1999
DOI: 10.1080/00927879908826712
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The derived subgroup of a group with commutators of bounded order can be non-periodic

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Cited by 11 publications
(6 citation statements)
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“…This result was proved in Brandl [3], and again in Shumyatsky [8,Lemma 3.2]. The analogue for infinite groups is true in some cases (see [8]), but not in general (see [4]). It is still open whether (infinite) groups in which [x, y] 3 = 1 for all x, y ∈ G are always soluble (see [7,Problem 12.4]).…”
Section: Introductionmentioning
confidence: 80%
“…This result was proved in Brandl [3], and again in Shumyatsky [8,Lemma 3.2]. The analogue for infinite groups is true in some cases (see [8]), but not in general (see [4]). It is still open whether (infinite) groups in which [x, y] 3 = 1 for all x, y ∈ G are always soluble (see [7,Problem 12.4]).…”
Section: Introductionmentioning
confidence: 80%
“…Moreover, if the assumption that G is residually finite is dropped from the hypothesis of Theorem 1.1, the derived group need not even be periodic. Deryabina and Kozhevnikov showed that for sufficiently big odd integers n there exist groups G in which all commutators have order dividing n such that G ′ has elements of infinite order [4]. Independently, this was also proved by Adian [1].…”
Section: Introductionmentioning
confidence: 91%
“…Moreover, if the assumption that G is residually finite is dropped from the hypothesis of Theorem 1.1, the derived group need not even be periodic. Deryabina and Kozhevnikov showed that for sufficiently large odd integers n there exist groups G in which all commutators have order dividing n such that G has elements of infinite order [4]. This was also proved independently by Adian [1].…”
Section: Introductionmentioning
confidence: 92%