# Fitting subgroups of infinite groups

Fitting subgroup is defined as the subgroup generated by all normal and nilpotent subgroups of a group G. If G is a finite group, we have that Fitting subgroup is nilpotent. If G is infinite not necessarily this occurs. We are looking for examples of infinite groups such that Fitting subgroup is non-nilpotent. Or some criterion to Fitting subgroup to be nilpotent in infinite case. In general, any answer in this direction (possibly with references) is welcome!

• A direct product of finite groups $G_n$ for $n \ge 0$, where $G_n$ is nilpotent of class $n$ is an infinite group in which the Fitting subgroup (the whole group) is not nilpotent. – Derek Holt Nov 6 '19 at 8:16
• "Direct product" in Derek's comment can both be interpreted as "restricted direct product", and "unrestricted direct product". – YCor Nov 7 '19 at 15:19
• Thanks for the example, but how can i show that this direct product is not nilpotent? For instance, is true that $\gamma_{m}(G_{1} \times G_{2} \times \dots ) = \gamma_{m}(G_{1}) \times \gamma_{m}(G_{2}) \times \dots$ , where $\gamma_{m}$ is the mth term of lower central series? – Inácio Nov 8 '19 at 23:32

There's an obvious criterion: the Fitting subgroup of a group $$G$$ is nilpotent if and only if there exists $$s$$ such that every nilpotent normal subgroup $$N$$ of $$G$$ is $$n$$-step nilpotent (i.e., satisfies $$N^{s+1}=\{1\}$$).
• to find examples it's enough to produce groups with suitably prescribed nilpotent normal subgroups (as in Derek's comment). Alternatively, the group of upper unipotent (=upper triangular with 1 on diagonal) with entries indexed by $$\mathbf{N}$$ matrices, over an arbitrary nonzero commutative associative unital ring, has such normal subgroups.
• if for some reason it is known that there is a bound on the step of nilpotency of (normal) subgroups of $$G$$, then one gets that the Fitting subgroup of $$G$$ is nilpotent. This applies if $$G$$ is linear (i.e., isomorphic to a subgroup of $$\mathrm{GL}_d(K)$$ for some field $$K$$ and $$d<\infty$$).