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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 subgroups?

edited Oct 12 2010 at 2:35

asked Oct 12 2010 at 1:57

Jeff Strom
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Every subgroup of a free group is free, in particular non-perfect. But it certainly isn't nilpotent. So, yes, the two notions do indeed diverge. – HW Oct 12 2010 at 2:21

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When you say "nilpotent" do you mean "solvable"? The group of permutations of three objects is not nilpotent, but it has no perfect subgroups (unless you count the trivial subgroup). – S. Carnahan♦ Oct 12 2010 at 2:25

thanks: changed nilpotent to solvable. (and of course I don't count the trivial subgroup). – Jeff Strom Oct 12 2010 at 2:35

I don't know a name for it, but one way to think about the property in question is the following: Recursively define subgroups $G_\lambda\subset G$ for ordinals $\lambda$ by: $G_0=G$, and $G_{n+1}=DG_n$, and for a limit ordinal $\lambda$ take the intersection of what you have so far. By definition $G$ is solvable if $G_n=1$ for some natural number $n$. $G$ has no nontrivial perfect subgroup if $G_\lambda=1$ for some $\lambda$. In fact, I suppose this construction shows that every group has a unique maximal perfect subgroup, namely the first $G_\lambda$ such that $G_{\lambda+1}=G_\lambda$. – Tom Goodwillie Oct 12 2010 at 4:09

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@Tom: The groups with $G_\lambda=1$ are called hypoabelian. See, for example, Vovsi, S. M. Two notes on local properties of groups. Simon Stevin 55 (1981), no. 1-2, 27–35. Also: planetmath.org/… – Mark Sapir Oct 12 2010 at 9:59

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Mark Sapir · Accepted Answer · 2010-10-12 11:25:34Z UTC

In the infinite case, there is a close notion of "locally indicable group", i.e. a group where every finitely generated subgroup maps onto $\mathbb Z$ (see, for example, this paper). Locally indicable groups are left (right) orderable, hence important. Note that in that notion, not all subgroups are considered but only finitely generated, and "non-perfect" is replaced by a stronger property "maps onto $\mathbb Z$". But in the finite case all subgroups are finitely generated, and "maps onto $\mathbb Z$" is an infinite analog of "maps onto a finite cyclic group" (= non-perfect). So "locally indicable" is possibly the infinite analog of the property you consider.

Update: The groups without perfect subgroups are called hypoabelian. See this text.

Groups with no perfect subgroups — terminology?

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