TITLE: Characterizations of the $p$-Prüfer group
QUESTION [15 upvotes]: I'm an undergrad student fairly keen on algebra. Over the different algebra courses I've taken, I've often encountered the so-called $p$-Prüfer group on exercises but somehow never got around to them. Now I'm trying to take care of that, but there are some statements I've seen about this group which I don't know how to prove (maybe because I lack some more background in group theory, especially in the study of infinite abelian groups?)

Definition A $p$-group is a $p$-Prüfer group if it is isomorphic to $$C_{p^\infty}=\{e^{\frac{2k\pi i}{p^n}}:k\in \mathbb{Z}, n\in\mathbb{Z}^+\} \subset (\mathbb{C}^\times, \cdot)$$

What I'm having trouble to prove is:

The following are $p$-Prüfer groups:
1) An infinite $p$-group whose subgroups are totally ordered by inclusion,
2) An infinite $p$-group such that every finite subset generates a cyclic group,
3) An infinite abelian $p$-group such that $G$ is isomorphic to every proper quotient,
4) An infinite abelian $p$-group such that every subgroup is finite

Just for the record, what I (think I) could prove was that the following are $p$-Prüfer groups:

5) An injective envelope of $C_{p^n}$, for any $n\geq 1$,
6) A Sylow $p$-subgroup of $\frac{\mathbb{Q}}{\mathbb{Z}}$,
7) The direct limit of $0\subset C_p \subset C_{p^2}\subset ...$

Here $C_{p^n}$ denotes a cyclic group of order $p^n$.
Any other characterizations of the $p$-Prüfer group are welcome.

REPLY [7 votes]: Arturo already explained the hint 1⇒7, 2⇒7 (and 1⇒2 is easy enough, I trust you can do it).  I'll explain how to use pG = { pg : g in G } to understand G for 3 and 4.
If every nonzero quotient of G is isomorphic to G, then what does G/pG look like?  Well, it is an elementary abelian p-group, so a vector space over Z/pZ.  If it is nonzero, then it has a one-dimensional quotient: Z/pZ.  By the hypothesis on G, that would mean G itself is Z/pZ.  Unfortunately such a group is not infinite, and so is not a group as in 3.  Hence a group as in 3 must have G=pG.  As it is a p-group, it is divisible, so a direct sum of Prüfer p-groups.  However, such a direct sum always has a single Prüfer p-group as a quotient, and so G must itself be the Prüfer p-group.
If every proper subgroup of G is finite, then what does pG look like?  If it is finite, then G/pG is infinite, and so infinite dimensional.  Take a proper subspace.  The preimage of that subspace in G is a proper subgroup that is infinite. Oops. So pG cannot be finite!  So again G=pG, and G is a direct sum of Prüfer p-groups.  How many?  Well each summand is a subgroup, and so if there is more than one, then one has an infinite proper subgroup.  So G must itself be a single Prüfer p-group.