PROCEEDINGSOF THE
AMERICANMATHEMATICALSOCIETY
Volume 117, Number 4, April 1993
CORRIGENDUM AND ADDENDUM TO
"CLASSIFICATION OF FINITE GROUPS
WITH ALL ELEMENTS OF PRIME ORDER"
CHENG KAI NAH, M. DEACONESCU,LANG MONG LUNG, AND SHI WUJIE
(Communicated by Warren J. Wong)
Abstract.
It is shown that the groups in question are either p-groups of exponent p or Frobenius groups of particular type, or they are isomorphic to the
simple group A5 ; the misprints and mistakes of a previous paper of the second
author are corrected.
1. Introduction
Let &> be the class of finite groups having all (nontrivial) elements of prime
order. The aim of this note is to correct some mistakes and misprints of [2]
and to give a more compact description of these groups, which offers a better
insight into their structure.
The statement of the Main Theorem of [2] (shortly: M.T.) is incorrect: Case
11(a) does not occur and Case Il(a') \G\ = p"q, 3 < p < q, a>3,
\F(G)\ =
\G'\ = pa , is missing. The omission stemmed from the incorrect application of
Lemma 2.8 of [2] to Case 11(b) of the proof of the M.T.
Some misprints affect the statements of 2.8 (put H < G instead of H ^ G),
of 2.10 (put K < G instead of K ^ G), and the 11th line from the bottom of
page 628 of [2] (put CG(F(G)) < G instead of CG(F(G)) = G). Finally, the
statements of 2.6 and 2.7 of [2] are misquoted (put Nq((x)) instead of Cq(x)) .
Although these two misquoted results affect the proof of Case III of M.T., the
outcome is correct: A$ is the unique nonsolvable <^-group.
At the time [2] was written, the second author was not aware of the large
number of results concerning the same topic and the more general one of determining those finite groups that have all elements of prime power order (see [1,
5, 7-9]). We would like thus to do here justice to some of these nice results by
using them in order to obtain alternate proofs of parts of M.T. We shall prove
the following
Theorem. Let G be a 3?-group. Then
(I) G is nilpotent if and only if G is a p-group of exponent p.
Received by the editors June 5, 1991 and, in revised form, July 19, 1991.
1991 Mathematics Subject Classification. Primary 20E34; Secondary 20D10.
Key words and phrases, p-groups, soluble groups.
©1993 American Mathematical Society
0002-9939/93 $1.00+ $.25 per page
1205
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1206
K. N. CHENG, M. DEACONESCU,M. L. LANG, AND SHI WUJIE
(II) G is solvable and nonnilpotent if and only if G is a Frobenius group
with kernel P e Sylp(G), with P a p-group of exponent p and complement
Q e Syl,(G), with \Q\ = q. Moreover, if \G\ = pnq then G has a chief
series G = Co > P = Cx > C2 > ■■■ > Ck > Ck+X= 1 such that for every
1 < i < k one has C,/C,+i < Z(P/Ci+x), Q acts irreducibly on Ci/Ci+X, and
|C,/C,+i | = pb, where b is the exponent of p (mod q).
(III) G is nonsolvable if and only if G = A5.
All groups under discussion are finite. The notation is standard and conforms
to that of [4].
2. Proof
of Theorem
Before starting the proof, we list here some key results that very much sim-
plify the argument.
(2.1) Let G be a group, and let P be a regular Sylow p-subgroup of G such
that P = Nq(P) . Then G has a normal p-complement.
Proof. This is an immediate consequence of [6, Satz 8.1 IV].
(2.2)
[3] Let G be a minimal nonnilpotent group. Then there exist P e Sylp(C7)
and Q e Sylq(G) (p ^ q) such that G = [P]Q. If P\ is the maximal normal
subgroup of G properly contained in P, then \P : Px\ = pb, where b is the
exponent of p (mod\Q\).
