Do you know results about maximal normal subgroup among normal subgroups not containing a given element $x$ ? The problem can be reduce to the case of free groups.

First, such a sugroup exists thanks to Zorn lemma. Secondly, I think that if $x$ is a primitive element, then there is only one maximal normal subgroup among normal subgroups not containing $x$; otherwise, there is not unicity: if $x=[a,b]$, there is a one-to-one correspondance between our normal subgroup and non abelian two-generator groups whose proper quotients are abelian (eg. the dihedral group $D_3$ or the quaternion group $Q_8$).

But I have no idea about how construct such subgroups.

  • $\begingroup$ What is a primitive element? $\endgroup$ – Qiaochu Yuan Jul 16 '12 at 12:06
  • $\begingroup$ A primitive element is an element of a free basis. $\endgroup$ – Seirios Jul 16 '12 at 12:45

I don't think you get uniqueness, even in the "primitive" case.

Look at the simple case of $C_2\times C_2$, and consider the maximal subgroups that do not contain $(x,x)$, where $x$ is the nontrivial element: you have $C_2\times\{1\}$ and $\{1\}\times C_2$ as maximal subgroups not containing $(x,x)$.

In the free case, lift the example. Take $F$ free of rank $2$ generated by $x$ and $y$. Then $xy$ is primitive, an element of the basis $x,xy$ (for instance). There are at least two maximal normal subgroups that do not contain $xy$: $N_1=\langle x,y^2,[F,F]\rangle$ and $N_2 = \langle x^2,y,[F,F]\rangle$. As these subgroups have index $2$ in $F$, they are normal and maximal in $F$; they also do not contain $xy$, so they are maximal normal subgroups that do not contain $xy$.


I understand that what you did above was reduced to free groups, yet I can't understand how you use, in the most general case, Zorn's lemma here, as when dealing with a chain (inductively ordered set in the set with partial order) it may be the maximal element you take there is the whole group...

Besides the above, there are some groups which have no proper maximal-normal subgroups, e.g. the rationals $\,\Bbb Q\,$ or, for that matter, any other (abelian) divisible group.

I also don't know what a primitive element within this context is...perhaps you meant one of the elements in some free basis?

What I can say right now is that finitely generated groups always have subgroups which are maximal wrt being normal and proper. I once dealed with this while trying to find out something about infinite f.g. simple groups of a rather "sweeter" nature than say Tarski Monsters (e.g., the Higman group which appears in Appendix G of the second book of the english translation, by Hirsch, of Kurosh's "The Theory of Groups". Very nice book, btw).

But even with all the above said, I can't understand why exactly could you hope for unicity of a maximal normal group not-containing some primitive element $\,x\,$, but perhaps this is due to my ignorance of what primitive means here.

  • $\begingroup$ Formally, my problem is: Let $F$ be a free group, $x \in F$ and set $\mathcal{N}_x= \{ N \lhd F | x \notin N\}$. What is the maximal elements of $\mathcal{N}_x$ (for inclusion) ? So such an element cannot be the whole group, because it doesn't contain $x$ by definition. $\endgroup$ – Seirios Jul 16 '12 at 12:49
  • $\begingroup$ @DonAntonio: Suppose $\{H_i\}$ is a chain of (normal) subgroups such that each $H_i$ does not contain $x$. Then $\cup H_i$ is a subgroup (easy to verify: nonempty, closed under products and inverses), is normal, and cannot be the whole group: If $\cup H_i = G$, then $x\in \cup H_i$, hence $x\in H_i$ for some $i$; but the $H_i$ are all subgroups that do not contains $x$. While there are no absolute maximal subgroups of $\mathbb{Q}$, if you specify a particular rational, then you can find subgroups maximal among those that do not contain $x$. (cont) $\endgroup$ – Arturo Magidin Jul 16 '12 at 13:07
  • $\begingroup$ @DonAntonio: (cont) E.g., given $x=\frac{1}{2}$, the subgroup of all rationals which can be written in least terms with odd denominator is maximal among subgroups not containing $\frac{1}{2}$: given any other rational, $\frac{n}{2k}$, $\gcd(2k,n)=1$, we can obtain $\frac{1}{2k}$, and from there $\frac{1}{2}$. $\endgroup$ – Arturo Magidin Jul 16 '12 at 13:10
  • $\begingroup$ @ArturoMagidin, thanks for that. I know how to get (normal) subgroups not containing some given fixed element different from the unit using Zorn's lemma. My comment on Q was wrong as there anything is normal and I didn't focus, as I should, on sbgps. not containing some element. It still bugs me though the unicity the OP: in your example this is attainable, but in general? I'm not sure... $\endgroup$ – DonAntonio Jul 16 '12 at 13:55
  • $\begingroup$ @Seirios, I see your comment and I really don't know if there exists some characterization of such maximal subgroups... $\endgroup$ – DonAntonio Jul 16 '12 at 14:10

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