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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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=\...

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 ...