Primoz
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## 13 Answers

13 votes

Lie algebras over rings (Lie rings) are important in group theory. For instance, to every group $G$ one can associate a Lie ring $$L(G)=\bigoplus _{i=1}^\infty \gamma _i(G)/\gamma _{i+1}(G),$$ where ...

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11 votes

Tarski monsters provide examples of 2-generator noetherian groups that is not finitely presented. Edit (YCor): Tarski monsters, as defined in the link (infinite groups of prime exponent $p$ in which ...

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8 votes

There is another description of $H_2(G)$ due to Miller: Miller, Clair The second homology group of a group; relations among commutators. Proc. Amer. Math. Soc. 3, (1952). 588--595. In modern ...

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5 votes

There is a whole theory of functional equations (or functional identities) in algebras. It was used for instance to obtain solutions to some of Herstein's problems on Lie homomorphisms. An overview of ...

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5 votes

Every group of exponent 3 is nilpotent of class at most 3, and this bound is best possible. The question whether finitely generated groups of exponent $n$ are finite is also known as the Burnside ...

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3 votes

The answer is yes, $S$ is an inverse limit of its lower central quotients. As these have bounded derived length, the same goes for the Cartesian product of these groups. By the way, all pro-$p$ groups ...

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3 votes

This should follow along the lines of Robinson's book "Finiteness conditions and generalized soluble groups", Part 2, Section 9.2. I will try to sketch an argument using Robinson's terminology. Since $... View answer 3 votes The negative solution of Burnside's problem by Novikov and Adjan is an outstanding example of use of complicated induction. Quoting from Wikipedia's entry on Adjan: "The solution of the Burnside ... View answer 3 votes The 1967 book Varieties of Groups by Hanna Neumann has the following footnote (p. 21): The term metabelian will always mean solvable of length two in agreement with current English usage; note however ... View answer 2 votes Classification of$p$-groups by nilpotency class is hard in general. As pointed above, coclass seems to be a better invariant. On the other hand, Ahmad, Magidin and Morse recently finished a ... View answer 2 votes Given an arbitrary countable group$H$containing an element of large enough order but no involutions, there exists a two-generator simple group$G$such that$H$is a proper subgroup of$G$and$G=\...

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1 votes

L.-C. Kappe and M. Meriano dealt with a similar problem, see Meriano's Groups St. Andrews 2013 slides They defined certain subgroups ${}^*w_2(g)$ for a two-variable word $w$ which seem to be what ...

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1 votes

Try the book "Functional identities" by Bresar, Chebotar, and Martindale. It deals with functional equations in the realm of associative algebras, Lie algebras and Jordan algebras. As I mentioned in ...

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