# An analogue definition of nilpotency in terms of factor groups of nontrivial subgroups like for solvable groups

A finite group $G$ is called solvable if every nontrivial subgroup $U$ has a nontrivial abelian factor group (equivalently $U' < U$), or equivalently if every nontrivial subgroup $U$ has a non-trivial nilpotent factor group, i.e. there exists $N \unlhd U, N \ne U$ such that $U / N$ is nilpotent. This follows as nilpotency implies solvability.

Similarly, a finite group is called $p$-solvable if every nontrivial subgroup $U$ has a nontrivial $p$-closed factor group, i.e. there exists $N \unlhd U, N \ne U$ such that $U / N$ has a unique (hence normal) Sylow $p$-subgroup.

Now is there a similar characterisation of nilpotency in terms of subgroups?

The notions of solvable, and also $p$-solvable could be defined in terms of certain (characteristic-, subnormal- or normal-) series, but in the case of nilpotency we just have upper and lower central series, so here a different kind of series arise; and also the "normalizers grow"-definition is not in terms of subgroups of some fixed group, but in terms of groups that properly contain our group, so I see here are some differences, but anyway; is there a characterisation of nilpotent groups in terms of factor groups of nontrivial subgroups?

• If $N$ is nilpotent and $G/N'$ is nilpotent then $G$ is nilpotent. – mesel Nov 23 '15 at 23:42
• The factor group is taken by the commutator subgroup of $N$? – StefanH Nov 24 '15 at 0:00
• Yes it is. $N'$ is the commutater of $N$. – mesel Nov 24 '15 at 0:04