THOMSON PROBLEM AND TOMPSON CONJECTURE
*Wujie Shi, **Шлепкин А.К.
*Suzhou University, China
**Красноярский государственный аграрный университет, Красноярск,
Россия
Mathematics is kept alive, much more than by the activities of either
class, by the appearance of a succession of unsolved problems, both from
within mathematics itself and from the increasing number of disciplines
where it is applied.
from “Unsolved Problems in Number Theory”
Also, We have “Unsolved Problems in Group Theory”, Edited by Prof.
V.D. Mazurov and other professor.
Derivation and integration
When I was university student, 1960, I posed the following problem to my
teacher, the relation between the derivations of function and its inverse function is
simple as below:
y
'
x
1
 '
xy
Find problems through the comparison
What about the relation between the integrations of function and its inverse
function? My teacher told me: “I do not know, but you can try to find the relation”,
“I do not find the relation in some textbooks”. After 2 or 3 months, I found the
relation form definite integration first as follow:
Find problems through the comparison
From the link between definite integration and indefinite integration, we get
Let  f(x) dx = g(x) + C1,
Then  f -1(y) dy = y f -1(y)  g[f -1(y)] + C2 .
This is my first paper finished in 1961 but not published.
Find problems through the comparison
When I was postgraduate student, 1979, Prof. Zhongmu Chen told us, the
order |G| of finite group G is a very important quantity. For example,
Sylow theorem, odd order theorem, …
Except |G|, which quantity is important? Compare with |G|, we find the set
e(G) of all orders of elements in G is another very important quantitative set.
In 1987 I reported my problem to Prof. J.G. Thompson:
All finite simple groups S can be characterized uniformly using “two
orders” of S.
In our communication, Thompson posed his problem and conjecture about
the judgment of solvability of finite groups and the quantitative characterization of
all finite simple groups. In this talk we survey these problem and conjecture
although they do not solve all up to now.
§ 1. What is “two orders of G”
The order |G| of a group G is an integral property of groups. For the special
number |G| we can determine the group G, for example, |G| = p, p is a prime. That
is, there is only one group of order p which is an abelian simple group.
Now we associate with e(G) = (G), the set of all orders of elements in G
(the spectrum of G), and give the pure quantitative characterization for all finite
simple groups G (using only |G| and e(G)).
We have proved the following theorem:
Theorem 1.1 Let G be a group and S one of the following simple groups:
(1) A cyclic simple group Zp;
(2) An alternating group An, n  5;
(3) A sporadic simple group;
(4) A Lie type group except Bn, Cn, Dn (n even and q≠2); or
(5) A simple group with order <108.
Then G  S if and only if (a) e(G)=e(S), and (b) |G|=|S|.
Since we had not found any counterexample, we posed the following
conjecture (put forward in 1987).
Conjecture 1.1 Let G be a group and S a finite simple group. Then G  S
 (a) e(G) = e(S), and (b) |G| = |S|.
From the above view (two orders) we may say the simple groups are
simple.
That is, all finite simple groups may be determined by “two orders”.
However, it does not true for some very small groups. Examples:
1.
D8 (dihedral group of order 8) and Q8 (quaternion group) |D8| = |Q8| = 8 and
e(D8) = e(Q8) = {1, 2, 4},
but
D8 ≇ Q8 , they are different groups.
2.
G1 = Z3  Z3  Z3 and G2 = (Z3  Z3) ⋊ Z3 (semidirect product)
| G1| = | G2| and e(G1) = e(G2) = {1, 3},
but G1 and G2| are not isomorphism.
From 1987 to 1991, after I wrote some letters to Professor John G.
Thompson and reported the above conjecture (Conjecture 1.1). Thompson
pointed that, “Good luck with your conjecture about simple groups. I hope
you continue to work on it”, ...,
“This would certainly be a nice theorem.”…,
in his reply letters. Also, he posed the following one problem and one
conjecture.
§ 2. On Thompson’s problem
A problem related to Conjecture 1.1 is the following open problem put
forward by J.G. Thompson, which is appeared in the communication letter in 1987:
For each finite group G and each integer d  1, let
G(d) = xG | xd = 1.
Definition 2.1. G1 and G2 are of the same order type if and only if
|G1(d)|=|G2(d)|, d=1, 2, ..…. .
