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An Elementary Proof That Finite Groups Lack Unique Product Structures
JOURNAL OF ALGEBRA ARTICLE NO.
180, 206]207 Ž1996.
0061
An Elementary Proof That Finite Groups Lack Unique Product Structures Matthew Cushman* Department of Mathematics, Carnegie Mellon Uni¨ ersity, 5000 Forbes A¨ enue, Pittsburgh, Pennsyl¨ ania 15213 Communicated by Walter Feit Received March 6, 1995
A group G is said to have a unique m-element product structure if there is a subset S of G such that the product map f : S m ª G is a bijection. D. Dimovski Ž1992, J. Algebra 146, 205]209. proved using character theory that no nontrivial finite group has a unique m-element product structure for m G 2. We provide an elementary proof of this fact. Q 1996 Academic Press, Inc.
DEFINITION. Let m be a positive integer. We say that a group G has a unique m-element product structure if there is a subset S of G such that the product map f : S m ª G defined by f Ž a1 , a2 , . . . , a m . s a1 a2 ??? a m is a bijection. THEOREM. If G is a nontri¨ ial finite group and m G 2, then G does not ha¨ e a unique m-element product structure. Proof. Suppose we have S ; G for which the above f is a bijection. Clearly < G < s < S < m . Let p ) m be a prime and let X s Ž a1 , . . . , a p . g S p < a1 a2 ??? a p s e 4 . For each Ž a1 , . . . , a pym . g S py m there is a unique Ž a pymq1 , . . . , a p . g S m such that Ž a1 , . . . , a p . g X. Therefore, < X < s < S < py m . Notice that if Ž a1 , a2 , . . . , a p . g X, then Ž a p , a1 , . . . , a py1 . g X. Thus ² g :, the cyclic group of order p, acts on X by g Ž a1 , . . . , a p . s Ž a p , a1 , . . . , a py1 .. The size of each orbit divides p, and thus each orbit either is a singleton or has cardinality p. We now consider two cases: Case 1. There is an orbit with one element, say Ž a1 , . . . , a p .. Clearly, all of the a i are equal, so a1p s e. Since e f S, a1 / e. Thus, a1 is an element of order p, so p < < G <. Case 2. Every orbit has p elements. Then p < < S < py m , so p < < S <. Consequently, p < < G <. 206 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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In either case, p < < G <. Since this holds for infinitely many p, we have a contradiction.
ACKNOWLEDGMENTS The author acknowledges useful conversations with Gary Sherman of the Rose]Hulman Institute of Technology and Phil Bradley of Rice University.
REFERENCE 1. D. Dimovski, Groups with unique product structures, J. Algebra 146 Ž1992., 205]209.
180, 206]207 Ž1996.
0061
An Elementary Proof That Finite Groups Lack Unique Product Structures Matthew Cushman* Department of Mathematics, Carnegie Mellon Uni¨ ersity, 5000 Forbes A¨ enue, Pittsburgh, Pennsyl¨ ania 15213 Communicated by Walter Feit Received March 6, 1995
A group G is said to have a unique m-element product structure if there is a subset S of G such that the product map f : S m ª G is a bijection. D. Dimovski Ž1992, J. Algebra 146, 205]209. proved using character theory that no nontrivial finite group has a unique m-element product structure for m G 2. We provide an elementary proof of this fact. Q 1996 Academic Press, Inc.
DEFINITION. Let m be a positive integer. We say that a group G has a unique m-element product structure if there is a subset S of G such that the product map f : S m ª G defined by f Ž a1 , a2 , . . . , a m . s a1 a2 ??? a m is a bijection. THEOREM. If G is a nontri¨ ial finite group and m G 2, then G does not ha¨ e a unique m-element product structure. Proof. Suppose we have S ; G for which the above f is a bijection. Clearly < G < s < S < m . Let p ) m be a prime and let X s Ž a1 , . . . , a p . g S p < a1 a2 ??? a p s e 4 . For each Ž a1 , . . . , a pym . g S py m there is a unique Ž a pymq1 , . . . , a p . g S m such that Ž a1 , . . . , a p . g X. Therefore, < X < s < S < py m . Notice that if Ž a1 , a2 , . . . , a p . g X, then Ž a p , a1 , . . . , a py1 . g X. Thus ² g :, the cyclic group of order p, acts on X by g Ž a1 , . . . , a p . s Ž a p , a1 , . . . , a py1 .. The size of each orbit divides p, and thus each orbit either is a singleton or has cardinality p. We now consider two cases: Case 1. There is an orbit with one element, say Ž a1 , . . . , a p .. Clearly, all of the a i are equal, so a1p s e. Since e f S, a1 / e. Thus, a1 is an element of order p, so p < < G <. Case 2. Every orbit has p elements. Then p < < S < py m , so p < < S <. Consequently, p < < G <. 206 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
UNIQUE PRODUCT STRUCTURES
207
In either case, p < < G <. Since this holds for infinitely many p, we have a contradiction.
ACKNOWLEDGMENTS The author acknowledges useful conversations with Gary Sherman of the Rose]Hulman Institute of Technology and Phil Bradley of Rice University.
REFERENCE 1. D. Dimovski, Groups with unique product structures, J. Algebra 146 Ž1992., 205]209.