JOHN POLAND - Academia.edu (original) (raw)
Papers by JOHN POLAND
Proceedings of the Japan Academy, 1971
Presenting a Mathematics Play
Mathematics Magazine, 1987
In the eigenvector solution the teachers are ranked in descending order D, B, C, A, E; whereas th... more In the eigenvector solution the teachers are ranked in descending order D, B, C, A, E; whereas the LLSM solution ranks them B, D, C, A, E. The LLSM minimizes deviations over all the entries of the matrix. The principal eigenvector does not attempt to minimize anything, but ...
Journal of Pure and Applied Algebra, 1982
Glasgow Mathematical Journal, 1982
If G and H are elementarily equivalent groups (that is, no elementary statement of group theory d... more 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?
The number of conjugacy classes of non-normal subgroups in nilpotent groups
Communications in Algebra, 1996
ABSTRACT. In a recent paper, Rolf Brandl classified ell finite groups having exactly one conjugac... more ABSTRACT. In a recent paper, Rolf Brandl classified ell finite groups having exactly one conjugacy class of nonnormal subgroups, and conjectured that for a nilpotent group C of nilpotency clau, c = c(C), the number v(C) = u of conjugacy cl-of nonnormd subgroup sathfiea the ...
Canadian Journal of Mathematics, 1968
This paper presents a list of all finite groups having exactly six and seven conjugate classes an... more This paper presents a list of all finite groups having exactly six and seven conjugate classes and an outline of the background necessary for the proof, and gives, in particular, two results which may be of independent interest. In 1903 E. Landau (8) proved, by induction, that for each the equation * has only finitely many solutions over the positive integers.
Bulletin of the Australian Mathematical Society, 1970
If G is a finite group and P is a group-theoretic property, G will be called P-max-core if for ev... more If G is a finite group and P is a group-theoretic property, G will be called P-max-core if for every maximal subgroup M of G , M/Mg has property P where M G = H i x~ Mx) is the xeG core of M in G. In a joint paper with John D. Dixon and A.H. Rhemtulla, we showed that if p is an odd prime and G is (p-nilpotent)-max-core, then G is p-solvable, and then using the techniques of the theory of solvable groups, we characterized nilpotent-max-core groups as finite nilpotent-by-nilpotent groups. The proof of the first result used John G. Thompson's p-nilpotency criterion and hence required p > 2. In this paper I show that supersolvable-max-core groups (and hence (2-nilpotent)-max-core groups] need not be 2-solvable (that is, solvable). Also I generalize the second result, among others, and characterize (p-nilpotent)-max-core groups (for p an> odd prime) as finite nilpotent-by-(p-nilpotent) groups.
On lattice-isomorphic abelian groups
Archiv der Mathematik, 1992
Theorem A (mostly Baer [1]). Let A and B be abelian groups with isomorphic subgroup lattices. The... more Theorem A (mostly Baer [1]). Let A and B be abelian groups with isomorphic subgroup lattices. Then the following hold. (i) A is locally cyclic iff B is. In particular, if A is infinite cyclic, then so ~s B, and the lattice isomorphisms from A to B correspond to bijections from the ...
On verifying lattice isomorphisms between groups
Archiv der Mathematik, 1985
A lattice isomorphism between two groups is actually an isomorphism between their lattices of sub... more A lattice isomorphism between two groups is actually an isomorphism between their lattices of subgroups. How does one go about constructing lattice isomorphisms? Sup-pose we were given two groups and we have sufficient information about their subgroups that we can guess ...
Archiv der Mathematik, 1993
A Modern Fairy Tale?
The American Mathematical Monthly, 1987
292 JOHN Poland [March emphasis upon the general Arts and Science student taking mathematics: onl... more 292 JOHN Poland [March emphasis upon the general Arts and Science student taking mathematics: only the usual programsin physics, chemistry, computer science and mathematics require the introductory calculus course. It is just a standard, traditional ...
Journal of the Australian Mathematical Society, 1968
Let G be a finite group of order g having exactly k conjugate classes. Let π(G) denote the set of... more Let G be a finite group of order g having exactly k conjugate classes. Let π(G) denote the set of prime divisors of g. K. A. Hirsch [4] has shown that By the same methods we prove g ≡ k modulo G.C.D. {(p–1)2 p ∈ π(G)} and that if G is a p-group, g = h modulo (p−1)(p2−1). It follows that k has the form (n+r(p−1)) (p2−1)+pe where r and n are integers ≧ 0, p is a prime, e is 0 or 1, and g = p2n+e. This has been established using representation theory by Philip Hall [3] (see also [5]). If then simple examples show (for 6 ∤ g obviously) that g ≡ k modulo σ or even σ/2 is not generally true.