Problem 2.1.(J.G. Thompson) Suppose G1 and G2 are groups of the same
order type. Suppose also that G1 is solvable. Is it true that G2 is also necessarily
solvable?
In Thompson’s Private letter he pointed out that
“I have talked with several mathematicians concerning groups of the same
order type. The problem arose initially in the study of algebraic number fields,
and is of consider-able interest”. Also we know it stems from the need of
research in differential geometry.
Given a finite group G, how can judge the solvability of G? The famous Odd
Order Theorem proved that all finite groups of odd order are solvable.
For the group of even order, how judge them?
If the answer of the above Thompson’s problem is positive, then we may
judge the solvability for all finite groups G using the order equation of G.
In fact, the groups of the same order type are such groups whose “order
equations” are the same. That is, according to the element orders as the equivalent
relation, the group G is partitioned into the same order classes. The “order
equation” is
|G| = n1 + n2 + … + ns,
where ni ≥0 denotes the number of elements of order i in G. Obviously the same
order class is an union of several conjugacy classes and the partition of “order
equation” is rougher than the class equation.
Some authors call the groups of the same order type conformal groups.
That is, e(G1) = e(G2), and for all ne(G1) , the number of elements of order n in
G1 and G2 are equal. The conformality of groups is also called the Grassmann
equivalence. Considering the counter- example, the conformal groups which are
not isomorphic are, in particular, they can not been characterized only using by the
set e (G). So it is necessary to study such groups which are characterizable groups
(using by the set e (G)).
Thompson has given a nonsolvable example of conformal groups which
are not isomorphic:
G1 = 24 : A7 and G2 = L3(4) : 22,
both are maximal subgroups of M23.
Question 2.1. For which natural number n are any two same order type
groups of order n isomorphic?
If the conjugacy classes equation of G as follows
|G| = c1 + c2 + … + ct.
Thus we have t ≥ s. Then there exists a non-negative integer k such that k
= t – s, and we call this group co(k)-group. Clearly, finite co(0)-groups is the
groups posed by Syskin. That means the elements of same order are conjugate.
W. Feit and G.M. Seitz, J. P. Zhang had proved well-known Syskin problem
independently. The result of Syskin problem is that: If any elements of finite
group with same order are conjugate, then G  Si , i = 1, 2 or 3.
We have proved that
Theorem 2.1. Finite co(1)-groups are one of the following groups: A5,
L2(7), S5 or the solvable groups, S4, A4, D10, Z3, Z4,
Hol(Z5) = ‹a, b | a5 = b4 = 1, b -1ab = a2›, or dicyclic group
‹a, b | a3 = b4 = 1, b-1 ab = a2›.
Also we have classified the finite solvable co(2)-groups.
Theorem 2.2. If G is finite solvable co(2)-group, then G is isomorphic to
one of the following groups: Z2Z2, Z6, Z7:Z3, Q8, D8, D14, D18, Hol(Z7), Z15:Z4,
S3Z2, Q8:Z3, (Z3Z3):Z4, (Z3Z3):Q8.
If G1 and G2 are finite groups of the same order type, then e(G1) = e(G2)
and |G1| = |G2|.
Thus if Conjecture 1.1. is correct, then we have the following corollary:
Let G1 and G2 be finite groups of the same order type. If G1 is solvable, then
G2 is not simple.
If G1 and G2 are of the same order type, assume that G1 is nilpotent, we have
G2 is also nilpotent . It is obvious.
M. C. Xu proved that if G1 possess Sylow tower (or G1 is supersolvable),
then G2 also possess Sylow tower.
Let G be a finite group. The prime graph (G) of a group G is the graph
whose vertex set is the set (G) and two vertices p, r  (G) are adjacent if and
only if G contains an element of order pr. Lastly, Mr. Rulin Shen and I proved:
Let G1 be a solvable group whose GK(G1) is not connection. If G1 and G2
are the groups of same order type, then G2 is also solvable.
That is, we may judge the solvability for the groups of even order using the
order equation if the prime graph is not connection.
Whether exists or not a counterexample for Thompson’s problem?
We relax the condition of “same order type” to e(G1) = e(G2) and | G1| = |
G2|. Up to now we do not find such groups G1 and G2 satisfying the conditions
e(G1) = e(G2) and | G1| = | G2|, such that G1 is solvable and G2 is not.