(2.3)
A5 is the unique simple nonabelian ^-group.
Proof. This follows for example by checking the list of simple groups that have
only elements of prime power order; this list appears in [9].
(2.4) [5] Let G be a finite nonsolvable group all of whose elements have prime
power order. Then G has a normal series G > N > P > 1, where (i) P is
the greatest solvable normal subgroup of G and is a p-group for some prime p ;
(ii) N/P is the unique minimal normal subgroup of G/P and is a simple group
of composite order, (iii) G has no solvable subgroup whose order is divisible by
three different primes.
We are now in a position to start our proof. The proof of (I) being trivial,
assume first that G is a solvable and nonnilpotent ^-group. Then by 2.11 of [2]
there exist P e Sylp(G), Q e Sylq(G) (p / q) such that G = PQ, Op(G)£ 1,
P is of exponent p , and \Q\ = q .
Observe first that one must have P<G. Indeed, if P fi G, as P is a maximal
subgroup of G and we clearly have exp(P) = p, we infer from (2.1) that G
has Q as a normal p-complement. But this together with Op(G) # 1 implies
that there exists in G an element of order pq, contradicting the hypothesis.
Therefore P < G and, since Q acts fixed-point-freely on P, it follows that
G —[P]Q is a Frobenius group with kernel P and complement Q.
Conversely, let G = [P]Q be a Frobenius group, where P e Sylp(G) is of
exponent p and Q e Sylq(G) is of order q. Then clearly Q - NG(Q), so
the number of elements of G that lie in a Sylow subgroup of G is equal to
1^1+ (IGI~ 1)I-P|= I^IIGI= \G\■This shows that all nontrivial elements of G
have prime order.
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FINITE GROUPS WITH ALL ELEMENTSOF PRIME ORDER
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We now deal with the normal structure of G. Suppose that |G| = p"q and
let G = C0 > P = Cx > C2 > ■ > Ck > Ck+X= 1 be a chief series of G.
Set Gx :- CkQ and suppose that \GX\= pbq . We claim that Gx is a minimal
nonnilpotent group.
Suppose by contradiction that there exists a proper nonnilpotent subgroup
G2 of Gx . Then \G2\ must be of the form pcq with 1 < c < b, and we
can suppose that Q < G2. Observe that Z(P) <G because P <G. Now the
minimality of Ck together with Ck n Z(P) ^ 1 forces Ck < Z(P).
Consider the subgroup G2 n Ck . As both P and Q normalize G2 n Ck ,
it follows that G2n Ck < G = PQ. But clearly G2 n Ck < Ck, against the
minimality of Ck . The claim is proved.
Applying now (2.2) to the group Gx , one obtains that \Ck\ = pb, where b
is the exponent of p (mod q). The proof of the remaining assertion is now a
matter of easy induction; the proof of Case II of the theorem is complete.
Suppose now G is a nonsolvable ^-group.
We shall prove that G = A5
following the ideas of [1] and [2]. By (2.3) one can suppose that G is not
simple. Let 1 < P < N < G be the normal series of G given by (2.4). If
P = 1 then N is simple and N = A$ by (2.3). Moreover, since N is the
unique minimal normal subgroup of G, it follows that CG(N) = 1 . This
implies that G is isomorphic to a subgroup of Aat(N) = A\it(A$) that contains
an isomorphic copy of N = A5. This forces N = G = A$ and contradicts the
nonsimplicity of G.
Therefore one can assume that P / 1 . But this case is ruled out by the same
argument as that of the proof of Case IV of M.T. The proof is complete.
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(M. Deaconescu) Department
mania
of Mathematics, University of Timisoara, Timisoara, Ro-
(K. N. Cheng and M. L. Lang) Department
Singapore, Singapore 0511
(S. Wujie) Department of Mathematics,
qing 630715, People's Republic of China
of Mathematics,
Southwest
National
China Normal
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University
University,
of
Chong-