Characterizing the Prüfer $p$-group I've been trying to solve these questions for the past few hours with no luck:


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*If $G$ is an infinite abelian group all of whose proper subgroups are finite, then $G$ is a  Prüfer $p$-group for some prime number $p$.

*If $G$ is a group such that $G \equiv G/H$ for all proper normal subgroup $H$ of $G$, then $G$ is a  Prüfer $p$-group for some prime number $p$.
Any elementary idea for these questions (by using only the Chapter I of Hungerford's book). Thanks in advance.
 A: Hints:


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*First show that all elements have order a power of some fixed prime $p$. Then expose $G$ as a union of an ascending chain of cyclic subgroups.

*What if $G$ is simple?

Solution:


*

*First notice that $G$ can only contain elements of finite order; otherwise $G$ would contain a subgroup isomorphic to $\Bbb Z$. Furthermore all elements must have order a power of some fixed prime $p$ otherwise the group would be direct sum $G = \bigoplus_{p\in P} G_{p}$ where $G_{p}$ is the $p$-power torsion subgroup of elements that have order a power of $p$ and $p\in P$ ranges over all primes that occur this way, and there would be infinite subgroups. (Either because $P$ is finite and of the components $G_p$ is infinite; otherwise because there are an infinite number of components and we can leave one out.) Consider an arbitrary element $x\in G$ and consider $H = \langle x\rangle \leq G$. Consider $G/H$ which is an infinite group with the same properties. An arbitrary $y\in G/H$, not the identity element, will lift to an element $x_1\in G$ with the property that $\langle x\rangle < \langle x_1\rangle$. Continuing like this, we obtain a chain $$ \langle x\rangle < \langle x_1\rangle < \dots < \langle x_n\rangle \dots $$ of proper subgroups. Their union $\cup_i \langle x_i\rangle$ is itself a group containing an infinite number of elements, so it must be equal to $G$. All groups are cyclic groups and all elements have order a power of a prime $p$ so it is clear that "$G = \cup_i C_{p^i}$" (in fact a direct limit), i.e. $G$ is a Prüfer $p$-group.

*I think this is wrong... For instance if $G$ is simple then the only proper normal subgroup is $1$ and this will certainly hold.

