Search Results

The abelianization of the Higman group $G$ is trivial, because in the presence of commutativity the relations become $a = 1$, $b =1$, $c =1$, $d=1$. Therefore $G$ is not free.

Reference: Higman, Graham, A finitely generated infinite simple group, J. Lond. Math. Soc. 26, 61-64 (1951). ZBL0042.02201. It is shown that $G$ is infinite and has no proper normal subgroups of fini …

Finite groups are solvable if they have no nontrivial perfect subgroup. But I am sure that for infinite groups, the two notions diverge. Is there standard terminology for groups with no perfect sub …

The standard rearrangement theorem for conditionally convergent series says that the terms in a conditionally convergent series can be rearranged so that the new sum is any desired number, or $\pm\inf …

I would like to think about formation of the smallest subgroup (or monoid, or whatever) $H$ of $G$ containing two given subgroups $A$ and $B$ as an operation on subgroups, and I wonder if there is a …