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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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8  
Please tell us what you mean by "the normalizer condition". –  Geoff Robinson Dec 28 '11 at 7:19
    
@GeoffRobinson: The normalizer condition is that "normalizers grow" for proper subgroups (as in finite nilpotent groups). –  user641 Dec 28 '11 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 '11 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. –  Geoff Robinson Dec 28 '11 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? –  Geoff Robinson Dec 28 '11 at 8:24
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1 Answer

up vote 6 down vote accepted

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? –  Robert Dec 28 '11 at 21:59
    
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 $Z_r(P_n)$. –  Derek Holt Dec 29 '11 at 21:41
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