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Elementary equivalence and the commutator subgroup

Published online by Cambridge University Press:  18 May 2009

L. P. D. van den Dries ,
A. M. W. Glass ,
Angus Macintyre ,
Alan H. Mekler  and
John Poland
L. P. D. van den Dries
Affiliation:
Yale University, New Haven, Connecticut, U.S.A.
A. M. W. Glass
Affiliation:
Bowling Green State University, Bowling Green, Ohio, U.S.A.
Alan H. Mekler
Affiliation:
University of Western Ontario, London, Ontario, Canada
John Poland
Affiliation:
Carleton University, Ottawa, Ontario, Canada
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If G and H are elementarily equivalent groups (that is, no elementary statement of group theory distinguishes between G and H) then the definable subgroups of G are elementarily equivalent to the corresponding ones of H. But G′ of G, consisting of all products of commutators [a, b] = a−1b−1ab of elements a and b of G, may not be definable. Must G′ and H′ be elementarily equivalent?

Information

Type
Research Article
Information
Glasgow Mathematical Journal , Volume 23 , Issue 2 , July 1982 , pp. 115 - 117
Copyright
Copyright © Glasgow Mathematical Journal Trust 1982

References

REFERENCES

1.Mal'cev, A. I., Two remarks on nilpotent groups, Mat. Sb. (N.S.) 37(79) (1955), 567–572 (in Russian).Google Scholar
2.Rogers, Pat, Preservation of saturation and stability in a variety of nilpoent groups, J. Symbolic Logic 46 (1981), 499–512.CrossRefGoogle Scholar