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My motivation for this question is from Universal Algebra: A congruence of an arbitrary algebra $A$ is said to be principal, if it is generated by a single element. In the case of rings, this is just the notion of principal ideal and for groups it is a normal subgroup which is the normal closure of a single element, more precisely:

A normal subgroup of the form $\langle x^G\rangle$ is called a principal subgroup of the group $G$. We say that $G$ is a principal group, if every normal subgroup of $G$ is principal.

Is there any classification of principal groups? Is there at least a classification of nilpotent (solvable) principal groups?

The same notion can be defined for Lie algebras and also the same questions for Lie algebras arise.

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Let us consider the case when $G$ is a finite $p$-group. It has been shown above that all the normal subgroups of $G$ are principal if and only if $G$ is cyclic.

One might try and relax the assumption, and do not require $G$ itself to be principal. Then $G / G_{2}$ can also be elementary abelian of order $p^{2}$. (I am writing $G_{2} = [G, G]$, and $G_{i+1} = [G_{i}, G]$ for the terms of the lower central series.) Now the nontrivial quotients $G_{i}/G_{i+1}$ are cyclic, as noted above, and also elementary abelian, since $G/G_{2}$ is. Thus they are cyclic of order $p$, and $G$ is a $p$-group of maximal class, as defined by Blackburn.

Conversely, all proper normal subgroups of a $p$-group $G$ of maximal class are principal. Recall that the proper normal subgroups of such a group $G$ are the terms $G_{i}$ of the lower central series, and the maximal subgroups. Now if $x \in G_{i} \setminus G_{i+1}$, then $\langle x \rangle^{G}$ is a normal subgroup of $G$ which is contained in $G_{i}$, but not in $G_{i+1}$. Thus $G_{i} = \langle x^{G} \rangle$. If $M$ is a maximal subgroup of $G$, take $m \in M \setminus G_{2}$, and take $s \in G \setminus M$ to be an element such that $[G_{i}, s] = G_{i+1}$ for all $i$ (this is known to exist by the theory of such groups). Now $[m, s] \in G_{2} \setminus G_{3}$, and commuting further with $s$ we find generators for all $G_{i}/G_{i+1}$. It follows that $M = \langle m \rangle^{G}$.

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@Andreas Carant: this is a nice point. Now we can conclude that if $G$ is a finite nilpotent group with all proper normal subgroups principal, then $G$ is cyclic or a $p$-group of maximal class. –  M. Shahryari Dec 13 '12 at 11:52
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My initial answer was, it turns out, weak. So I'm going to add in some of the observations made in comments, and turn this answer into community wiki, so as not to get credit.

  1. If G is principle, then the factors of its upper central series, lower central series and derived series are cyclic. There are many other immediate consequences of definition: If G is principal, then so are its quotients and its characteristic subgroups. Such a group has the max-n property. If it is hypercentral, then it is nilpotent. If solvable, then super-solvable.

  2. For infinite groups, the observation of the previous bullet implies that Baumslag and Blackburn's paper ``Groups with Cyclic Upper Central Factors'' is relevant.

  3. For finite $p$-groups, observe that any principal p-group P is cyclic as the Frattini quotient $P/[P,P]∗P^p$ is one dimensional (again, this follows from the first bullet point). This implies, in particular, that a finite principal nilpotent group is cyclic.

  4. As for finite principal solvable groups, well, things are less clear. Note that $S_3$ is principal.

  5. If $G$ is a finite $p$-group such that its proper normal subgroups are principal ($G$ itself is not principal), then $G$ has maximal class. The converse is also true.

  6. If $G$ is finite nilpotent group such that its proper normal subgroups are principal, then $G$ is cyclic or it is a $p$-group of maximal class.

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Any principal $p$-group $P$ is cyclic as the Frattini quotient $P/[P,P]*P^p$ is one dimensional. –  JSpecter Dec 12 '12 at 17:22
    
@Nick Gill: please send a copy for me mshahryari@tabrizu.ac.ir thank you so much for comment. –  M. Shahryari Dec 12 '12 at 17:59
    
For finite principal $p$-groups $P$, you can also see that $P/[P,P]$ is cyclic, so $P/\Phi(P)$ is certainly cyclic- this is just equivalent to what JSpecter says. –  Geoff Robinson Dec 12 '12 at 18:04
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@Nick Gill: much more is true: If $G$ is principle, then the factors of its upper central series, lower central series and derived series are cyclic. There are many other immediate consequences of definition: If $G$ is principal, then so is its quotients and its characteristic subgroups. Such a group has max-$n$ property. –  M. Shahryari Dec 12 '12 at 19:50
    
@JSpecter and Geoff Robinson: By your comment, if $G$ is a finite nilpotent principal group, then it must be cyclic. For proof, let $G=P_1\times \cdots\times P_n$, where $P_i$'s are Sylow subgroups of $G$. Then every $P_i$ is principal (since it is characteristic) and so $P_i$ is cyclic. So $G$ is cyclic. I don't know is the same true for infinite nilpotent principal groups or not. Also for finite solvable groups, can we obtain a similar result? –  M. Shahryari Dec 12 '12 at 19:58
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