Characterizations of the $p$-Prüfer group 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.
 A: How one procedes to prove these probably depends a bit on which description of the $p$-Prüfer group one is most comfortable with. So my particular arguments may not seem very natural to you, in which case this will probably not help much.


*

*Because the subgroups are ordered by inclusion, given any $x,y\in G$, either $x\in\langle y\rangle$ or $y\in\langle x\rangle$. Take an element $x_1\in G$. The subgroup it generates cannot be all of $G$ (it is finite, since $G$ is a $p$-group), so there exists $x_2\in G$ with $x_2\notin\langle x_1\rangle$. Therefore, $x_1\in\langle x_2\rangle$. But $x_2$ cannot generate all of $G$. So there exists $x_3\in G$, $x_3\notin \langle x_2\rangle$.
Continuing this way, you obtain a (countably infinite) collection of elements of $G$, $x_1,x_2,\ldots,$ such that $x_n\in\langle x_{n+1}\rangle$, $x_{n+1}\notin\langle x_n\rangle$. By introducing suitable elements into the sequence, we may assume that $[\langle x_{n+1}\rangle\colon\langle x_n\rangle]=p$. 
Now prove that the subgroup generated by all the $x_i$ is isomorphic to the $p$-Prüfer group, and that it is equal to $G$.

*Let $x_1\in G$. It does not generate all of $G$, so let $y\in G$ that is not in $\langle x_1\rangle$. Then $\langle x_1,y\rangle$ is cyclic, strictly larger than $\langle x_1\rangle$; let $x_2$ be a generator of this cyclic group. Since $x_2$ does not generate all of $G$, there exists $y\in G$ not in $\langle x_2\rangle$; but $\langle x_2,y\rangle$ is cyclic, so let $x_3$ be a generator. Lather, rinse, repeat, to get an infinite sequence as above, adding elements if necessary to get each step to increase the size of the cyclic group by $p$; that is, $x_n$ is of order $p^n$.
Again, prove the group generated by the $x_i$ is the $p$-Prüfer group and equal to all of $G$.
I'd written suitable arguments for 3 and 4, but Jack has posted them as well and they are essentially the same, so I'll defer to him on them.
A: 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.
