30B39,916,80019,958,400 + 19,958,400
31A39,916,80039,916,800
31B239,916,80039,916,800
31C439,916,80039,916,800
31D839,916,80039,916,800
31E1639,916,80039,916,800
37A39,916,80039,916,800
37B39,916,80039,916,800
40A39,916,80019,958,400 + 19,958,400
40B39,916,80019,958,400 + 19,958,400
42A39,916,80019,958,400 + 19,958,400
42B39,916,80019,958,400 + 19,958,400
67A39,916,80039,916,800
67B239,916,80039,916,800
67C439,916,80039,916,800

Proof.

Let xX, and set z=txC, C a G-conjugacy class. We recall that CG(t)2.Alt(11), so |CG(t)|=39,916,800=283452711.

Suppose that CG(z)=z. Then if z has odd order, CG(t)CG(y)=1 for all yXC, and if z has even order, |CG(t)CG(y)|=2 for all yXC. This deals with XC for

C{18A,24A,24B,24C,25C,28A,31A,31B2,31C4,31D8,31E16,37A,37B,40A,40B,42A,42B,67A,67B2,67C4}.

Also if z has odd order and [CG(z):z]=2, then |CG(t)CG(y)|=2 for all yXC. This observation yields that X21A and X21B both break into two CG(t)-orbits.

If z has even order, then, as G has only one involution conjugacy class, we may regard CG(t)CG(x) as a subgroup of 2.Alt(11). So we may calculate in 2.Alt(11), using the permutation representation of degree 5,040 in [24]. This approach allows us to break XC into CG(t)-orbits for

C{2A,4A,6A,6B,6C,8A,8B,10A,10B,12A,12B,14A,20A}.

Note that, for z7A, CG*(z)=CG*(w), where w2=z and w14A, so X7A is a CG(t)-orbit just as X14A is.

At this point, it remains to examine XC for

C{3A,3B,5A,5B,9A,15A,15B,15C}.

For zC=3A, CG*(z)3.McL:2 and CCG(z)(t)M11 (see [10]), and therefore X3A is a CG(t)-orbit. Suppose z has order 15. For z15B, we have CG*(z)=CG(w) for w30B and w2=z, so X15B breaks into two CG(t)-orbits just as X30B does. If z15A15C, then z53A, and we may calculate in McL:2, using the permutation representation of McL:2 of degree 275. Suppose z9A. Then [CG(z):z]=6. Since G has no elements of order 18, CG(z)/z cannot be cyclic, and so CG(z)/zSym(3). If y is an involution inverting z, then y must induce inner automorphisms on CG(z)/z. Since CCG(z)(y)z=1, we conclude that CCG(z)(y)=2 or 6; whence it follows that X9A had two CG(t)-orbits of sizes 19,958,400 and 6,652,800.

We are now left with the cases C{3B,5A,5B}. The structure of CG*(z) in these cases is more intricate, so we use the procedure discussed in Section 1.

(2.2.1) C=3B. For different choices of x1,x2X3B, we obtain

LiCG(t)CG(xi),

where |L1|=2234, |L2|=2332. In fact, Li=CG(t)CG(xi) for i=1,2, and we explain how this was established for i=1, i=2 being similar. Using the LMGRadicalQuotient command in Magma on CG(t)2.Alt(11), we obtain a map from CG(t) to CG(t)/tAlt(11), with the latter given as a permutation group of degree 11. Let L1¯ be the projection of L1 into Alt(11) (and note L1t=1). Because |L1¯|=3422, the only possible maximal subgroups of Alt(11) containing L1¯ are isomorphic to either Sym(9) or Alt(10). Within Alt(10), the only proper overgroups might be Alt(9), and then within Alt(9) only 33:Sym(4). Within Sym(9), the only proper subgroup that could contain L1¯ has shape 33:(Sym(4)×2). So if CG(t)CG(x1)>L1, we must have 23 dividing |CG(t)CG(x1)|. Selecting a Sylow 2-subgroup of L1¯, S1¯, we calculate NCG(t)/t(S1¯) and then take its full inverse image in CG(t), say K. Since we have |K|=263, we may quickly enumerate CK(x1) finding that |CK(x1)|=22. So 23 does not divide |CG(t)CG(x1)|, and we conclude that

CG(t)CG(x1)=L1.

This gives a 2652711=123,200 size CG(t)-orbit of X3B.

