# Two order related questions for subgroups of infinite groups

For finite groups what I am asking is really trivial. In finite groups if we pick a non trivial subgroup $H$ of $G$ we can always find (there could be multiple choices) a subgroup that is an immediate predecessor (with respect to inclusion) and a subgroup that is an immediate successor of $H$.

1) Can we choose for each non trivial subgroup in an infinite group an immediate predecessor/successor?

2) Choose $H<G$. Can we define a procedure that allows to determine a total order including $H$ ?

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1) No. Take for example $G=\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ and $H=0\oplus\mathbb{Z}/2\mathbb{Z}$. For every subgroup K with H < KG, there is some subgroup L with H < L < K, so H has no immediate successor. (This is equivalent to saying $\mathbb{Z}$ has no minimal nonzero subgroups.)

2) Yes, if I understand what you want. This is called the Hausdorff maximal principle. Every totally ordered subset (including a single guy, H) is contained in a maximal totally ordered subset. However, this maximal totally ordered subset could be very weird, and need not be discrete.

Classification in abelian case: Suppose H < KG are subgroups. If there is no subgroup L with H < L < K, then we say H is a maximal subgroup of K and that K covers H in G. The trivial subgroup has no maximal subgroups, and the full group G has no covers in G.

Call a subgroup "maximal good" if it has a maximal subgroup or if it is the identity subgroup. Call a subgroup "cover good in G" if it is the whole group G or if it is a maximal subgroup of a subgroup of G. Call a subgroup of G "good" if it is both maximal good and cover good. Call a group G wholesome if every subgroup is good in G.

For instance finite groups, Tarski monsters, and torsion DSC groups are wholesome, but groups with a normal infinite cyclic subgroup are not wholesome.

For abelian groups one gets the nice classification:

An abelian group is wholesome if and only if it is torsion and reduced.

A torsion abelian group is one in which every element has finite order, and a reduced abelian group is one that has no direct summand isomorphic to a direct summand of $\mathbb{R}/\mathbb{Z}$, a so called divisible group.

Proof of classification in abelian case:

Lemma on covers and maximal subgroups in torsion abelian groups: Suppose K is torsion and H is a maximal subgroup of K. Then there is some prime p such that pKH < K. Conversely, if pK < K, then there is some maximal subgroup H of K with pKH < K. In particular, for K to be maximal good it is necessary and sufficient for pK < K, and for H to be cover good in G it is necessary and sufficient for $H < (H:p)_G$ where $(H:p)_G = \{ g \in G : pg \in H \}$.

Wholesome G must be torsion: Assume by way of contradiction that G is not torsion. Let Z be an infinite cyclic subgroup, and H a subgroup maximal with respect to ZH = 0 (which exists by Zorn's lemma or Hausdorff's maximal principal). Since H does not contain all of Z, HG, and so let KG properly contain H. By maximality, KZ ≠ 0, and so let z ≠ 0 be a nonzero element of K ∩ Z. Consider the subgroup L generated by H and 2z. Since Z is infinite cyclic and z ≠ 0, 2z ≠ 0, and so H < LK. However, z cannot be written as a sum of h + 2nz lest h = (1−2n)z lie in H ∩ Z = 0. Hence H < L < K, and there is no subgroup covering H.

Wholesome G must be reduced: Let H be a divisible summand of G. Then pH = H, so H is not maximal good unless H = 0. Hence G is reduced.

The converse also works:

G is reduced implies every subgroup of G is maximal good: Let H be a subgroup of G. Since G is reduced, H is not divisible, and so there is some prime p with pH < H, and hence H has a maximal subgroup.

G is torsion implies every subgroup of G is cover good in G: Let H be a proper subgroup of G. Then $G/H$ is a nonzero torsion group, and so has an element gH of prime order (take a nonzero element and raise it to the right power). Then H is a maximal subgroup of KG, the subgroup generated by H and g.

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Proof that $(\mathbb{Q}, +)$ has no maximal subgroups:

Suppose $M < \mathbb{Q}$ is maximal. Since $\mathbb{Q}$ is abelian, $M$ is a normal subgroup. Thus $\mathbb{Q}/M$ has no proper subgroups ($M$ is maximal), and therefore $\mathbb{Q}/M$ is cyclic of prime order $p$.

We have $[G:M] = p$, so using the left coset action we can find a homomorphism $\phi: \mathbb{Q} \rightarrow S_p$ with $Ker(\phi) \leq M$. Now $S_p$ is finite with order $d = p!$. Then for every $q \in \mathbb{Q}$, we have $\phi(q) = \phi(d \cdot \frac{q}{d}) = \phi(\frac{q}{d} + \cdots + \frac{q}{d}) = \phi(\frac{q}{d})^d = 1$. Thus $Ker(\phi) = \mathbb{Q}$, and $M = \mathbb{Q}$.

This is a contradiction, and thus no such $M$ can exist.

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How did you conclude that $\mathbb Q \cong S_p$ from $\ker \phi = \mathbb Q$? – JSchlather Dec 18 '11 at 23:19
Once you state that $\mathbb{Q}/M$ is cyclic of prime order $p$, you know there is something wrong because which $p$ should be singled out for this task (except perhaps 2)? Why would one prime be better than another? Sure, this is no proof, but it was my immediate reaction. – lhf Dec 19 '11 at 0:11
@JacobSchlather: Yes, that is wrong.. I have no idea what I was thinking there. I have edited my post. This was my original idea, but for some reason I changed it to the wrong one.. – Mikko Korhonen Dec 19 '11 at 11:06