# Groups satisfying the normalizer condition

I have two questions related to nilpotent groups:

1. Is the class of groups satisfying the normalizer condition closed under taking quotients?

2. Are there examples of (infinite) groups satisfying the normalizer condition but not solvable?

Thanks.

EDIT: Here 'the normalizer condition' is the condition that the normalizer of any proper subgroup properly contains it.

• Please tell us what you mean by "the normalizer condition". Dec 28, 2011 at 7:19
• @GeoffRobinson: The normalizer condition is that "normalizers grow" for proper subgroups (as in finite nilpotent groups).
– user641
Dec 28, 2011 at 8:15
• 1) is true, because the normalizer condition is equivalent to every subgroup being an "infinite" version of subnormal: that is, every subgroup has an ascending series going from it to the whole group. There are definitely examples for 2), but I can't think of any. The normalizer condition (groups with it are usually called N-groups) implies local nilpotence, hence local solvability. It also implies there is an ascending series for the group with abelian factors, much like a solvable group.
– user641
Dec 28, 2011 at 8:19
• @Steve: Sure that happens in nilpotent groups, and is a well-known property, but the OP should have told us which normalizer condition he meant, there could be others. Dec 28, 2011 at 8:20
• @Steve: Is 1. really clear? It works in nilpotent groups, but in general why couldn't you have an infinite strictly ascending chain of subgroups which never reaches the whole group? Dec 28, 2011 at 8:24

As for 2, let's fix a prime $p$, and let $P_n$ be a sequence of finite $p$-groups of unbounded derived length. For example, we could take iterated wreath products. Now let $P$ be the direct product of the $P_n$. Then $P$ is not solvable, since there is no bound on its derived length.
To show that $P$ satisfies the normalizer condition, let $Q$ be a subgroup of $P$. If $Q$ does not contain $Z(P)$, which is the direct product of the $Z(P_n)$, then $H$ is properly contained in $QZ(P) \le N_P(Q)$. So $Z(P) \le Q$. But now if $P$ does not contain the second centre $Z_2(P)$ of $P$, then $Q$ is properly contained in $QZ_2(P) \le N_P(Q)$. So by induction we get $Z_r(P) \le Q$ for all $r$, but the union of the $Z_r(P)$ is $P$, so $Q=P$.
• Yes, prove that each of the individual direct factors $P_n$ is in the union of $Z_r(P_n)$ and hence in the union of $Z_r(P_n)$. Dec 29, 2011 at 21:41