Proceedings of the Japan Academy, 1971
Presenting a Mathematics Play
Mathematics Magazine, 1987
In the eigenvector solution the teachers are ranked in descending order D, B, C, A, E; whereas th... more In the eigenvector solution the teachers are ranked in descending order D, B, C, A, E; whereas the LLSM solution ranks them B, D, C, A, E. The LLSM minimizes deviations over all the entries of the matrix. The principal eigenvector does not attempt to minimize anything, but ...
Journal of Pure and Applied Algebra, 1982
Glasgow Mathematical Journal, 1982
If G and H are elementarily equivalent groups (that is, no elementary statement of group theory d... more 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?
The number of conjugacy classes of non-normal subgroups in nilpotent groups
Communications in Algebra, 1996
ABSTRACT. In a recent paper, Rolf Brandl classified ell finite groups having exactly one conjugac... more ABSTRACT. In a recent paper, Rolf Brandl classified ell finite groups having exactly one conjugacy class of nonnormal subgroups, and conjectured that for a nilpotent group C of nilpotency clau, c = c(C), the number v(C) = u of conjugacy cl-of nonnormd subgroup sathfiea the ...
Canadian Journal of Mathematics, 1968
This paper presents a list of all finite groups having exactly six and seven conjugate classes an... more This paper presents a list of all finite groups having exactly six and seven conjugate classes and an outline of the background necessary for the proof, and gives, in particular, two results which may be of independent interest. In 1903 E. Landau (8) proved, by induction, that for each the equation * has only finitely many solutions over the positive integers.
Bulletin of the Australian Mathematical Society, 1970
If G is a finite group and P is a group-theoretic property, G will be called P-max-core if for ev... more If G is a finite group and P is a group-theoretic property, G will be called P-max-core if for every maximal subgroup M of G , M/Mg has property P where M G = H i x~ Mx) is the xeG core of M in G. In a joint paper with John D. Dixon and A.H. Rhemtulla, we showed that if p is an odd prime and G is (p-nilpotent)-max-core, then G is p-solvable, and then using the techniques of the theory of solvable groups, we characterized nilpotent-max-core groups as finite nilpotent-by-nilpotent groups. The proof of the first result used John G. Thompson's p-nilpotency criterion and hence required p > 2. In this paper I show that supersolvable-max-core groups (and hence (2-nilpotent)-max-core groups] need not be 2-solvable (that is, solvable). Also I generalize the second result, among others, and characterize (p-nilpotent)-max-core groups (for p an> odd prime) as finite nilpotent-by-(p-nilpotent) groups.
On lattice-isomorphic abelian groups
Archiv der Mathematik, 1992
Theorem A (mostly Baer [1]). Let A and B be abelian groups with isomorphic subgroup lattices. The... more Theorem A (mostly Baer [1]). Let A and B be abelian groups with isomorphic subgroup lattices. Then the following hold. (i) A is locally cyclic iff B is. In particular, if A is infinite cyclic, then so ~s B, and the lattice isomorphisms from A to B correspond to bijections from the ...
On verifying lattice isomorphisms between groups
Archiv der Mathematik, 1985
A lattice isomorphism between two groups is actually an isomorphism between their lattices of sub... more A lattice isomorphism between two groups is actually an isomorphism between their lattices of subgroups. How does one go about constructing lattice isomorphisms? Sup-pose we were given two groups and we have sufficient information about their subgroups that we can guess ...
Archiv der Mathematik, 1993
A Modern Fairy Tale?
The American Mathematical Monthly, 1987
292 JOHN Poland [March emphasis upon the general Arts and Science student taking mathematics: onl... more 292 JOHN Poland [March emphasis upon the general Arts and Science student taking mathematics: only the usual programsin physics, chemistry, computer science and mathematics require the introductory calculus course. It is just a standard, traditional ...
Journal of the Australian Mathematical Society, 1968
Let G be a finite group of order g having exactly k conjugate classes. Let π(G) denote the set of... more Let G be a finite group of order g having exactly k conjugate classes. Let π(G) denote the set of prime divisors of g. K. A. Hirsch [4] has shown that By the same methods we prove g ≡ k modulo G.C.D. {(p–1)2 p ∈ π(G)} and that if G is a p-group, g = h modulo (p−1)(p2−1). It follows that k has the form (n+r(p−1)) (p2−1)+pe where r and n are integers ≧ 0, p is a prime, e is 0 or 1, and g = p2n+e. This has been established using representation theory by Philip Hall [3] (see also [5]). If then simple examples show (for 6 ∤ g obviously) that g ≡ k modulo σ or even σ/2 is not generally true.