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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.

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    Please tell us what you mean by "the normalizer condition".2011-12-28
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    @GeoffRobinson: The normalizer condition is that "normalizers grow" for proper subgroups (as in finite nilpotent groups).2011-12-28
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    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.2011-12-28
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    @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.2011-12-28
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    @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?2011-12-28
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    @GeoffRobinson: How big could such a chain be? Surely it has to be bounded by a function of $|G|$, and so the chain must have two neighbors equal to one another at some ordinal, which only happens with the whole group $G$.2011-12-28
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    @Geoff: I'm very sorry! I meant what Steve said: the normalizer of any proper subgroup properly contains it. (I thought the words nilpotent and the article "the" were sufficient).2011-12-28
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    @Steve: Could you please elaborate on your last comment? I don't follow you.2011-12-28
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    @Robert: About how big chains can be? What about it is troubling you?2011-12-28
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    @Steve,Robert: I suppose you can use some form of choice: take a maximal ascending chain of groups in which your original subgroup is subnormal, and take the union of that chain. If it is not the whole group, it is strictly contained in its own normalizer. But the original subgroup is still subnormal in that normalizer, contrary to maximality.2011-12-28
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    @Steve: First, it is not clear what "ascending series" means. Strictly ascending sequence of subgroups, each normal in the following?2011-12-28
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    @Geoff: By 'maximal ascending chain' you mean a chain that we cannot insert more subgroups, right? What is the original subgroup you are referring to? I'm sorry but I don't understand what you are proving.2011-12-28
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    @Robert: First of all, no need to apologise really: it would have been better to define what you meant by the "normalizer condition", since some people might be unsure what you meant. I was trying to prove that if we start with a subgroup $H$ of $G,$ and $G$ satisfies the normalizer condition, then (as Steve D asserted) there is a chain of groups, with each normal in the next, starting with $H$ and ending with $G.$ I was trying to do it using Zorn's Lemma.2011-12-28
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    An ascending series is a collection of subgroups, indexed by a well-ordered set $A$, such that for any $\alpha\in A$, $H_\alpha$ is normal in $H_{\alpha+1}$.2011-12-28
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    If I have a subgroup $H$, I define $H_0=H$, and for any ordinal $\alpha$, I define $H_{\alpha+1}=N_G(H_\alpha)$. For limit ordinals $\beta$, I do the usual trick of taking the union over all $\alpha<\beta$. If this never reached $G$, we would get a collection of distinct subgroups for any ordinal. That means $|G|$ is bigger than any cardinal, including $2^{|G|}$, which is absurd. So there is some $\alpha$ with $H_\alpha=H_{\alpha+1}$, so that $H_\alpha=G$.2011-12-28
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    @Geoff, Steve: Thank you for the comments!2011-12-28

1 Answers 1

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Isn't the answer to 1 clearly yes, because the normalizer of the inverse image of a subgroup of a quotient properly contains it?

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 $Q=P$.

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    Derek, Thank you for answering! Yes, 1 was that easy. About 2, can you give me a hint to see why P is the union of its higher centers?2011-12-28
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    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 $Z_r(P_n)$.2011-12-29