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 $H$ ? 
 A: Short answer:
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 < K ≤ G, 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 < K ≤ G 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 pK ≤ H < K.  Conversely, if pK < K, then there is some maximal subgroup H of K with pK ≤ H < 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 Z ∩ H = 0 (which exists by Zorn's lemma or Hausdorff's maximal principal).  Since H does not contain all of Z, H ≠ G, and so let K ≤ G properly contain H. By maximality, K ∩ Z ≠ 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 < L ≤ K.  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 K ≤ G, the subgroup generated by H and g.
A: 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.