(2.2.2) C=5A. Selecting a random xX5A, we use the procedure in Section 1 and obtain LCG(t)CG(x) with L52:2. When taking the image of L (as for X3B), say L¯, in CG(t)/tAlt(11), we find

L¯K¯(Alt(5)×Alt(5)).4.

Taking the full inverse image of K¯, say K, in CG(t) (which has order 283252), we calculate that |KCG(x)|=2352. Projecting KCG(x) back into Alt(11), we see that KCG(x)=CG(t)CG(x). Consequently, X5A consists of a single CG(t)-orbit.

(2.2.3) C=5B. Again selecting a random xX5B and employing the procedure in Section 1, we locate x1,x2X5B with

L1CG(t)CG(x1),L2CG(t)CG(x2),

L1Dih(10) and L2Dih(30). Let L1¯ be the image of L1 in CG(t)/t, and let N2¯ and N5¯ be the normalizers in Alt(11) of respectively a Sylow 2- and 5-subgroup of L1¯. If N2 and N5 are respective full inverse images of N2¯ and N5¯, we have |N2|=21,152 and |N5|=2100; whence we quickly check that

|N2CG(x)|=2and|N5CG(x)|=10.

Hence we conclude that CG(t)CG(x1)=L1Dih(10). Very similar arguments show that CG(t)CG(x2)=L2Dih(30), and therefore X5B breaks into two CG(t)-orbits of sizes 27345711 and 27335711.

Since we have now dealt with all the non-empty subsets XC, the proof of Theorem 2.2 is complete. ∎

2.2 Calculations in E6(2)

We now look further at the calculation in E6(2), mentioned in Section 1. Repeating the calculation for the given t and s 100 times, in each case using 10,000 random elements, gave the following five outcomes with the following frequencies.

|S| obtained2123712232112103210
Number of times6920245

On the other hand, using 12,000 random elements from E6(2) in each of 100 experiments always gave |S|=21237.

2.3 Calculations in E8(2)

We again go to [24] for our copy of G=E8(2), using the 2-generator version there. So G=a,b with a and b being 248×248 matrices over GF(2). Let V denote the 248-dimensional GF(2)G-module. The order of a is 30 and the order of b is 10. We shall employ the naming system in [1] (which follows Atlas conventions) for the conjugacy classes of G. So a152D and b52B (because dimCV(a15)=128 and dimCV(b5)=156). In each of the following cases, we did 10 experiments each using 40,000 random elements from E8(2). (i) t=a15 and s=tg, where g=b3ab-1aba11. Then ts has order 17 with dimCV(ts)=32, so, using [1], ts17AB with CG(ts)17×Ω8-(2).

|S|85774673910
|S|285263293293263283293243283293

(ii) t=b5 and s=tg, where g=ab3a7b4a19. This time ts has order 8.

|S|78787968925375637491
|S|218218221223230222223221219226

In case we are getting too rosy a picture, we have the following.

  1. t=b5 and s=tg, g=ab3a7. Here ts has order 5, dimCV(ts)=68. So ts5A with CG(ts)5×Ω12-(2). Each of the 10 experiments yielded no elements of S at all. Moreover, note that, by Lemma 2.1, CG(t)CG(s) must be non-trivial.

2.4 Centralizers in a 670×670 matrix group

Using the GF(2) representation of O10+(2) of degree 670 given in [24], we may regard G=O10+(2) as a subgroup of GL670(2). In the notation of [24], G=a,b, where a has order 2 and b has order 20. Set H=a,ab,a(bab3). We have that HL3(2), and we seek to calculate CG(H). First employing the procedure in Section 1, we calculate CG(a,ab) (which has order 273357). Then using random elements selected from CG(a,ab), we repeat the procedure to calculate elements in CCG(a,ab)(a(bab3)) (and note that a(bab3)CG(a,ab)), thus delivering a subgroup of CG(H) of order 2232. There is of course the perennial problem of knowing whether we have obtained all of CG(H) – though there are instances where just having some elements of CG(H) may be helpful. As it happens, |CG(H)|=2232 (which may be checked by working in the 496 degree permutation representation for O10+(2) available in [24]). Finally, we remark that the standard Magma functions could not calculate CG(H).


Communicated by Michael Giudici


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Received: 2019-03-22
Published Online: 2019-11-20
Published in Print: 2020-03-01

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