Furthermore, we consider such finite groups G1 and G2 satisfying e(G1) =
e(G2) only, then there exist an example such that e(G1) = e(G2) = {1, 2, 3, 4, 5,
6}, and G1 = [P5]Q, where Q  SL(2,3) and P5 = Z 52t , G2= S5. G1 is solvable and G2
is non-solvable, but | G1|  | G2|.
§ 3. On Thompson’s conjecture
Another Thompson’s conjecture, which is also aim at characterizing all
finite simple groups by a quantity set, is posed in 1988, which is appeared in
another communication letter:
If G is a finite group, set N(G) = n Z+ | G has a conjugacy class C with |C|
= n
Conjecture 3.1(J.G. Thompson) If G and M are finite groups and N(G) =
N(M), and if in addition, M is a non-Abelian simple group while the center of G is
1, then G and M are isomorphic.
On Conjecture 3.1, G.Y. Chen has dealt with the situation that M is a finite
simple group having a non-connected prime graph in his Ph.D. thesis, and
published the case of t(M)  3, where t(M) is the number of connected components
of prime graph of M. Also Chen has set up two important lemmas.
Lemma 3.1. Let G be a finite group with Z(G) = 1 and M a non-connected
prime graph non-Abelian simple group such that N(G)=N(M). Then |G|=|M|.
Lemma 3.2. Let G and M be finite groups having same orders. If
N(G)=N(M), then the number of components of prime graph of G is equal to that of
M. And the prime graph components of G are equal to those of M, where the prime
graph components are viewed as the sets of vertices.
What is the components of prime graph of G?
Denote by t(G) the number of connected components of (G), by i = i(G)
for i = 1,2,…, t(G) the connected components of (G). If |G| is even, then suppose
2 is a vertex of 1. The following lemma is published in J.S. Williams’ paper.
Lemma 3.3. If G is a finite group whose prime graph has more than one
component, then G has one of the following structures: (a) Frobenius or 2-
Frobenius; (b) simple; (c) an extension of a 1-group by a simple group; (d)
simple by 1; or (e) 1 by simple by 1.
This lemma plays an important role in our proof.
By using above lemmas Chen has proved that Thompson’s conjecture holds
for all non-Abelian simple groups with non-connected prime graphs. Hence
Thompson's conjecture holds for all non-Abelian simple groups except An(q),
Bn(q), Cn(q), Dn(q), 2An(q), 2Dn(q), E7(q) and the alternating groups An, and the
whole result is written in his Ph. D. thesis. Some of them published in several
papers.
Definition 3.1. Let G be a finite group and 1, 2, … , t the connected
components of prime graph (G). Then |G| can expressed as a product of m1,
m2,…, mt, where mi are positive integers with (mi) = i. These mi are called the
order components of G.
Theorem 3.1. Let G be a finite group and M one of the following simple
groups: A sporadic simple group, G2(q), q  0(mod 3), E8(q), a Suzuki-Ree groups,
L2(q), L3(q), U3(q), q > 5, F4(q), q even, Ap, where p and p  2 are primes.
Then G  M if and only if G and M have the same order components.
The conditions “e(G)=e(M), and |G|=|M| ”  “G and M have the same
order components ”.
Also the conditions “|G|=|M| and N(G) = N(M) ”  “ G and M have the
same order components ” .
So the conditions “Z(G) = 1 and M is a non-connected prime graph group
such that N(G) = N(M) ”  “G and M have the same order components ”.
Question 3.1. Let G be a finite group and M a non-Abelian simple group.
Suppose G and M have the same order components, for which simple group M, we
have G  M ?
For many simple groups M which order components number is greater than
1, the answer is affirmative. However, this answer is negative for the case of the
order components number being 1. In particle, Bn(q), Cn(q), q odd, have the same
order components, but they are not isomorphic.
Question 3.2. For Thompson’s conjecture (Conjecture 3.1), how deal with
the case of connected prime graph, for example, the case of M = A10 (the smallest
simple group of connected prime graph)?
References
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Wujie Shi, A new characterization of the sporadic simple groups, Group
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Wujie Shi, A new characterization of some simple groups of Lie type,
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